Let the probability of passing the test be denoted by \(p = 0.3\) and the probability of failing be \(1 - p = 0.7\). We are looking for the probability that at least three out of five friends pass the test. This can be calculated using the binomial probability formula:

\(P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)\)

Using the binomial formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n = 5\):

\(P(X = 3) = \binom{5}{3} (0.3)^3 (0.7)^2 = 10 \times 0.027 \times 0.49 = 0.1323\)

\(P(X = 4) = \binom{5}{4} (0.3)^4 (0.7)^1 = 5 \times 0.0081 \times 0.7 = 0.02835\)

\(P(X = 5) = \binom{5}{5} (0.3)^5 (0.7)^0 = 1 \times 0.00243 \times 1 = 0.00243\)

Adding these probabilities gives:

\(P(X \geq 3) = 0.1323 + 0.02835 + 0.00243 = 0.16308\)

Let the probability of rating the service as 'poor' or 'satisfactory' be the sum of the probabilities of these two ratings:

\(p = 0.12 + 0.36 = 0.48\).

We need to find the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory'. This is equivalent to finding \(P(3 \leq X \leq 7)\) where \(X\) is a binomial random variable with parameters \(n = 8\) and \(p = 0.48\).

Using the complement rule, we have:

\(P(3 \leq X \leq 7) = 1 - P(X = 0, 1, 2, 8)\)

Calculate \(P(X = 0, 1, 2, 8)\):

\(P(X = 0) = \binom{8}{0} (0.48)^0 (0.52)^8 = 0.00534597\)

\(P(X = 1) = \binom{8}{1} (0.48)^1 (0.52)^7 = 0.039478\)

\(P(X = 2) = \binom{8}{2} (0.48)^2 (0.52)^6 = 0.127544\)

\(P(X = 8) = \binom{8}{8} (0.48)^8 (0.52)^0 = 0.0028179\)

Thus,

\(P(X = 0, 1, 2, 8) = 0.00534597 + 0.039478 + 0.127544 + 0.0028179 = 0.1752\)

Therefore,

\(P(3 \leq X \leq 7) = 1 - 0.1752 = 0.8248\)

Thus, the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory' is approximately 0.825.

The probability of obtaining a total less than 4 when throwing two dice is given by the outcomes (1,1), (1,2), and (2,1). The probability for each of these outcomes is:

\(\frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{3}{36} = \frac{1}{12}\)

We need to find the probability of obtaining a total less than 4 on at least three throws out of 10. This is a binomial probability problem where:

\(n = 10, \quad p = \frac{1}{12}\)

We need to calculate:

\(1 - P(0, 1, 2)\)

Using the binomial probability formula:

\(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\)

Calculate for 0, 1, and 2 successes:

\(P(X = 0) = \binom{10}{0} \left(\frac{1}{12}\right)^0 \left(\frac{11}{12}\right)^{10}\)

\(P(X = 1) = \binom{10}{1} \left(\frac{1}{12}\right)^1 \left(\frac{11}{12}\right)^9\)

\(P(X = 2) = \binom{10}{2} \left(\frac{1}{12}\right)^2 \left(\frac{11}{12}\right)^8\)

Calculate each:

\(P(X = 0) = 0.418904\)

\(P(X = 1) = 0.380822\)

\(P(X = 2) = 0.155791\)

Sum these probabilities:

\(P(0, 1, 2) = 0.418904 + 0.380822 + 0.155791 = 0.955517\)

Therefore, the probability of obtaining a total less than 4 on at least three throws is:

\(1 - 0.955517 = 0.0445\)

Let the probability that a student plays at least one musical instrument be 0.72 (since 28% do not play any instrument).

We need to find the probability that more than 9 students play at least one instrument in a sample of 12 students. This is equivalent to finding:

\(P(X > 9) = P(10) + P(11) + P(12)\)

Using the binomial probability formula:

\(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\)

where \(n = 12\), \(p = 0.72\), and \(1-p = 0.28\).

Calculate each probability:

\(P(10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Substitute and calculate:

\(P(10) = 66 \times 0.1934917632 \times 0.0784 = 0.193725\)

\(P(11) = 12 \times 0.1393140695 \times 0.28 = 0.0905726\)

\(P(12) = 1 \times 0.1934917632 \times 1 = 0.0190484\)

Sum these probabilities:

\(P(X > 9) = 0.193725 + 0.0905726 + 0.0190484 = 0.303346\)

Thus, the probability that more than 9 students play at least one musical instrument is approximately 0.304.

(a) To find \(a\), calculate the probability of obtaining exactly 1 head:

\(a = 0.7 \times (0.5)^3 + 0.3 \times (0.5)^3 = \frac{1}{5}\).

For \(b\), calculate the probability of obtaining exactly 2 heads:

\(b = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 \times 3 = \frac{3}{8}\).

For \(c\), calculate the probability of obtaining exactly 3 heads:

\(c = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 = \frac{3}{10}\).

(b) The expected value \(E(X)\) is calculated as:

\(E(X) = \frac{3 \times 0 + 16 \times 1 + 30 \times 2 + 24 \times 3 + 7 \times 4}{80} = \frac{176}{80} = 2.2\).

(c) The probability of obtaining exactly one head on fewer than 3 occasions is:

\(P(0, 1, 2) = \binom{10}{0} 0.2^0 0.8^{10} + \binom{10}{1} 0.2^1 0.8^9 + \binom{10}{2} 0.2^2 0.8^8\).

Calculating gives \(0.107374 + 0.268435 + 0.301989 = 0.678\).

(d) The probability that Jacob obtains exactly one head for the first time on the 7th or 8th time is:

\(0.8^6 \times 0.2 + 0.8^7 \times 0.2 = 0.0524288 + 0.041943 = 0.09437\).

(a) Let the random variable X be the number of days with more than 10 cm of rain in a 7-day period. X follows a binomial distribution with parameters n = 7 and p = 0.18. We need to find P(X > 2).

\(P(X > 2) = 1 - P(X \leq 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)).\)

Using the binomial probability formula, we calculate:

\(P(X = 0) = \binom{7}{0} (0.18)^0 (0.82)^7 = 0.249285,\)

\(P(X = 1) = \binom{7}{1} (0.18)^1 (0.82)^6 = 0.383048,\)

\(P(X = 2) = \binom{7}{2} (0.18)^2 (0.82)^5 = 0.252251.\)

\(Thus, P(X \leq 2) = 0.249285 + 0.383048 + 0.252251 = 0.884584.\)

\(Therefore, P(X > 2) = 1 - 0.884584 = 0.115.\)

(b) Let Y be the number of 7-day periods with at least one day of more than 10 cm of rain. The probability of at least one day of rain in a 7-day period is 1 - P(X = 0) = 1 - 0.82^7 = 0.7507.

\(Y follows a binomial distribution with parameters n = 3 and p = 0.7507. We need to find P(Y = 2).\)

\(P(Y = 2) = \binom{3}{2} (0.7507)^2 (1 - 0.7507)^1 = 3 \times 0.7507^2 \times 0.2493 = 0.421.\)

(a) The sum of probabilities must equal 1: \(p + q + 0.65 = 1\). Therefore, \(p + q = 0.35\).

Using the expected value: \(E(X) = 0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\).

This simplifies to \(p + 2q + 0.15 = 0.55\), so \(p + 2q = 0.4\).

Solving the equations \(p + q = 0.35\) and \(p + 2q = 0.4\), we find \(q = 0.05\) and \(p = 0.3\).

(c) The probability that \(X = 1\) in at least 3 of 12 turns is \(1 - P(0, 1, 2)\).

\(P(0, 1, 2) = \binom{12}{0} (0.3)^0 (0.7)^{12} + \binom{12}{1} (0.3)^1 (0.7)^{11} + \binom{12}{2} (0.3)^2 (0.7)^{10}\).

Calculating these gives \(0.01384 + 0.07118 + 0.16779 = 0.25299\).

Thus, the probability is \(1 - 0.25299 = 0.747\).

(d) The probability that Jim first succeeds on the 9th turn is \((0.95)^8 \times 0.05\).

This evaluates to \(0.0332\).

Let the random variable \(X\) represent the number of wet days in a 10-day period. \(X\) follows a binomial distribution with parameters \(n = 10\) and \(p = 0.16\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.16)^0 (0.84)^{10} = 0.17490\)

\(P(X = 1) = \binom{10}{1} (0.16)^1 (0.84)^9 = 0.333145\)

\(P(X = 2) = \binom{10}{2} (0.16)^2 (0.84)^8 = 0.28555\)

Summing these probabilities gives:

\(P(X < 3) = 0.17490 + 0.333145 + 0.28555 = 0.794\)

Let the probability that a household has no broadband service be 0.35. We are choosing 10 households, so this is a binomial distribution with parameters \(n = 10\) and \(p = 0.35\).

We need to find the probability that fewer than 3 households have no broadband service, i.e., \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.35)^0 (0.65)^{10} = 1 \times 1 \times 0.013463 = 0.013463\)

\(P(X = 1) = \binom{10}{1} (0.35)^1 (0.65)^9 = 10 \times 0.35 \times 0.072492 = 0.072492\)

\(P(X = 2) = \binom{10}{2} (0.35)^2 (0.65)^8 = 45 \times 0.1225 \times 0.178565 = 0.17565\)

Add these probabilities:

\(P(X < 3) = 0.013463 + 0.072492 + 0.17565 = 0.261605\)

Rounding to three decimal places, the probability is \(0.262\).

(a) The probability that the flight does not arrive late is the sum of the probabilities that it arrives early or on time:

\(P( ext{not late}) = 0.15 + 0.55 = 0.7\)

The probability that on each of 3 days the flight does not arrive late is:

\((0.7)^3 = 0.343\)

(b) We need to find the probability that the flight arrives early at least 3 times out of 9 days. This is a binomial probability problem where \(n = 9\), \(p = 0.15\), and we want \(P(X \geq 3)\).

First, calculate \(P(X < 3)\):

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 0) = \binom{9}{0} (0.15)^0 (0.85)^9 = 0.231617\)

\(P(X = 1) = \binom{9}{1} (0.15)^1 (0.85)^8 = 0.367862\)

\(P(X = 2) = \binom{9}{2} (0.15)^2 (0.85)^7 = 0.259667\)

Thus,

\(P(X < 3) = 0.231617 + 0.367862 + 0.259667 = 0.859146\)

Therefore,

\(P(X \geq 3) = 1 - P(X < 3) = 1 - 0.859146 = 0.140854\)

Rounding to three decimal places, the probability is 0.141.

Let the random variable \(X\) represent the number of adults who travel to work by car in a sample of 12. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.6\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

Calculate \(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

\(P(X = 10) = \binom{12}{10} (0.6)^{10} (0.4)^2 = 0.063852\)

\(P(X = 11) = \binom{12}{11} (0.6)^{11} (0.4)^1 = 0.017414\)

\(P(X = 12) = \binom{12}{12} (0.6)^{12} (0.4)^0 = 0.0021768\)

Thus, \(P(X \geq 10) = 0.063852 + 0.017414 + 0.0021768 = 0.083443\).

Therefore, \(P(X < 10) = 1 - 0.083443 = 0.917\).

The problem involves a binomial distribution where the probability of obtaining a 5 on a single spin is \(p = \frac{1}{5} = 0.2\), and the probability of not obtaining a 5 is \(1 - p = 0.8\). George spins the spinner 10 times, so \(n = 10\).

We need to find the probability of obtaining a 5 more than 4 times but fewer than 8 times, i.e., \(P(5 \leq X \leq 7)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate \(P(X = 5)\), \(P(X = 6)\), and \(P(X = 7)\):

\(P(X = 5) = \binom{10}{5} (0.2)^5 (0.8)^5 = 0.02642\)

\(P(X = 6) = \binom{10}{6} (0.2)^6 (0.8)^4 = 0.005505\)

\(P(X = 7) = \binom{10}{7} (0.2)^7 (0.8)^3 = 0.0007864\)

Sum these probabilities:

\(P(5 \leq X \leq 7) = 0.02642 + 0.005505 + 0.0007864 = 0.0327\)

Let the probability that a student does not like any sports be \(p = 0.3\). The number of students chosen is \(n = 10\). We need to find the probability that at least 3 students do not like any sports, which is \(P(X \geq 3)\).

