Past Exam Questions

(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 \(0.111\).