This can be calculated as \(1 - P(X < 3)\), where \(X\) is a binomial random variable with parameters \(n = 10\) and \(p = 0.3\).

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 0.028248\)

\(P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 0.121061\)

\(P(X = 2) = \binom{10}{2} (0.3)^2 (0.7)^8 = 0.233474\)

Thus, \(P(X < 3) = 0.028248 + 0.121061 + 0.233474 = 0.382985\).

Therefore, \(P(X \geq 3) = 1 - 0.382985 = 0.617015\).

Rounding to three decimal places, the probability is \(0.617\).

Let the random variable \(X\) represent the number of days the train is late out of 7 days. \(X\) follows a binomial distribution with parameters \(n = 7\) and \(p = 0.35\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\)

Calculate each probability:

\(P(X = 0) = \binom{7}{0} (0.35)^0 (0.65)^7 = 0.65^7 = 0.049022\)

\(P(X = 1) = \binom{7}{1} (0.35)^1 (0.65)^6 = 7 \times 0.35 \times 0.65^6 = 0.184776\)

\(P(X = 2) = \binom{7}{2} (0.35)^2 (0.65)^5 = 21 \times 0.35^2 \times 0.65^5 = 0.29848\)

Sum these probabilities:

\(P(X < 3) = 0.049022 + 0.184776 + 0.29848 = 0.532\)

(a) The probability of not getting a 4 in a single throw is \(\frac{5}{6}\). The probability of getting a 4 for the first time on the 6th throw is \(\left( \frac{5}{6} \right)^5 \times \frac{1}{6}\). Therefore, the probability of obtaining a 4 in fewer than 6 throws is:

\(1 - \left( \frac{5}{6} \right)^5\)

Calculating this gives:

\(1 - \left( \frac{5}{6} \right)^5 = 1 - \frac{3125}{7776} = \frac{4651}{7776}\)

(b) The probability of getting a 4 in a single throw is \(\frac{1}{6}\). We use the binomial distribution to find the probability of getting at least 3 fours in 10 throws:

\(1 - P(0, 1, 2) = 1 - \left( \binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} + \binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 + \binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 \right)\)

Calculating each term:

\(\binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} = 0.1615056\)

\(\binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 = 0.3230111\)

\(\binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 = 0.290710\)

Summing these gives:

\(0.1615056 + 0.3230111 + 0.290710 = 0.7752267\)

Thus, the probability of getting at least 3 fours is:

\(1 - 0.7752267 = 0.2247733\)

Rounding to three decimal places gives \(0.225\).

(a) The probability that Kayla takes more than 6 throws to achieve a success is given by the geometric distribution formula:

\(P(X > 6) = (0.75)^6\)

Calculating this gives:

\(P(X > 6) = 0.178\)

(b) To find the probability that Kayla achieves at least 3 successes in 10 throws, we use the binomial distribution:

\(1 - P(0, 1, 2) = 1 - \left( (0.75)^{10} + \binom{10}{1} (0.25)^1 (0.75)^9 + \binom{10}{2} (0.25)^2 (0.75)^8 \right)\)

Calculating each term:

\((0.75)^{10} = 0.0563135\)

\(\binom{10}{1} (0.25)^1 (0.75)^9 = 0.1877117\)

\(\binom{10}{2} (0.25)^2 (0.75)^8 = 0.2815676\)

Summing these gives:

\(0.0563135 + 0.1877117 + 0.2815676 = 0.5256\)

Thus,

\(1 - 0.5256 = 0.474\)

(a) The probability that a student owns a car is 0.22. The probability that all 3 students own a car is given by:

\(0.22^3 = 0.0106\)

(b) We use the binomial probability formula to find the probability that the number of students who own a car is at least 2 and at most 4 out of 16 students. The probability that a student does not own a car is 0.78.

The probability is given by:

\(P(2, 3, 4) = \binom{16}{2} (0.22)^2 (0.78)^{14} + \binom{16}{3} (0.22)^3 (0.78)^{13} + \binom{16}{4} (0.22)^4 (0.78)^{12}\)

Calculating each term:

\(\binom{16}{2} (0.22)^2 (0.78)^{14} = 0.179205\)

\(\binom{16}{3} (0.22)^3 (0.78)^{13} = 0.235877\)

\(\binom{16}{4} (0.22)^4 (0.78)^{12} = 0.216221\)

Adding these probabilities gives:

\(0.179205 + 0.235877 + 0.216221 = 0.631\)

(a) We use the binomial distribution with parameters \(n = 12\) and \(p = 0.72\). We need to find \(P(X \leq 9)\), which is \(1 - P(X \geq 10)\).

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\)

\(P(X = 10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(X = 11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(X = 12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Calculating these probabilities:

\(P(X = 10) = 0.19372\)

\(P(X = 11) = 0.09057\)

\(P(X = 12) = 0.01941\)

\(P(X \geq 10) = 0.19372 + 0.09057 + 0.01941 = 0.30304\)

\(P(X \leq 9) = 1 - 0.30304 = 0.696\)

(b) The probability that Moena does not message Pasha for 3 consecutive days and then messages on the 4th day is:

\((0.28)^3 \times 0.72 = 0.0158\)

Let the random variable \(X\) represent the number of boxes containing more jellies than chocolates. \(X\) follows a binomial distribution with parameters \(n = 10\) and \(p = 0.64\).

We need to find \(P(X \leq 7)\), which is equivalent to \(1 - P(X \geq 8)\).

Calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

Using the binomial probability formula:

\(P(X = 8) = \binom{10}{8} (0.64)^8 (0.36)^2\)

\(P(X = 9) = \binom{10}{9} (0.64)^9 (0.36)^1\)

\(P(X = 10) = \binom{10}{10} (0.64)^{10}\)

Calculate each probability:

\(P(X = 8) = 0.164156\)

\(P(X = 9) = 0.064852\)

\(P(X = 10) = 0.11529\)

Sum these probabilities: \(P(X \geq 8) = 0.164156 + 0.064852 + 0.11529 = 0.344298\)

Therefore, \(P(X \leq 7) = 1 - 0.344298 = 0.759\).

Let the random variable \(X\) represent the number of adults who own a car in a sample of 8. \(X\) follows a binomial distribution with parameters \(n = 8\) and \(p = 0.7\).

We need to find \(P(X < 6)\), which is equivalent to \(1 - P(X \geq 6)\).

Calculate \(P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8)\).

\(P(X = 6) = \binom{8}{6} (0.7)^6 (0.3)^2\)

\(P(X = 7) = \binom{8}{7} (0.7)^7 (0.3)^1\)

\(P(X = 8) = \binom{8}{8} (0.7)^8 (0.3)^0\)

\(P(X \geq 6) = \binom{8}{6} (0.7)^6 (0.3)^2 + \binom{8}{7} (0.7)^7 (0.3)^1 + \binom{8}{8} (0.7)^8 (0.3)^0\)

\(P(X \geq 6) = 0.55177\)

Thus, \(P(X < 6) = 1 - 0.55177 = 0.448\).

(a) Let the probability of a person being a Notes supporter be 0.3. We use the binomial distribution with parameters 6 (trials) and 0.3 (success probability).

The probability that no more than 2 people are Notes supporters is:

\(P(0, 1, 2) = \binom{6}{0} (0.3)^0 (0.7)^6 + \binom{6}{1} (0.3)^1 (0.7)^5 + \binom{6}{2} (0.3)^2 (0.7)^4\)

Calculating each term:

\(\binom{6}{0} (0.3)^0 (0.7)^6 = 0.1176\)

\(\binom{6}{1} (0.3)^1 (0.7)^5 = 0.3025\)

\(\binom{6}{2} (0.3)^2 (0.7)^4 = 0.3241\)

Adding these probabilities gives:

\(0.1176 + 0.3025 + 0.3241 = 0.744\)

(b) The probability that a person supports neither choir is:

\(1 - (0.3 + 0.45) = 0.25\)

The probability that none of the 6 people support either choir is:

\(P(6 \text{ support neither choir}) = (0.25)^6\)

Calculating this gives:

\((0.25)^6 = 0.000244\)

The problem involves a binomial distribution where the probability of success (a household being satisfied) is 0.66, and the number of trials is 10. We need to find the probability that at least 8 households are satisfied, which means finding the probability of 8, 9, or 10 successes.

The probability of exactly 8 successes is given by:

\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2\)

The probability of exactly 9 successes is given by:

\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1\)

The probability of exactly 10 successes is given by:

\(P(X = 10) = \binom{10}{10} (0.66)^{10}\)

Thus, the probability of at least 8 successes is:

\(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\)

Calculating each term:

\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2 = 45 \times 0.66^8 \times 0.34^2\)

\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1 = 10 \times 0.66^9 \times 0.34\)

\(P(X = 10) = \binom{10}{10} (0.66)^{10} = 1 \times 0.66^{10}\)

Summing these probabilities gives:

\(P(X \geq 8) = 0.284\)

The probability of obtaining a score of 8 or more with two dice is calculated first. The possible outcomes for scores of 8 or more are: (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 15 favorable outcomes out of 36 total outcomes, so the probability is:

\(P(\text{score} \geq 8) = \frac{15}{36} = \frac{5}{12}\)

We need to find the probability of obtaining a score of 8 or more on fewer than 3 occasions in 8 throws. This is a binomial distribution problem where:

\(n = 8, \quad p = \frac{5}{12}\)

We calculate the probability of obtaining a score of 8 or more on 0, 1, or 2 occasions:

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each term:

\(P(X = 0) = \binom{8}{0} \left(\frac{5}{12}\right)^0 \left(\frac{7}{12}\right)^8 = 0.01341\)

\(P(X = 1) = \binom{8}{1} \left(\frac{5}{12}\right)^1 \left(\frac{7}{12}\right)^7 = 0.07661\)

\(P(X = 2) = \binom{8}{2} \left(\frac{5}{12}\right)^2 \left(\frac{7}{12}\right)^6 = 0.1915\)

Summing these probabilities gives:

\(P(X < 3) = 0.01341 + 0.07661 + 0.1915 = 0.28152\)

Rounding to three decimal places, the probability is:

\(\boxed{0.282}\)

Let the probability that a customer rates the logo as good be \(p = 0.42\). The number of customers rating the logo as good follows a binomial distribution \(B(n, p)\) where \(n = 10\).

We need to find the probability that fewer than 8 customers rate the logo as good, i.e., \(P(X < 8)\).

This is equivalent to \(1 - P(X \geq 8)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

\(P(X = 8) = \binom{10}{8} (0.42)^8 (0.58)^2\)

\(P(X = 9) = \binom{10}{9} (0.42)^9 (0.58)^1\)

\(P(X = 10) = \binom{10}{10} (0.42)^{10} (0.58)^0\)

Thus, \(P(X \geq 8) = 10C_8 (0.42)^8 (0.58)^2 + 10C_9 (0.42)^9 (0.58)^1 + 10C_{10} (0.42)^{10}\).

Therefore, \(P(X < 8) = 1 - (10C_8 (0.42)^8 (0.58)^2 + 10C_9 (0.42)^9 (0.58)^1 + 10C_{10} (0.42)^{10})\).

Calculating this gives \(P(X < 8) = 0.983\).

Let the probability that a person is a man be \(p = 0.34\). The number of men in a sample of 14 people follows a binomial distribution \(B(14, 0.34)\).

We need to find \(P(X \leq 2)\), where \(X\) is the number of men.

\(P(X = 0) = \binom{14}{0} (0.34)^0 (0.66)^{14} = (0.66)^{14}\)

\(P(X = 1) = \binom{14}{1} (0.34)^1 (0.66)^{13} = 14 \times 0.34 \times (0.66)^{13}\)

\(P(X = 2) = \binom{14}{2} (0.34)^2 (0.66)^{12} = \frac{14 \times 13}{2} \times (0.34)^2 \times (0.66)^{12}\)

Calculating each term:

\(P(X = 0) = 0.0029758\)

\(P(X = 1) = 0.02146239\)

\(P(X = 2) = 0.071866\)

Adding these probabilities gives:

\(P(X \leq 2) = 0.0029758 + 0.02146239 + 0.071866 = 0.0963\)

We model this situation using a binomial distribution with parameters: number of trials \(n = 10\) and probability of success \(p = 0.35\).

The probability that Janice buys at most 7 items is given by:

\(P(X \leq 7) = 1 - P(X = 8) - P(X = 9) - P(X = 10)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 8) = \binom{10}{8} (0.35)^8 (0.65)^2\)

\(P(X = 9) = \binom{10}{9} (0.35)^9 (0.65)^1\)

\(P(X = 10) = \binom{10}{10} (0.35)^{10} (0.65)^0\)

Substituting these into the equation:

\(P(X \leq 7) = 1 - (0.004281 + 0.0005123 + 0.00002759)\)

\(P(X \leq 7) = 1 - 0.00482089\)

\(P(X \leq 7) = 0.995\)

Let the random variable \(X\) represent the number of children who own a bicycle. \(X\) follows a binomial distribution with parameters \(n = 15\) and \(p = 0.65\).

We need to find \(P(X > 12) = P(X = 13) + P(X = 14) + P(X = 15)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 13) = \binom{15}{13} (0.65)^{13} (0.35)^2\)

\(P(X = 14) = \binom{15}{14} (0.65)^{14} (0.35)^1\)

\(P(X = 15) = \binom{15}{15} (0.65)^{15} (0.35)^0\)

Summing these probabilities gives:

\(P(X > 12) = 0.0617\)

Let the probability that a customer shops online be \(p = 0.35\). The number of customers chosen is \(n = 6\). We need to find the probability that more than three customers shop online, i.e., \(P(X > 3)\) where \(X\) is a binomial random variable with parameters \(n = 6\) and \(p = 0.35\).

We calculate \(P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 4) = \binom{6}{4} (0.35)^4 (0.65)^2\)

\(P(X = 5) = \binom{6}{5} (0.35)^5 (0.65)^1\)

\(P(X = 6) = \binom{6}{6} (0.35)^6 (0.65)^0\)

Calculating each term:

\(P(X = 4) = 15 \times 0.35^4 \times 0.65^2\)

\(P(X = 5) = 6 \times 0.35^5 \times 0.65\)

\(P(X = 6) = 1 \times 0.35^6\)

Summing these probabilities gives \(P(X > 3) = 0.117\).

Let the random variable \(X\) represent the number of days Jake completes the puzzle in 5 days. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = 0.75\).

We need to find \(P(X \geq 3)\), which is \(P(X = 3) + P(X = 4) + P(X = 5)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\):

\(P(X = 3) = \binom{5}{3} (0.75)^3 (0.25)^2 = 10 \times 0.421875 \times 0.0625 = 0.26367\)

\(P(X = 4) = \binom{5}{4} (0.75)^4 (0.25)^1 = 5 \times 0.31640625 \times 0.25 = 0.39551\)

\(P(X = 5) = \binom{5}{5} (0.75)^5 (0.25)^0 = 1 \times 0.2373046875 \times 1 = 0.23730\)

Adding these probabilities gives:

\(P(X \geq 3) = 0.26367 + 0.39551 + 0.23730 = 0.896\)

The problem involves a binomial distribution where the probability of passing, \(p\), is 0.8, and the number of trials, \(n\), is 9. We need to find the probability that at most 6 students pass, which is \(P(X \leq 6)\).

This can be calculated as:

\(P(X \leq 6) = 1 - (P(7) + P(8) + P(9))\)

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate \(P(7)\), \(P(8)\), and \(P(9)\):

\(P(7) = \binom{9}{7} (0.8)^7 (0.2)^2\)

\(P(8) = \binom{9}{8} (0.8)^8 (0.2)^1\)

\(P(9) = \binom{9}{9} (0.8)^9 (0.2)^0\)

Substitute these into the equation:

\(P(X \leq 6) = 1 - (0.3019899 + 0.3019899 + 0.1342177)\)

\(P(X \leq 6) = 1 - 0.7381975 = 0.2618025\)

Rounding to three decimal places, the probability is \(0.262\).

Let the random variable \(X\) represent the number of people who choose a mobile phone made by Company A. \(X\) follows a binomial distribution with parameters \(n = 13\) and \(p = 0.6\).

We need to find \(P(X < 11)\), which is equivalent to \(1 - P(X \geq 11)\).

Using the complement rule, \(P(X \geq 11) = P(X = 11) + P(X = 12) + P(X = 13)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 11) = \binom{13}{11} (0.6)^{11} (0.4)^2\)

\(P(X = 12) = \binom{13}{12} (0.6)^{12} (0.4)^1\)

\(P(X = 13) = \binom{13}{13} (0.6)^{13} (0.4)^0\)

Thus, \(P(X \geq 11) = \binom{13}{11} (0.6)^{11} (0.4)^2 + \binom{13}{12} (0.6)^{12} (0.4)^1 + \binom{13}{13} (0.6)^{13}\).

Calculate \(P(X < 11) = 1 - P(X \geq 11)\).

\(P(X < 11) = 1 - (\binom{13}{11} (0.6)^{11} (0.4)^2 + \binom{13}{12} (0.6)^{12} (0.4)^1 + \binom{13}{13} (0.6)^{13})\)

\(P(X < 11) = 0.942\).

Let S be the event that a vehicle goes straight, L be the event that a vehicle turns left, and R be the event that a vehicle turns right.

The probability that one vehicle goes straight and the other two turn left is given by:

\(P(SLL) = (0.3)(0.55)(0.55) = 0.09075\)

The probability that one vehicle goes straight and the other two turn right is given by:

\(P(SRR) = (0.3)(0.15)(0.15) = 0.00675\)

There are 3 ways to arrange SLL (SSL, SLL, LSL) and 3 ways to arrange SRR (SSR, SRR, RSR), so the total probability is:

\(\binom{3}{1} \times P(SLL) + \binom{3}{1} \times P(SRR) = 3 \times 0.09075 + 3 \times 0.00675 = 0.27225 + 0.02025 = 0.2925\)

Thus, the probability is approximately 0.293.

Let the probability of a family owning a dishwasher be \(p = 0.22\). The number of families is \(n = 15\).

We need to find \(P(4 \, \text{to} \, 6)\), which is the probability that between 4 and 6 families own a dishwasher.

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate for \(k = 4, 5, 6\):

\(P(4) = \binom{15}{4} (0.22)^4 (0.78)^{11}\)

\(P(5) = \binom{15}{5} (0.22)^5 (0.78)^{10}\)

\(P(6) = \binom{15}{6} (0.22)^6 (0.78)^9\)

Sum these probabilities:

\(P(4, 5, 6) = \binom{15}{4} (0.22)^4 (0.78)^{11} + \binom{15}{5} (0.22)^5 (0.78)^{10} + \binom{15}{6} (0.22)^6 (0.78)^9\)

\(= 0.398\)

Let the probability of selling less than 16 kg of grapes on any day be denoted by \(p = 0.1\). We are interested in finding the probability that less than 16 kg of grapes are sold on more than 2 out of 12 days. This is equivalent to finding \(1 - P(0, 1, 2)\), where \(P(0, 1, 2)\) is the probability of selling less than 16 kg on 0, 1, or 2 days.

Using the binomial distribution formula, we have:

\(P(0, 1, 2) = \binom{12}{0} (0.1)^0 (0.9)^{12} + \binom{12}{1} (0.1)^1 (0.9)^{11} + \binom{12}{2} (0.1)^2 (0.9)^{10}\)

Calculating each term:

\(\binom{12}{0} (0.1)^0 (0.9)^{12} = 1 \times 1 \times 0.2824 = 0.2824\)

\(\binom{12}{1} (0.1)^1 (0.9)^{11} = 12 \times 0.1 \times 0.3138 = 0.3766\)

\(\binom{12}{2} (0.1)^2 (0.9)^{10} = 66 \times 0.01 \times 0.3487 = 0.2301\)

Thus, \(P(0, 1, 2) = 0.2824 + 0.3766 + 0.2301 = 0.8891\).

Therefore, the probability that less than 16 kg of grapes are sold on more than 2 days is:

\(1 - 0.8891 = 0.1109\)

Rounding to three significant figures, the probability is \(0.111\).

Let \(X\) be the number of students who own a car out of the five chosen. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = \frac{1}{4}\).

We need to find \(P(X \geq 4) = P(X = 4) + P(X = 5)\).

Using the binomial probability formula:

\(P(X = 4) = \binom{5}{4} \left( \frac{1}{4} \right)^4 \left( \frac{3}{4} \right)^1\)

\(P(X = 5) = \binom{5}{5} \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0\)

Calculating each term:

\(P(X = 4) = 5 \times \left( \frac{1}{4} \right)^4 \times \frac{3}{4} = 0.014648\)

\(P(X = 5) = 1 \times \left( \frac{1}{4} \right)^5 = 0.00097656\)

Adding these probabilities gives:

\(P(X \geq 4) = 0.014648 + 0.00097656 = 0.0156\)

Thus, the probability that at least four of these students own a car is \(0.0156\) or \(\frac{1}{64}\).

Let the probability that a person agrees to complete the survey be 0.2. Therefore, the probability that a person does not agree is 0.8.

The probability that none of the first three people agree to complete the survey is given by:

\((0.8)^3 = 0.512\)

Thus, the probability that at least one person agrees is:

\(1 - (0.8)^3 = 1 - 0.512 = 0.488\)

Alternatively, using the binomial probability formula, the probability that exactly one, two, or three people agree can be calculated and summed:

\(P(1, 2, 3) = \binom{3}{1}(0.2)(0.8)^2 + \binom{3}{2}(0.2)^2(0.8) + (0.2)^3\)

\(= 3(0.2)(0.8)^2 + 3(0.2)^2(0.8) + (0.2)^3\)

\(= 0.384 + 0.096 + 0.008 = 0.488\)

(i) The probability that the die lands on 4 is \(\frac{1}{4}\). The probability of getting exactly 2 heads in 4 coin tosses is given by the binomial probability formula:

\(P(4, 2H) = \frac{1}{4} \times \binom{4}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^2\)

\(= \frac{1}{4} \times 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{2}{27}\)

(ii) The probability that the die lands on 3 is \(\frac{1}{4}\). The probability of getting exactly 3 heads in 3 coin tosses is:

\(P(3, 3H) = \frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)

(iii) We need to find the probability that the number the die lands on is the same as the number of heads. This includes the following cases:

• Die lands on 1 and 1 head: \(\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
• Die lands on 2 and 2 heads: \(\frac{1}{4} \times \binom{2}{2} \left( \frac{1}{3} \right)^2 = \frac{1}{36}\)
• Die lands on 3 and 3 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)
• Die lands on 4 and 4 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^4 = \frac{1}{324}\)

Summing these probabilities gives:

\(\frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \frac{1}{324} = \frac{10}{81}\)

(i) The two conditions required for a binomial distribution are:

1. Constant probability of success on each trial.
2. Independent trials/events.

(ii) To find the probability that Hebe completes at least 5 puzzles in a week, we use the binomial distribution with parameters:

\(= 7 (number of trials),\)

\(= 0.7 (probability of success).\)

We need to calculate:

\(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\)

\(= \binom{7}{5}(0.7)^5(0.3)^2 + \binom{7}{6}(0.7)^6(0.3)^1 + (0.7)^7\)

\(= 0.647\)

(iii) To find the probability that Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks, we first find the probability of completing 4 or fewer puzzles in a week:

\(P(X \leq 4) = 1 - P(X \geq 5) = 1 - 0.647 = 0.3529\)

Now, use the binomial distribution with parameters:

\(= 10 (weeks),\)

\(= 0.3529 (probability of completing 4 or fewer puzzles in a week).\)

We need to calculate:

\(P(Y = 3) = \binom{10}{3}(0.3529)^3(0.6471)^7\)

\(= 0.251\)

Let the probability that Khalid rides his skateboard on a given day be 0.4. We need to find the probability that he rides on his skateboard on at least 2 out of 10 days.

This is a binomial probability problem where the number of trials is 10, and the probability of success (riding the skateboard) is 0.4.

The probability of riding on the skateboard on 0 or 1 day is:

\(P(0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = (0.6)^{10}\)

\(P(1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 10 \times 0.4 \times (0.6)^9\)

Thus, the probability of riding on the skateboard on at least 2 days is:

\(1 - (P(0) + P(1)) = 1 - [(0.6)^{10} + 10 \times 0.4 \times (0.6)^9]\)

Calculating these values:

\((0.6)^{10} \approx 0.0060466176\)

\(10 \times 0.4 \times (0.6)^9 \approx 0.0403936\)

So:

\(1 - (0.0060466176 + 0.0403936) = 1 - 0.04644 \approx 0.954\)

(i) The probability that Annabel eats pasta on exactly 2 days in a week of 7 days can be modeled using the binomial distribution. The probability of eating pasta on a given day is 0.1, and not eating pasta is 0.9. We use the binomial formula:

\(P(2) = \binom{7}{2} (0.1)^2 (0.9)^5\)

Calculating this gives:

\(P(2) = 21 \times 0.01 \times 0.59049 = 0.124\)

(ii) For the second part, we need to find the probability that Annabel eats rice on 2 days, potato on 1 day, and pasta on 2 days in a period of 5 days. The probabilities are 0.75 for rice, 0.15 for potato, and 0.1 for pasta. The number of ways to arrange these days is given by:

\(\frac{5!}{2!1!2!}\)

The probability is then:

\((0.75)^2 (0.15)^1 (0.1)^2 \times \frac{5!}{2!1!2!}\)

Calculating this gives:

\(0.5625 \times 0.15 \times 0.01 \times 30 = 0.0253\)

(i) Let the probability of landing on a 3 be \(p = \frac{1}{3}\). We need to find the probability of getting at least two 3s in four spins. This is given by:

\(P(\geq 2) = 1 - P(0, 1) = 1 - \left(\frac{2}{3}\right)^4 - 4 \cdot \binom{4}{1} \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^3\)

Alternatively, calculate directly:

\(P(2, 3, 4) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 + \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right) + \left(\frac{1}{3}\right)^4\)

Thus, \(P(\geq 2) = \frac{11}{27}\) or 0.407.

(ii) To find the probability that the sum of the four scores is 5, consider the combination (1, 1, 1, 2). There are 4 permutations of this combination:

\(P(\text{sum is 5}) = P(1, 1, 1, 2) \times 4 = \left(\frac{1}{3}\right)^4 \times 4\)

Thus, \(P(\text{sum is 5}) = \frac{4}{81}\) or 0.0494.

The probability that Kersley is either asleep or studying at noon on any given day is the sum of the probabilities of these two independent events:

\(P(A) = 0.2 + 0.6 = 0.8\).

We need to find the probability that Kersley is either asleep or studying on at least 6 days out of 7. This is a binomial probability problem with parameters \(n = 7\) and \(p = 0.8\).

The probability of Kersley being asleep or studying on exactly 6 days is given by:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1\).

The probability of Kersley being asleep or studying on all 7 days is:

\(P(7) = (0.8)^7\).

Thus, the probability that Kersley is either asleep or studying on at least 6 days is:

\(P(6 \text{ or } 7) = P(6) + P(7)\).

Calculating these:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1 = 7 \times 0.262144 \times 0.2 = 0.3670016\).

\(P(7) = (0.8)^7 = 0.2097152\).

Therefore,

\(P(6 \text{ or } 7) = 0.3670016 + 0.2097152 = 0.5767168\).

Rounding to three decimal places, the probability is \(0.577\).

(i) We use the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

where \(n = 12\), \(p = 0.65\), and \(1-p = 0.35\).

Calculate for \(r = 8, 9, 10\):

\(\binom{12}{8} (0.65)^8 (0.35)^4 + \binom{12}{9} (0.65)^9 (0.35)^3 + \binom{12}{10} (0.65)^{10} (0.35)^2\)

\(= 0.541\)

(ii) The probability that the fourth passenger is the first to carry a backpack is calculated by considering the first three passengers do not carry a backpack and the fourth does:

\(P(\overline{R}\overline{R}\overline{R}R) = (0.35)^3 \times 0.65\)

\(= 0.0279\)

First, find the probability of throwing a 4. Since the probabilities for scores 5 and 6 are 0.2 each, the total probability for scores 1 to 4 is:

\(1 - 0.4 = 0.6\)

Since scores 1 to 4 have equal probabilities, the probability of throwing a 4 is:

\(\frac{0.6}{4} = 0.15\)

Let \(p = 0.15\) be the probability of throwing a 4, and \(q = 1 - p = 0.85\) be the probability of not throwing a 4.

We need to find the probability of getting a score of 4 on not more than 1 of the 3 throws. This is the sum of the probabilities of getting a score of 4 on exactly 0 or 1 throw:

\(P(0) = q^3 = (0.85)^3\)

\(P(1) = \binom{3}{1} p q^2 = 3 \times 0.15 \times (0.85)^2\)

Thus, the total probability is:

\(P(0) + P(1) = (0.85)^3 + 3 \times 0.15 \times (0.85)^2\)

Calculating these values:

\((0.85)^3 = 0.614125\)

\(3 \times 0.15 \times (0.85)^2 = 0.325875\)

Adding these gives:

\(0.614125 + 0.325875 = 0.939\)

Let the probability that the score is greater than 2 be denoted by \(p\). The score is greater than 2 if the difference between the numbers on the two spinners is 3 or 4. The possible pairs are (1,4), (1,5), (2,5), (4,1), (5,1), (5,2). There are 6 such pairs out of a total of 25 possible outcomes (since each spinner has 5 sides), so \(p = \frac{6}{25}\).

We need to find the probability that the score is greater than 2 on at least 3 occasions out of 9 spins. This is a binomial probability problem where \(X \sim \text{Binomial}(9, \frac{6}{25})\).

The probability that the score is greater than 2 on at least 3 occasions is:

\(P(X \geq 3) = 1 - P(X = 0, 1, 2)\)

Calculate \(P(X = 0, 1, 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{9}{0} \left(\frac{6}{25}\right)^0 \left(\frac{19}{25}\right)^9\)

\(P(X = 1) = \binom{9}{1} \left(\frac{6}{25}\right)^1 \left(\frac{19}{25}\right)^8\)

\(P(X = 2) = \binom{9}{2} \left(\frac{6}{25}\right)^2 \left(\frac{19}{25}\right)^7\)

Substitute and calculate:

\(P(X \geq 3) = 1 - (0.08459 + 0.2404 + 0.3037)\)

\(P(X \geq 3) = 1 - 0.62869 = 0.37131\)

Thus, the probability that the score is greater than 2 on at least 3 occasions is approximately 0.371.

Let the probability that a water pistol does not work properly be denoted by \(p = 0.08\). The probability that a water pistol works properly is \(1 - p = 0.92\).

We are dealing with a binomial distribution where \(n = 19\) and \(p = 0.08\). We need to find the probability that at most 2 do not work properly, i.e., \(P(X \leq 2)\).

The probability \(P(X \leq 2)\) is given by:

\(P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 0) = \binom{19}{0} (0.08)^0 (0.92)^{19} = (0.92)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.08)^1 (0.92)^{18} = 19 \times 0.08 \times (0.92)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.08)^2 (0.92)^{17} = \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Summing these probabilities gives:

\(P(X \leq 2) = (0.92)^{19} + 19 \times 0.08 \times (0.92)^{18} + \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Calculating this expression yields \(P(X \leq 2) = 0.809\).

Let the probability that a car is fitted with satellite navigation equipment be denoted by \(p = 0.76\).

The number of cars fitted with the equipment follows a binomial distribution with parameters \(n = 11\) and \(p = 0.76\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

Calculate \(P(X \geq 10) = P(X = 10) + P(X = 11)\).

Using the binomial probability formula:

\(P(X = 10) = \binom{11}{10} (0.76)^{10} (0.24)^{1}\)

\(P(X = 11) = \binom{11}{11} (0.76)^{11} (0.24)^{0}\)

\(P(X = 10) = 11 \times (0.76)^{10} \times (0.24)\)

\(P(X = 11) = (0.76)^{11}\)

\(P(X \geq 10) = 11 \times (0.76)^{10} \times (0.24) + (0.76)^{11}\)

\(P(X \geq 10) = 0.219\)

Thus, \(P(X < 10) = 1 - 0.219 = 0.781\).

The problem involves a binomial distribution where the number of trials is 10, and the probability of success (rolling a six) is \(\frac{1}{6}\).

We need to find the probability of getting between 3 and 5 sixes inclusive, i.e., \(P(3 \leq X \leq 5)\).

This can be calculated as:

\(P(3, 4, 5) = \binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 + \binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 + \binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5\)

Calculating each term:

\(\binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 = 0.155045\)

\(\binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 = 0.054253\)

\(\binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5 = 0.012924\)

Adding these probabilities gives:

\(0.155045 + 0.054253 + 0.012924 = 0.222\)

The problem involves a binomial distribution where the probability of success (a household having a printer) is 0.68, and the number of trials is 8. We need to find the probability that 5, 6, or 7 households have a printer.

The probability of exactly k successes in a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

where \(n = 8\), \(p = 0.68\), and \(1-p = 0.32\).

Calculate the probabilities for 5, 6, and 7 successes:

\(P(X = 5) = \binom{8}{5} (0.68)^5 (0.32)^3\)

\(P(X = 6) = \binom{8}{6} (0.68)^6 (0.32)^2\)

\(P(X = 7) = \binom{8}{7} (0.68)^7 (0.32)^1\)

Sum these probabilities:

\(P(5, 6, 7) = \binom{8}{5} (0.68)^5 (0.32)^3 + \binom{8}{6} (0.68)^6 (0.32)^2 + \binom{8}{7} (0.68)^7 (0.32)\)

Using the binomial coefficients and calculations:

\(P(5, 6, 7) = 0.722\)

(i) The greatest number of books that could be read is 12. The probability that a member reads this number of books is given by \(P(12) = (0.7)^{12}\).

Calculating this gives \(P(12) = 0.0138\).

(ii) To find the probability that a member reads fewer than 10 books, we calculate \(1 - P(10, 11, 12)\).

Using the binomial probability formula, we have:

\(P(10) = \binom{12}{10} (0.7)^{10} (0.3)^2\)

\(P(11) = \binom{12}{11} (0.7)^{11} (0.3)^1\)

\(P(12) = (0.7)^{12}\)

Thus, \(P(10, 11, 12) = \binom{12}{10} (0.7)^{10} (0.3)^2 + \binom{12}{11} (0.7)^{11} (0.3) + (0.7)^{12}\).

Calculating these probabilities gives \(P(10, 11, 12) = 0.2528\).

Therefore, the probability that a member reads fewer than 10 books is \(1 - 0.2528 = 0.747\).

The problem can be modeled using a binomial distribution where the probability of success (owning a cell phone) is given by \(p = 0.75\), and the number of trials \(n = 8\).

We need to find the probability that the number of successes (adults owning a cell phone) is between 4 and 6 inclusive, i.e., \(P(4 \leq X \leq 6)\).

This can be calculated as:

\(P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)\)

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), we calculate each term:

\(P(X = 4) = \binom{8}{4} (0.75)^4 (0.25)^4\)

\(P(X = 5) = \binom{8}{5} (0.75)^5 (0.25)^3\)

\(P(X = 6) = \binom{8}{6} (0.75)^6 (0.25)^2\)

Calculating each:

\(P(X = 4) = 70 \times (0.75)^4 \times (0.25)^4\)

\(P(X = 5) = 56 \times (0.75)^5 \times (0.25)^3\)

\(P(X = 6) = 28 \times (0.75)^6 \times (0.25)^2\)

Summing these probabilities gives:

\(P(4 \leq X \leq 6) = 0.606\)

(i) To find the probability that the numbers on four dice add up to 5, consider the combinations: (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1). Each die has a probability of \(\frac{1}{6}\) for a specific number. The probability for one combination is \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{1296}\). There are 4 such combinations, so the total probability is \(4 \times \frac{1}{1296} = \frac{1}{324}\).

(ii) Let \(p = \frac{1}{324}\) be the probability from part (i) and \(q = 1 - p = \frac{323}{324}\). We use the binomial probability formula for exactly 1 or 2 successes in 7 trials: \(P(1,2) = \binom{7}{1} p^1 q^6 + \binom{7}{2} p^2 q^5\). Calculate each term: \(\binom{7}{1} \times \frac{1}{324} \times \left(\frac{323}{324}\right)^6 + \binom{7}{2} \times \left(\frac{1}{324}\right)^2 \times \left(\frac{323}{324}\right)^5 = 0.0214\).

Let the random variable \(X\) represent the number of houses with solar heating. \(X\) follows a binomial distribution with parameters \(n = 19\) and \(p = 0.12\), i.e., \(X \sim B(19, 0.12)\).

We need to find \(P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 0) = \binom{19}{0} (0.12)^0 (0.88)^{19} = (0.88)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.12)^1 (0.88)^{18} = 19 \times (0.12) \times (0.88)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.12)^2 (0.88)^{17} = \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17}\)

\(P(X = 3) = \binom{19}{3} (0.12)^3 (0.88)^{16} = \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Summing these probabilities gives:

\(P(X < 4) = (0.88)^{19} + 19 \times (0.12) \times (0.88)^{18} + \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17} + \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Calculating these values, we find \(P(X < 4) = 0.813\).

(i) The three conditions for a binomial distribution are:

1. The probability of success, denoted as \(p\), is constant for each trial.
2. The trials are independent.
3. There are a fixed number of trials, and each trial has only two possible outcomes (success or failure).

(ii) Let \(X\) be the number of months George buys shares in a small company. \(X\) follows a binomial distribution with parameters \(n = 18\) and \(p = 0.15\).

We need to find \(P(X \geq 3)\), which is equivalent to \(1 - P(X \leq 2)\).

Calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\) using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

\(P(X = 0) = \binom{18}{0} (0.15)^0 (0.85)^{18} = (0.85)^{18}\)

\(P(X = 1) = \binom{18}{1} (0.15)^1 (0.85)^{17} = 18 \times (0.15) \times (0.85)^{17}\)

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16} = \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}\)

Thus, \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(P(X \geq 3) = 1 - P(X \leq 2)\)

Substitute the values:

\(P(X \geq 3) = 1 - [(0.85)^{18} + 18 \times (0.15) \times (0.85)^{17} + \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}]\)

\(P(X \geq 3) = 0.520\)

The probability that a mango is medium or large is given by:

\(p = 0.70 + 0.15 = 0.85\)

We need to find the probability that fewer than 12 mangoes are medium or large. This is equivalent to finding:

\(P(X < 12) = 1 - P(X \geq 12)\)

Where \(X\) is the number of medium or large mangoes out of 14. We use the binomial distribution:

\(P(X \geq 12) = P(X = 12) + P(X = 13) + P(X = 14)\)

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 12) = \binom{14}{12} (0.85)^{12} (0.15)^2\)

\(P(X = 13) = \binom{14}{13} (0.85)^{13} (0.15)^1\)

\(P(X = 14) = \binom{14}{14} (0.85)^{14} (0.15)^0\)

Summing these gives:

\(P(X \geq 12) = 0.6479\)

Thus,

\(P(X < 12) = 1 - 0.6479 = 0.352\)

(b) The probability of getting a leopard in one magazine is \(p = 0.2\). We need to find the probability of getting more than two leopards in 12 magazines. This is given by:

\(1 - P(0, 1, 2) = 1 - \left( \binom{12}{0} (0.2)^0 (0.8)^{12} + \binom{12}{1} (0.2)^1 (0.8)^{11} + \binom{12}{2} (0.2)^2 (0.8)^{10} \right)\)

\(= 1 - (0.06872 + 0.20615 + 0.28347)\)

\(= 0.442\)

(c) The probability that after 5 weeks Sahim has exactly one of each animal is given by:

\((0.2)^5 \times 5!\)

\(= 0.0384 = \frac{24}{625}\)

The problem involves a binomial distribution where the probability of being an office worker is given as 0.22. The probability of not being an office worker is therefore 1 - 0.22 = 0.78.

We need to find the probability that between 4 and 6 people inclusive are office workers out of 8 people. This is calculated using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

where \(n = 8\), \(p = 0.22\), and \(r\) is the number of office workers.

Calculate for \(r = 4, 5, 6\):

\(P(4) = \binom{8}{4} (0.22)^4 (0.78)^4\)

\(P(5) = \binom{8}{5} (0.22)^5 (0.78)^3\)

\(P(6) = \binom{8}{6} (0.22)^6 (0.78)^2\)

Sum these probabilities:

\(P(4, 5, 6) = P(4) + P(5) + P(6)\)

\(= \binom{8}{4} (0.22)^4 (0.78)^4 + \binom{8}{5} (0.22)^5 (0.78)^3 + \binom{8}{6} (0.22)^6 (0.78)^2\)

\(= 0.0763\)

(i) The number of fireworks that fail follows a binomial distribution with parameters \(n = 20\) and \(p = 0.05\). We need to find \(P(X > 1)\).

\(P(X > 1) = 1 - P(X = 0) - P(X = 1)\)

\(P(X = 0) = (0.95)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.95)^{19} (0.05)^1\)

\(P(X > 1) = 1 - (0.95)^{20} - \binom{20}{1} (0.95)^{19} (0.05)\)

\(P(X > 1) = 1 - 0.3585 - 0.3770 = 0.264\)

(ii) The profit if 19 or 20 fireworks work is \(450 \times 10 - 480 = 4020\).

If more than 1 firework fails, the profit is \(-480\).

The expected profit is:

\(4020 \times (1 - 0.264) - 480 \times 0.264\)

\(= 4020 \times 0.736 - 480 \times 0.264\)

\(= 2960.64 - 126.72 = 2833.92\)

Rounding gives \(\$2830\).

The total number of pairs of shoes is 20, and the number of pairs with designer labels is 8. Therefore, the probability of choosing a pair with designer labels is \(p = \frac{8}{20} = 0.4\).

The probability of choosing a pair without designer labels is \(1 - p = 0.6\).

We need to find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of 7. This is a binomial probability problem where \(n = 7\), \(p = 0.4\), and we want \(P(X \leq 4)\).

Using the complement rule, \(P(X \leq 4) = 1 - P(X = 5) - P(X = 6) - P(X = 7)\).

Calculate each probability:

\(P(X = 5) = \binom{7}{5} (0.4)^5 (0.6)^2 = 21 \times 0.01024 \times 0.36 = 0.0774144\).

\(P(X = 6) = \binom{7}{6} (0.4)^6 (0.6)^1 = 7 \times 0.004096 \times 0.6 = 0.0172032\).

\(P(X = 7) = \binom{7}{7} (0.4)^7 (0.6)^0 = 1 \times 0.0016384 \times 1 = 0.0016384\).

Thus, \(P(X \leq 4) = 1 - (0.0774144 + 0.0172032 + 0.0016384) = 1 - 0.096256 = 0.903744\).

Rounding to three decimal places, the probability is \(0.904\).

(i) Let the random variable \(X\) represent the number of days with more than 20 cm of snow in a 7-day period. \(X\) follows a binomial distribution with parameters \(n = 7\) and \(p = 0.21\).

We need to find \(P(X < 5)\), which is \(1 - P(X \geq 5)\).

Calculate \(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\).

\(P(X = 5) = \binom{7}{5} (0.21)^5 (0.79)^2\)

\(P(X = 6) = \binom{7}{6} (0.21)^6 (0.79)^1\)

\(P(X = 7) = \binom{7}{7} (0.21)^7\)

\(P(X \geq 5) = 21(0.21)^5(0.79)^2 + 7(0.21)^6(0.79) + (0.21)^7\)

\(P(X \geq 5) = 0.006 + 0.0003 + 0.0000008 \approx 0.0063\)

Thus, \(P(X < 5) = 1 - 0.0063 = 0.994\).

(ii) Let \(Y\) be the number of 7-day periods with at least 1 day of more than 20 cm of snow. The probability of at least 1 day with more than 20 cm of snow in a 7-day period is \(1 - (0.79)^7 = 0.808\).

\(Y\) follows a binomial distribution with parameters \(n = 4\) and \(p = 0.808\).

We need to find \(P(Y = 3)\).

\(P(Y = 3) = \binom{4}{3} (0.808)^3 (0.192)^1\)

\(P(Y = 3) = 4(0.808)^3(0.192)\)

\(P(Y = 3) = 4(0.527)(0.192) \approx 0.405\).

First, calculate the probability of a person being group O+ given they are Rhesus +. The probability of being O+ is 37%, and the probability of being Rhesus + is the sum of A+, B+, AB+, and O+, which is 35% + 8% + 3% + 37% = 83%.

Thus, the probability of being O+ given Rhesus + is:

\(P(O \text{ given } +) = \frac{0.37}{0.83} = 0.4458\)

Now, use the binomial distribution to find the probability that fewer than 3 out of 9 people are group O+.

Let \(X\) be the number of people who are group O+. Then \(X \sim \text{Binomial}(9, 0.4458)\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

\(P(X = 0) = \binom{9}{0} (0.4458)^0 (0.5542)^9\)

\(P(X = 1) = \binom{9}{1} (0.4458)^1 (0.5542)^8\)

\(P(X = 2) = \binom{9}{2} (0.4458)^2 (0.5542)^7\)

Calculating these:

\(P(X = 0) = (0.5542)^9\)

\(P(X = 1) = 9 \times (0.4458) \times (0.5542)^8\)

\(P(X = 2) = 36 \times (0.4458)^2 \times (0.5542)^7\)

Summing these probabilities gives:

\(P(X < 3) = 0.156\)

(i) The probability that a boy chooses Music is 0.15 (since 75% choose Games and 10% choose Drama, leaving 15% for Music). We use the binomial distribution with parameters: number of trials = 6, probability of success (choosing Music) = 0.15.

The probability that fewer than 3 boys choose Music is:

\(P(0, 1, 2) = (0.85)^6 + (0.15)(0.85)^5 \binom{6}{1} + (0.15)^2(0.85)^4 \binom{6}{2}\)

Calculating each term:

\((0.85)^6 = 0.377149515625\)

\((0.15)(0.85)^5 \binom{6}{1} = 0.2679075\)

\((0.15)^2(0.85)^4 \binom{6}{2} = 0.308\)

Summing these probabilities gives \(0.953\).

(ii) The probability that a student is a Drama student is:

\(P(D) = 0.6 \times 0.1 + 0.4 \times 0.55 = 0.28\)

The probability that a Drama student is a boy is:

\(P(B|D) = \frac{P(B \cap D)}{P(D)} = \frac{0.06}{0.28} = 0.2143\)

The probability that at least 1 of the 5 Drama students is a boy is:

\(P(>1) = 1 - P(0) = 1 - (0.7857)^5\)

\(P(0) = (0.7857)^5 = 0.7078\)

Thus, \(P(>1) = 1 - 0.7078 = 0.2922\)

Therefore, the probability is \(0.701\).

(ii) The mean of \(X\) is given by \(E(X) = 14 \times 0.75 = 10.5\). We evaluate the probabilities for \(X = 10\) and \(X = 11\):

\(P(10) = \binom{14}{10} (0.75)^{10} (0.25)^4 = 0.220\)

\(P(11) = \binom{14}{11} (0.75)^{11} (0.25)^3 = 0.240\)

The value of \(X\) with the highest probability is 11.

(iii) We first find the probability that Sue completes more than 11 puzzles in a month:

\(P(>11) = \binom{14}{12} (0.75)^{12} (0.25)^2 + \binom{14}{13} (0.75)^{13} (0.25)^1 + (0.75)^{14} = 0.281\)

Now, we find the probability that in exactly 3 of the next 5 months, Sue completes more than 11 puzzles:

\(P(3) = \binom{5}{3} (0.281)^3 (0.7189)^2 = 0.115\)

(i) The three conditions for a binomial distribution are:

1. Constant probability of success.
2. Independent trials.
3. Fixed number of trials with only two outcomes (success or failure).

(ii) Let the random variable \(X\) represent the number of days Julie's train is late out of 9 days. \(X\) follows a binomial distribution with parameters \(n = 9\) and \(p = 0.3\).

We need to find \(P(X > 7) + P(X < 2)\).

\(P(X > 7) = P(X = 8) + P(X = 9)\)

\(P(X = 8) = \binom{9}{8} (0.3)^8 (0.7)^1\)

\(P(X = 9) = \binom{9}{9} (0.3)^9\)

\(P(X < 2) = P(X = 0) + P(X = 1)\)

\(P(X = 0) = \binom{9}{0} (0.3)^0 (0.7)^9\)

\(P(X = 1) = \binom{9}{1} (0.3)^1 (0.7)^8\)

Calculating these probabilities:

\(P(X = 8) = 9 \times (0.3)^8 \times (0.7) = 0.000015\)

\(P(X = 9) = (0.3)^9 = 0.00000019683\)

\(P(X = 0) = (0.7)^9 = 0.0403536\)

\(P(X = 1) = 9 \times (0.3) \times (0.7)^8 = 0.121060821\)

Adding these probabilities:

\(P(X > 7) + P(X < 2) = 0.000015 + 0.00000019683 + 0.0403536 + 0.121060821 = 0.196\)

This problem can be modeled using a binomial distribution where the probability of success (Martin playing games when his friend phones) is 0.25, and the probability of failure is 0.75. We are looking for the probability of exactly 2 successes in 8 trials.

The probability mass function for a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.

Substituting the given values:

\(P(X = 2) = \binom{8}{2} (0.25)^2 (0.75)^6\)

Calculating each part:

\(\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28\)

\((0.25)^2 = 0.0625\)

\((0.75)^6 = 0.17803125\)

Therefore,

\(P(X = 2) = 28 \times 0.0625 \times 0.17803125 = 0.311\)

Let the random variable \(X\) represent the number of buses in the sample of 11 vehicles. Since 16% of the vehicles are buses, the probability of selecting a bus is \(p = 0.16\).

We need to find \(P(X < 3) = P(0) + P(1) + P(2)\).

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), where \(n = 11\) and \(p = 0.16\):

\(P(0) = \binom{11}{0} (0.16)^0 (0.84)^{11} = (0.84)^{11} = 0.1469\)

\(P(1) = \binom{11}{1} (0.16)^1 (0.84)^{10} = 11 \times 0.16 \times (0.84)^{10} = 0.30782\)

\(P(2) = \binom{11}{2} (0.16)^2 (0.84)^9 = 55 \times (0.16)^2 \times (0.84)^9 = 0.2931\)

Therefore, \(P(X < 3) = 0.1469 + 0.30782 + 0.2931 = 0.748\).

The problem involves a binomial distribution where the probability of success (a resident being in favour) is 0.8, and the number of trials is 20. We need to find the probability that more than 17 residents are in favour, i.e., \(P(X > 17)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

We calculate:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20}\)

Summing these probabilities gives:

\(P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20)\)

Calculating each term:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2 = 0.13691\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1 = 0.05765\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20} = 0.01153\)

Adding these gives:

\(P(X > 17) = 0.13691 + 0.05765 + 0.01153 = 0.20609\)

Rounding to three decimal places, the probability is 0.206.

The probability of rolling an odd number (1, 3, or 5) is given by:

\(P(\text{odd}) = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{2}{3}\)

We need to find the probability of getting at least 7 odd numbers in 8 throws, which is \(P(7) + P(8)\).

Using the binomial probability formula:

\(P(7) = \binom{8}{7} \left( \frac{2}{3} \right)^7 \left( \frac{1}{3} \right)^1 = 8 \times \left( \frac{2}{3} \right)^7 \times \frac{1}{3} = 0.156\)

\(P(8) = \binom{8}{8} \left( \frac{2}{3} \right)^8 = \left( \frac{2}{3} \right)^8 = 0.0390\)

Thus, the probability of obtaining at least 7 odd numbers is:

\(P(7 \text{ or } 8) = 0.156 + 0.0390 = 0.195\)

The probability of throwing a 5 is given as 0.75. Therefore, the probability of not throwing a 5 is:

\(1 - 0.75 = 0.25\)

We need to find the probability of getting at least 8 fives in 10 throws. This is a binomial probability problem where:

\(n = 10, \ p = 0.75\)

The probability of getting exactly \(r\) fives is given by the binomial formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

We need to calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

Calculating each term:

\(P(X = 8) = \binom{10}{8} (0.75)^8 (0.25)^2\)

\(P(X = 9) = \binom{10}{9} (0.75)^9 (0.25)^1\)

\(P(X = 10) = \binom{10}{10} (0.75)^{10}\)

Substituting the values:

\(P(X = 8) = 45 \times (0.75)^8 \times (0.25)^2\)

\(P(X = 9) = 10 \times (0.75)^9 \times (0.25)\)

\(P(X = 10) = 1 \times (0.75)^{10}\)

Summing these probabilities gives:

\(P(X \geq 8) = 0.526\)

Let \(X\) be the number of years out of 15 that have New Year's Day on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 15\) and \(p = \frac{1}{7}\).

We need to find \(P(X \geq 3)\), which is equivalent to \(1 - P(X \leq 2)\).

Calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\):

\(P(X = 0) = \left( \frac{6}{7} \right)^{15}\)

\(P(X = 1) = \binom{15}{1} \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14}\)

\(P(X = 2) = \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

Thus, \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\).

\(P(X \leq 2) = \left( \frac{6}{7} \right)^{15} + 15 \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14} + \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

\(P(X \leq 2) \approx 0.0990 + 0.2476 + 0.2889 = 0.6355\)

Therefore, \(P(X \geq 3) = 1 - 0.6355 = 0.3645\).

Thus, the probability that at least 3 of these years have New Year's Day on a Saturday is approximately 0.365 (accept 0.364).

To solve this problem, we use the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

where \(n\) is the total number of trials, \(k\) is the number of successful trials, and \(p\) is the probability of success on a single trial.

(i) For equal numbers of large and small bands, we need 10 large and 10 small bands. The probability of a band being large is 0.4, and small is 0.6. Thus,

\(P(10 \text{ large and } 10 \text{ small}) = \binom{20}{10} (0.6)^{10} (0.4)^{10}\)

Calculating this gives:

\(= 0.117\)

(ii) For more than 17 small bands, we consider 18, 19, or 20 small bands. The probability of a band being small is 0.6. Thus,

\(P(18, 19, 20 \text{ small}) = \binom{20}{18} (0.6)^{18} (0.4)^{2} + \binom{20}{19} (0.6)^{19} (0.4)^{1} + (0.6)^{20}\)

Calculating each term:

\(= 0.003087 + 0.000487 + 0.00003635\)

Summing these gives:

\(= 0.00361\)

Let the probability that an adult does not wear a watch be 0.09. We are dealing with a binomial distribution where the number of trials is 14 and the probability of success (an adult not wearing a watch) is 0.09.

We need to find the probability that more than 2 adults do not wear a watch, which is equivalent to finding:

\(P(X > 2) = 1 - P(X \leq 2)\)

Calculate \(P(X \leq 2)\) as follows:

\(P(X = 0) = (0.91)^{14}\)

\(P(X = 1) = \binom{14}{1} (0.09)^1 (0.91)^{13}\)

\(P(X = 2) = \binom{14}{2} (0.09)^2 (0.91)^{12}\)

Thus,

\(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

Substitute the values:

\(P(X = 0) = 0.2670\)

\(P(X = 1) = 0.3698\)

\(P(X = 2) = 0.2377\)

Therefore,

\(P(X \leq 2) = 0.2670 + 0.3698 + 0.2377 = 0.8745\)

Finally,

\(P(X > 2) = 1 - 0.8745 = 0.1255\)

Rounding to three decimal places, the probability is 0.126.

(i) The two conditions for a binomial distribution are:

1. There are a fixed number of trials.
2. Each trial is independent.
3. There are only two possible outcomes for each trial.
4. The probability of success is constant for each trial.

(ii) Let the random variable \(X\) represent the number of cars made by Ford. \(X\) follows a binomial distribution with parameters \(n = 14\) and \(p = 0.28\).

We need to find \(P(X < 4)\), which is the sum of probabilities \(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 0) = \binom{14}{0} (0.28)^0 (0.72)^{14} = 0.72^{14} = 0.0101\)

\(P(X = 1) = \binom{14}{1} (0.28)^1 (0.72)^{13} = 14 \times 0.28 \times 0.72^{13} = 0.0548\)

\(P(X = 2) = \binom{14}{2} (0.28)^2 (0.72)^{12} = 91 \times 0.28^2 \times 0.72^{12} = 0.1385\)

\(P(X = 3) = \binom{14}{3} (0.28)^3 (0.72)^{11} = 364 \times 0.28^3 \times 0.72^{11} = 0.2154\)

Summing these probabilities gives:

\(P(X < 4) = 0.0101 + 0.0548 + 0.1385 + 0.2154 = 0.419\)

Let the random variable \(X\) represent the number of damaged tapes in a sample of 15. The probability of a tape being damaged is \(p = 0.2\), and the probability of a tape not being damaged is \(1 - p = 0.8\).

We use the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) where \(n = 15\).

Calculate \(P(X = 0)\):

\(P(0) = (0.8)^{15} = 0.03518\)

Calculate \(P(X = 1)\):

\(P(1) = \binom{15}{1} (0.2)^1 (0.8)^{14} = 15 \times 0.2 \times (0.8)^{14} = 0.1319\)

Calculate \(P(X = 2)\):

\(P(2) = \binom{15}{2} (0.2)^2 (0.8)^{13} = 105 \times (0.2)^2 \times (0.8)^{13} = 0.2309\)

The probability that at most 2 tapes are damaged is:

\(P(X \leq 2) = P(0) + P(1) + P(2) = 0.03518 + 0.1319 + 0.2309 = 0.398\)

Let the probability of getting the required picture be 0.05 (since there is 1 desired picture out of 20).

1. The probability that Richard does not complete his collection is calculated by not getting the required picture in all 5 attempts:

   \((0.95)^5 = 0.774\)

2. The probability that he has the required picture exactly once is calculated using the binomial probability formula:

   \((0.95)^4 \times (0.05)^1 \times \binom{5}{1} = 0.204\)

3. The probability that he completes his collection with the third picture he looks at is calculated by not getting the picture in the first two attempts and getting it on the third:

   \((0.95)^2 \times (0.05) = 0.0451\)

Let the probability of a bag being underweight be \(p = 0.1\). The number of underweight bags follows a binomial distribution \(X \sim B(10, 0.1)\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.1)^0 (0.9)^{10} = 0.348678\)

\(P(X = 1) = \binom{10}{1} (0.1)^1 (0.9)^9 = 0.38742\)

\(P(X = 2) = \binom{10}{2} (0.1)^2 (0.9)^8 = 0.19371\)

Sum these probabilities:

\(P(X < 3) = 0.348678 + 0.38742 + 0.19371 = 0.929808\)

Rounding to three decimal places, the probability is \(0.930\).

Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.9\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 10) = \binom{12}{10} (0.9)^{10} (0.1)^2 = 0.230128\)

\(P(X = 11) = \binom{12}{11} (0.9)^{11} (0.1)^1 = 0.376573\)

\(P(X = 12) = \binom{12}{12} (0.9)^{12} (0.1)^0 = 0.282430\)

\(P(X \geq 10) = 0.230128 + 0.376573 + 0.282430 = 0.889131\)

Thus, \(P(X < 10) = 1 - 0.889131 = 0.110869\).

Therefore, the probability that fewer than 10 callers are connected immediately is approximately 0.111.

(i) Let the probability of throwing a 4 be denoted by \(p\). The expected number of 4s in 30 throws is given by \(30p\). We are given that this average is 6.21, so:

\(30p = 6.21\)

Solving for \(p\), we get:

\(p = \frac{6.21}{30} = 0.207\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\). Here, \(n = 30\) and \(p = 0.207\), so:

\(\text{Variance} = 30 \times 0.207 \times (1 - 0.207)\)

\(= 30 \times 0.207 \times 0.793\)

\(= 4.92\)

(iii) For 15 throws, the probability of obtaining 2 or more 4s can be found using the complement rule. First, find the probability of obtaining fewer than 2 4s (i.e., 0 or 1 4s) using the binomial probability formula:

\(P(X = 0) = \binom{15}{0} (0.207)^0 (0.793)^{15} = (0.793)^{15}\)

\(P(X = 1) = \binom{15}{1} (0.207)^1 (0.793)^{14} = 15 \times 0.207 \times (0.793)^{14}\)

Calculate these probabilities and sum them:

\(P(X < 2) = (0.793)^{15} + 15 \times 0.207 \times (0.793)^{14}\)

Finally, the probability of obtaining 2 or more 4s is:

\(P(X \geq 2) = 1 - P(X < 2) = 0.848\)

(i) The probability of selecting a non-orange disc is \(\frac{200}{300} = \frac{2}{3}\). Since the selection is with replacement, the probability that no orange discs are selected in 5 draws is \(\left( \frac{2}{3} \right)^5\).

(ii) The probability of selecting a disc with a number ending in 6 is \(\frac{10}{100} = \frac{1}{10}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3\).

(iii) The probability of selecting an orange disc with a number ending in 6 is \(\frac{10}{300} = \frac{1}{30}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{30} \right)^2 \left( \frac{29}{30} \right)^3\).

(iv) The probability of selecting a pink disc is \(\frac{100}{300} = \frac{1}{3}\). For a binomial distribution with \(n = 5\) and \(p = \frac{1}{3}\), the mean is \(np = 5 \times \frac{1}{3} = \frac{5}{3}\) and the variance is \(np(1-p) = 5 \times \frac{1}{3} \times \frac{2}{3} = \frac{10}{9}\).

Let the number of plants per box be denoted by \(n\), and the probability of a plant producing yellow flowers be \(p\). The number of plants producing yellow flowers follows a binomial distribution \(B(n, p)\).

The mean of a binomial distribution is given by \(np = 11\) and the variance is \(np(1-p) = 4.95\).

From \(np = 11\), we have \(p = \frac{11}{n}\).

Substitute \(p\) into the variance equation:

\(np(1-p) = 4.95\)

\(n \left(\frac{11}{n}\right) \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 - \frac{121}{n} = 4.95\)

\(11 - 4.95 = \frac{121}{n}\)

\(6.05 = \frac{121}{n}\)

\(n = \frac{121}{6.05} \approx 20\)

Thus, there are 20 plants per box.

For part (b), we need to find the probability that exactly 12 plants produce yellow flowers, i.e., \(P(X = 12)\).

Using \(n = 20\) and \(p = \frac{11}{20} = 0.55\), the probability mass function of a binomial distribution is:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

\(P(X = 12) = \binom{20}{12} (0.55)^{12} (0.45)^{8}\)

\(P(X = 12) \approx 0.162\)

Let the number of days Khalid rides his bicycle be a binomial random variable with parameters 45 and 0.6. The variance is given by:

\(\text{Var}(B) = 45 \times 0.6 \times 0.4\)

Similarly, let the number of days Khalid rides his skateboard be a binomial random variable with parameters 45 and 0.4. The variance is given by:

\(\text{Var}(S) = 45 \times 0.4 \times 0.6\)

Since \(\text{Var}(B) = \text{Var}(S)\), the variances are the same.

(i) The probability that a visitor sees all the animals is the product of the individual probabilities:

\(0.9 \times 0.95 \times 0.85 \times 0.1 = 0.0727\).

(ii) Let the random variable \(X\) be the number of visitors who see lions. \(X\) follows a binomial distribution \(X \sim B(12, 0.1)\). We need to find \(P(X < 3)\), which is \(P(0) + P(1) + P(2)\).

\(P(0) = (0.9)^{12}\)

\(P(1) = \binom{12}{1} (0.1)(0.9)^{11}\)

\(P(2) = \binom{12}{2} (0.1)^2(0.9)^{10}\)

Thus, \(P(X < 3) = (0.9)^{12} + \binom{12}{1} (0.1)(0.9)^{11} + \binom{12}{2} (0.1)^2(0.9)^{10} = 0.889\).

(iii) Let \(Y\) be the number of people who see zebras. \(Y\) follows a binomial distribution \(Y \sim B(50, 0.85)\). The mean \(E(Y)\) is given by \(np\) and the variance \(Var(Y)\) is given by \(np(1-p)\).

Mean: \(50 \times 0.85 = 42.5\)

Variance: \(50 \times 0.85 \times 0.15 = 6.375\).

(i) Let the probability of a screw being faulty be \(p\). Given the mean number of faulty screws per packet is 1.2, we have:

\(1.2 = 15p\)

Solving for \(p\), we get:

\(p = \frac{1.2}{15} = 0.08\)

The variance \(\text{Var} = npq\), where \(q = 1 - p\).

\(\text{Var} = 15 \times 0.08 \times 0.92 = 1.104\)

(ii) The probability that a packet contains at most 2 faulty screws is:

\(P(0, 1, 2) = (0.92)^{15} + 15C_1(0.08)(0.92)^{14} + 15C_2(0.08)^2(0.92)^{13}\)

Calculating each term:

\((0.92)^{15} = 0.275\)

\(15 \times 0.08 \times (0.92)^{14} = 0.359\)

\(15C_2 \times (0.08)^2 \times (0.92)^{13} = 0.253\)

Adding these probabilities gives:

\(P(0, 1, 2) = 0.275 + 0.359 + 0.253 = 0.887\)

(iii) The probability that there is at least 1 faulty screw in a packet is:

\(P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15}\)

\(= 1 - 0.275 = 0.7137\)

The probability that there are exactly 7 packets with at least 1 faulty screw is:

\(P(\text{exactly 7 packets}) = 8C_7(0.7137)^7(0.2863)\)

\(= 0.216\)

(i) Let the probability that a random integer is less than or equal to 4 be p. Since there are 4 integers (1, 2, 3, 4) less than or equal to 4 out of 9, p = 4/9. The probability that at least 2 of the 5 integers are less than or equal to 4 is given by:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - \left(\frac{5}{9}\right)^5 - \binom{5}{1} \left(\frac{4}{9}\right) \left(\frac{5}{9}\right)^4\)

\(= 0.735\)

(ii) We know that the mean \(np = 96\) and the variance \(npq = 32\), where \(q = 1 - p\).

From \(np = 96\), we have \(p = \frac{96}{n}\).

From \(npq = 32\), we have \(np(1-p) = 32\).

Substituting \(p = \frac{96}{n}\) into \(npq = 32\), we get:

\(n \left(\frac{96}{n}\right) \left(1 - \frac{96}{n}\right) = 32\)

\(96 \left(1 - \frac{96}{n}\right) = 32\)

\(96 - \frac{9216}{n} = 32\)

\(\frac{9216}{n} = 64\)

\(n = 144\)

Since \(p = \frac{96}{144} = \frac{2}{3}\), \(k\) is such that \(P(\leq k) = \frac{2}{3}\). Therefore, \(k = 6\).

The average number of 5s in 20 throws is given as 4.8. This implies that the expected value for the number of 5s is 4.8, which can be expressed as:

\(20p = 4.8\)

Solving for \(p\), we get:

\(p = \frac{4.8}{20} = 0.24\)

We need to find the probability that the number of 5s is less than 3 in 20 throws. This is a binomial distribution problem where \(n = 20\) and \(p = 0.24\).

The probability of getting exactly \(k\) successes in a binomial distribution is given by:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

We need to calculate \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Calculating each term:

\(P(X = 0) = \binom{20}{0} (0.24)^0 (0.76)^{20} = (0.76)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.24)^1 (0.76)^{19} = 20 \times 0.24 \times (0.76)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.24)^2 (0.76)^{18} = \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Summing these probabilities:

\(P(X < 3) = (0.76)^{20} + 20 \times 0.24 \times (0.76)^{19} + \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Calculating the above gives:

\(P(X < 3) = 0.109\)

Let the probability of a biscuit being broken be denoted by \(p\). Since the mean number of broken biscuits is 2.7, we have:

\(18p = 2.7\)

Solving for \(p\), we get:

\(p = \frac{2.7}{18} = 0.15\)

We need to find the probability that the number of broken biscuits \(X\) is between 2 and 4 inclusive, i.e., \(P(2 \leq X \leq 4)\).

\(X\) follows a binomial distribution \(B(18, 0.15)\). Therefore,

\(P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16}\)

\(P(X = 3) = \binom{18}{3} (0.15)^3 (0.85)^{15}\)

\(P(X = 4) = \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Summing these probabilities gives:

\(P(2 \leq X \leq 4) = \binom{18}{2} (0.15)^2 (0.85)^{16} + \binom{18}{3} (0.15)^3 (0.85)^{15} + \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Calculating each term:

\(\binom{18}{2} (0.15)^2 (0.85)^{16} \approx 0.231\)

\(\binom{18}{3} (0.15)^3 (0.85)^{15} \approx 0.287\)

\(\binom{18}{4} (0.15)^4 (0.85)^{14} \approx 0.137\)

Adding these gives:

\(P(2 \leq X \leq 4) = 0.231 + 0.287 + 0.137 = 0.655\)

(i) We use the binomial distribution where the probability of going to the park on any day is 0.6. We need to find the probability of going to the park more than 5 times in 7 days, i.e., 6 or 7 times.

The probability of going exactly 6 times is given by:

\(\binom{7}{6} (0.6)^6 (0.4)^1 = 7 \times 0.046656 \times 0.4 = 0.1306\)

The probability of going exactly 7 times is given by:

\(\binom{7}{7} (0.6)^7 = 1 \times 0.279936 = 0.02799\)

Thus, the probability of going more than 5 times is:

\(0.1306 + 0.02799 = 0.159\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\).

Here, \(n = 30\) and \(p = 0.6\).

So, the variance is:

\(30 \times 0.6 \times 0.4 = 7.2\)

Let the number of defective batteries in a pack be a binomial random variable, denoted by \(X\), with parameters \(n = 20\) and probability of defect \(p\).

Given the mean number of defective batteries is 1.6, we have:

\(20p = 1.6\)

Solving for \(p\), we get:

\(p = \frac{1.6}{20} = 0.08\)

We need to find \(P(X > 2)\). Using the complement rule:

\(P(X > 2) = 1 - P(X \leq 2)\)

Calculate \(P(X \leq 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{20}{0} (0.08)^0 (0.92)^{20} = (0.92)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.08)^1 (0.92)^{19} = 20 \times 0.08 \times (0.92)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.08)^2 (0.92)^{18} = \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Thus,

\(P(X \leq 2) = (0.92)^{20} + 20 \times 0.08 \times (0.92)^{19} + \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Substituting the values:

\(P(X \leq 2) = 0.1887 + 0.3281 + 0.2711 = 0.7879\)

Therefore,

\(P(X > 2) = 1 - 0.7879 = 0.212\)

The probability that a customer rates the logo as good is 0.42. Therefore, the probability that a customer does not rate the logo as good is 0.58.

The probability that none of the n customers rate the logo as good is given by:

\((0.58)^n\)

We want the probability that at least one person rates the logo as good to be greater than 0.995. Therefore, we have:

\(1 - (0.58)^n > 0.995\)

Rearranging gives:

\((0.58)^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.58) < \log(0.005)\)

Solving for n gives:

\(n > \frac{\log(0.005)}{\log(0.58)}\)

Calculating the right-hand side:

\(n > 9.727\)

Since n must be an integer, the smallest value of n is 10.

The probability that a water pistol works properly is 92% or 0.92.

The probability that all n water pistols work properly is given by:

\(P(0) = (0.92)^n\)

We want the probability that at least one does not work properly to be greater than 0.9:

\(1 - P(0) > 0.9\)

Substituting for \(P(0)\):

\(1 - (0.92)^n > 0.9\)

Rearranging gives:

\((0.92)^n < 0.1\)

Taking logarithms on both sides:

\(n \log(0.92) < \log(0.1)\)

Solving for \(n\):

\(n > \frac{\log(0.1)}{\log(0.92)}\)

Calculating gives:

\(n > 27.6\)

Since \(n\) must be an integer, the smallest possible value is:

28

The probability that a mango is not small is 0.85 (since 15% are small).

The probability that none of the n mangoes is small is given by (0.85)^n.

We need to find the largest n such that (0.85)^n \geq 0.1.

Taking logarithms on both sides, we have:

\(n \leq \frac{\log(0.1)}{\log(0.85)} \approx 14.2\)

\(Since n must be an integer, the largest possible value is n = 14.\)

The probability that a student does not have blue eyes is given by:

\(P( ext{not blue eyes}) = 1 - \frac{1}{5} = \frac{4}{5} = 0.8\)

For n students, the probability that none of them has blue eyes is:

\((0.8)^n\)

We need this probability to be less than 0.001:

\((0.8)^n < 0.001\)

Taking logarithms on both sides:

\(n \log(0.8) < \log(0.001)\)

Solving for n:

\(n > \frac{\log(0.001)}{\log(0.8)}\)

Calculating the values:

\(n > \frac{-3}{-0.09691} \approx 30.9\)

Since n must be an integer, the least possible value of n is 31.

(i) The integers between 7 and 21 that are multiples of 5 are 10, 15, and 20. There are 3 such numbers out of 15 total numbers (7 to 21 inclusive), so the probability \(p\) that a randomly chosen integer is a multiple of 5 is \(\frac{3}{15} = 0.2\). Therefore, \(X \sim \text{Bin}(12, 0.2)\).

(ii) We need to calculate \(P(3 \leq X \leq 5)\). This is given by:

\(P(X = 3, 4, 5) = \binom{12}{3} (0.2)^3 (0.8)^9 + \binom{12}{4} (0.2)^4 (0.8)^8 + \binom{12}{5} (0.2)^5 (0.8)^7\)

Calculating each term:

\(\binom{12}{3} (0.2)^3 (0.8)^9 = 0.23622\)

\(\binom{12}{4} (0.2)^4 (0.8)^8 = 0.13287\)

\(\binom{12}{5} (0.2)^5 (0.8)^7 = 0.05315\)

Adding these probabilities gives \(0.23622 + 0.13287 + 0.05315 = 0.422\).

(iii) We want \(P(X = 0) < 0.01\). For \(X \sim \text{Bin}(n, 0.2)\), \(P(X = 0) = (0.8)^n\). We solve:

\((0.8)^n < 0.01\)

Taking logarithms, \(n \log(0.8) < \log(0.01)\).

\(n > \frac{\log(0.01)}{\log(0.8)} \approx 20.65\)

Thus, the smallest integer \(n\) is 21.

(i) Let the random variable \(X\) represent the number of customers who rate the food as ‘good’. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.65\). We need to find \(P(2 < X < 12)\).

\(P(2 < X < 12) = 1 - P(X \leq 2)\)

\(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(P(X = 0) = (0.35)^{12}\)

\(P(X = 1) = \binom{12}{1} (0.65)^1 (0.35)^{11}\)

\(P(X = 2) = \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = (0.35)^{12} + \binom{12}{1} (0.65)^1 (0.35)^{11} + \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = 0.0065359\)

\(P(2 < X < 12) = 1 - 0.0065359 = 0.993\)

(ii) Let \(Y\) be the number of customers who rate the food as ‘poor’. \(Y\) follows a binomial distribution with parameters \(n\) and \(p = 0.13\). We need \(P(Y \geq 1) > 0.95\).

\(P(Y \geq 1) = 1 - P(Y = 0)\)

\(P(Y = 0) = (0.87)^n\)

\(1 - (0.87)^n > 0.95\)

\((0.87)^n < 0.05\)

Solving \((0.87)^n < 0.05\) gives \(n = 22\).

The probability that Sue completes all n puzzles correctly is given by \((0.75)^n\).

We need to find the smallest n such that \((0.75)^n < 0.06\).

Taking logarithms on both sides, we have:

\(\log((0.75)^n) < \log(0.06)\)

\(n \cdot \log(0.75) < \log(0.06)\)

\(n > \frac{\log(0.06)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 9.78\).

Since n must be an integer, the smallest possible value is \(n = 10\).

(i) The probability of not getting a 3 on a single die is \(\frac{5}{6}\). Therefore, the probability of not getting a 3 on all 9 dice is \(\left( \frac{5}{6} \right)^9\). The probability of getting at least one 3 is:

\(1 - \left( \frac{5}{6} \right)^9 = 1 - 0.1938 = 0.806\)

(ii) The probability of getting at least one 3 when \(n\) dice are thrown is \(1 - \left( \frac{5}{6} \right)^n\). We need this probability to be greater than 0.9:

\(1 - \left( \frac{5}{6} \right)^n > 0.9\)

\(\left( \frac{5}{6} \right)^n < 0.1\)

Taking logarithms on both sides:

\(n \log \left( \frac{5}{6} \right) < \log(0.1)\)

\(n > \frac{\log(0.1)}{\log \left( \frac{5}{6} \right)}\)

Calculating gives \(n > 12.6\), so the smallest integer \(n\) is 13.

(i) The probability of performing the routine correctly is 0.65, and incorrectly is 0.35. We use the binomial probability formula:

\(P(X = 5) = (0.65)^5 \times (0.35)^2 \times \binom{7}{5}\)

\(= (0.65)^5 \times (0.35)^2 \times 21\)

\(= 0.298\)

(iii) The expected number of correct performances is given by \(np\), where \(p = 0.65\). We need \(np \geq 8\).

\(0.65n \geq 8\)

\(n \geq \frac{8}{0.65} \approx 12.31\)

The smallest integer \(n\) is 13.

The probability that a tape is not damaged is 0.8 (since 1 in 5 are damaged, so 4 in 5 are not).

The probability that all n tapes in a sample are not damaged is given by (0.8)^n.

The probability that at least one tape is damaged is therefore 1 - (0.8)^n.

We need this probability to be at least 0.85:

1 - (0.8)^n \geq 0.85

(0.8)^n \leq 0.15

\(By trial and error or using logarithms, we find that n = 9 satisfies this inequality.\)

The probability that Janice does not buy an item online in any week is given by:

\(1 - 0.35 = 0.65\)

The probability that Janice does not buy any item online in n weeks is:

\((0.65)^n\)

The probability that Janice buys at least one item online in n weeks is greater than 0.99, so:

\(1 - (0.65)^n > 0.99\)

Rearranging gives:

\((0.65)^n < 0.01\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.65)} \approx 10.69\)

Since n must be an integer, the smallest possible value of n is 11.

\(The probability that a customer does not shop online is 0.65 (since 1 - 0.35 = 0.65).\)

The probability that none of the n customers shop online is given by:

(0.65)^n

We want the probability that at least one customer shops online to be greater than 0.95:

1 - (0.65)^n > 0.95

Rearranging gives:

(0.65)^n < 0.05

Taking logarithms on both sides:

n \log(0.65) < \log(0.05)

Solving for n gives:

\(n > \frac{\log(0.05)}{\log(0.65)} \approx 6.95\)

Therefore, the least integer value of n is 7.

The probability that a phone is not made by Company B is 0.65 (since 35% are made by Company B).

The probability that none of the n phones is made by Company B is given by \((0.65)^n\).

We want the probability that at least one phone is made by Company B to be more than 0.98, so:

\(1 - (0.65)^n > 0.98\)

\((0.65)^n < 0.02\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.02)\)

\(n > \frac{\log(0.02)}{\log(0.65)}\)

Calculating the right-hand side gives \(n > 9.08\).

Since n must be an integer, the smallest possible value is \(n = 10\).

The probability that a student does not own a car is \(1 - \frac{1}{4} = \frac{3}{4} = 0.75\).

The probability that none of the \(n\) students owns a car is \(0.75^n\).

We need the probability that at least one student owns a car to be greater than 0.995:

\(1 - 0.75^n > 0.995\)

\(0.75^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.75) < \log(0.005)\)

\(n > \frac{\log(0.005)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 18.4\).

Therefore, the least possible integer value of \(n\) is 19.

(i) The mean number of cracked eggs is 1.4, so the probability of a cracked egg, p, is given by:

\(p = \frac{1.4}{20} = 0.07\)

The probability of exactly 2 cracked eggs in a box of 20 is calculated using the binomial distribution:

\(P(2) = \binom{20}{2} (0.07)^2 (0.93)^{18}\)

Calculating this gives:

\(P(2) = 0.252\)

(ii) The probability of at least 1 cracked egg is:

\(P(\text{at least 1 cracked egg}) = 1 - P(0)\)

Where:

\(P(0) = (0.93)^{20}\)

Thus:

\(P(\text{at least 1 cracked egg}) = 1 - (0.93)^{20} = 0.766\)

(iii) We need to find the smallest n such that:

\((0.766)^n < 0.01\)

Taking logarithms:

\(n \log(0.766) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.766)} \approx 17.8\)

Thus, the smallest integer n is 18.

The probability that a person takes more than t minutes is 0.12. Therefore, the probability that a person takes t minutes or less is 0.88.

The probability that none of the n people takes more than t minutes is given by:

\((0.88)^n < 0.003\)

Taking logarithms on both sides, we have:

\(n \log(0.88) < \log(0.003)\)

Solving for n gives:

\(n > \frac{\log(0.003)}{\log(0.88)}\)

Calculating the right-hand side:

\(n > 45.4\)

Since n must be an integer, the smallest possible value is:

\(n = 46\)

(i) The probability that a person buys something is 0.66 (since 34% do not buy anything). We model this situation with a binomial distribution: \(X \sim B(15, 0.66)\).

The probability that at least 14 people buy something is \(P(X \geq 14) = P(X = 14) + P(X = 15)\).

Using the binomial probability formula:

\(P(X = 14) = \binom{15}{14} (0.66)^{14} (0.34)^1\)

\(P(X = 15) = \binom{15}{15} (0.66)^{15}\)

Calculating these gives:

\(P(X = 14) = 15 \times (0.66)^{14} \times 0.34\)

\(P(X = 15) = (0.66)^{15}\)

Adding these probabilities: \(P(X \geq 14) = 0.0171\).

(ii) The probability that a person spends less than $50 is 0.87. We want \((0.87)^n < 0.04\).

Taking logarithms: \(n \log(0.87) < \log(0.04)\).

Solving for \(n\):

\(n > \frac{\log(0.04)}{\log(0.87)} \approx 23.5\).

The smallest integer \(n\) is 24.

The probability that a person takes a holiday in their own country is 0.65. For a group of n people, the probability that all take a holiday in their own country is given by:

\((0.65)^n < 0.002\)

Taking logarithms on both sides, we have:

\(n \cdot \log(0.65) < \log(0.002)\)

Solving for n, we get:

\(n > \frac{\log(0.002)}{\log(0.65)}\)

Calculating the right-hand side:

\(\log(0.002) \approx -2.69897\)

\(\log(0.65) \approx -0.18709\)

Thus:

\(n > \frac{-2.69897}{-0.18709} \approx 14.43\)

Since n must be an integer, the smallest possible value of n is 15.