Past Exam Questions

First, use the fact that the sum of probabilities must equal 1:

\(0.12 + p + q + 0.16 + 0.3 = 1\)

Simplifying gives:

\(p + q = 0.42\)

Next, use the given expected value \(E(X) = 0.28\):

\(-2(0.12) - 1(p) + 0.5(q) + 1(0.16) + 2(0.3) = 0.28\)

This simplifies to:

\(-0.24 - p + 0.5q + 0.16 + 0.6 = 0.28\)

\(-p + 0.5q = -0.24\)

Now solve the system of equations:

1. \(p + q = 0.42\)
2. \(-p + 0.5q = -0.24\)

Add the two equations to eliminate \(p\):

\((p + q) + (-p + 0.5q) = 0.42 - 0.24\)

\(1.5q = 0.18\)

\(q = 0.12\)

Substitute \(q = 0.12\) back into \(p + q = 0.42\):

\(p + 0.12 = 0.42\)

\(p = 0.3\)

(a) The sum of probabilities must equal 1:

\(p + q + 0.6 + 0.05 = 1\)

\(p + q + 0.65 = 1\)

\(p + q = 0.35\) (1)

Using the expected value \(E(X) = 0.55\):

\(0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\)

\(p + 2q + 0.15 = 0.55\)

\(p + 2q = 0.4\) (2)

Solving equations (1) and (2):

From (1): \(p = 0.35 - q\)

Substitute into (2):

\(0.35 - q + 2q = 0.4\)

\(0.35 + q = 0.4\)

\(q = 0.05\)

Substitute \(q = 0.05\) into (1):

\(p + 0.05 = 0.35\)

\(p = 0.3\)

(b) The variance \(\text{Var}(X)\) is given by:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

Calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times 0.6 + 1^2 \times 0.3 + 2^2 \times 0.05 + 3^2 \times 0.05\)

\(E(X^2) = 0 + 0.3 + 0.2 + 0.45\)

\(E(X^2) = 0.95\)

Thus, \(\text{Var}(X) = 0.95 - 0.55^2\)

\(\text{Var}(X) = 0.95 - 0.3025\)

\(\text{Var}(X) = 0.6475\)

The sum of all probabilities must equal 1:

\(p + p + 0.1 + q + q = 1\)

Simplifying, we get:

\(2p + 0.1 + 2q = 1\)

Given that \(P(X \geq 0) = 3P(X < 0)\), we have:

\(0.1 + 2q = 3(2p)\)

Now, solve the system of equations:

1. \(2p + 0.1 + 2q = 1\)
2. \(0.1 + 2q = 6p\)

From equation (2), express \(q\) in terms of \(p\):

\(2q = 6p - 0.1\)

\(q = 3p - 0.05\)

Substitute \(q = 3p - 0.05\) into equation (1):

\(2p + 0.1 + 2(3p - 0.05) = 1\)

\(2p + 0.1 + 6p - 0.1 = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Substitute \(p = \frac{1}{8}\) back to find \(q\):

\(q = 3 \times \frac{1}{8} - 0.05\)

\(q = \frac{3}{8} - \frac{1}{20}\)

\(q = \frac{15}{40} - \frac{2}{40}\)

\(q = \frac{13}{40}\)

(i) The sum of all probabilities must equal 1:

\(\frac{1}{4} + p + p + \frac{3}{8} + 4p = 1\)

Simplifying, we get:

\(\frac{1}{4} + 2p + \frac{3}{8} + 4p = 1\)

\(\frac{1}{4} + \frac{3}{8} + 6p = 1\)

\(\frac{2}{8} + \frac{3}{8} + 6p = 1\)

\(\frac{5}{8} + 6p = 1\)

\(6p = 1 - \frac{5}{8}\)

\(6p = \frac{3}{8}\)

\(p = \frac{1}{16}\)

(ii) To find \(E(X)\):

\(E(X) = (-1) \times \frac{1}{4} + 0 \times \frac{1}{16} + 1 \times \frac{1}{16} + 2 \times \frac{3}{8} + 4 \times \frac{1}{4}\)

\(E(X) = -\frac{1}{4} + 0 + \frac{1}{16} + \frac{6}{8} + 1\)

\(E(X) = -\frac{1}{4} + \frac{1}{16} + \frac{6}{8} + 1\)

\(E(X) = \frac{-4}{16} + \frac{1}{16} + \frac{12}{16} + \frac{16}{16}\)

\(E(X) = \frac{25}{16}\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \sum (x^2 \cdot P(X = x)) - (E(X))^2\)

\(\text{Var}(X) = ((-1)^2 \times \frac{1}{4}) + (0^2 \times \frac{1}{16}) + (1^2 \times \frac{1}{16}) + (2^2 \times \frac{3}{8}) + (4^2 \times \frac{1}{4}) - \left(\frac{25}{16}\right)^2\)

\(\text{Var}(X) = \frac{1}{4} + 0 + \frac{1}{16} + \frac{12}{8} + \frac{16}{4} - \left(\frac{25}{16}\right)^2\)

\(\text{Var}(X) = \frac{1}{4} + \frac{1}{16} + \frac{12}{8} + 4 - \frac{625}{256}\)

\(\text{Var}(X) = \frac{863}{256} \; \text{or} \; 3.37\)

(iii) Given that \(X > 0\), find \(P(X = 2)\):

\(P(X > 0) = P(X = 1) + P(X = 2) + P(X = 4) = \frac{1}{16} + \frac{3}{8} + \frac{1}{4}\)

\(P(X > 0) = \frac{1}{16} + \frac{6}{16} + \frac{4}{16} = \frac{11}{16}\)

\(P(X = 2 | X > 0) = \frac{P(X = 2)}{P(X > 0)} = \frac{\frac{3}{8}}{\frac{11}{16}}\)

\(P(X = 2 | X > 0) = \frac{6}{11} \; \text{or} \; 0.545\)

(i) The sum of probabilities must equal 1: \(0.2 + 0.1 + p + 0.1 + q = 1\). Simplifying gives \(p + q = 0.6\).

Given \(E(X) = 1.7\), we have \(\Sigma px = -0.4 + 0 + p + 0.3 + 4q = 1.7\). Simplifying gives \(p + 4q = 1.8\).

Solving the simultaneous equations \(p + q = 0.6\) and \(p + 4q = 1.8\), we find \(p = 0.2\) and \(q = 0.4\).

(ii) The variance is given by \(\text{Var}(X) = \Sigma x^2p - (E(X))^2\).

Calculate \(\Sigma x^2p = 4 \times 0.2 + 1 \times p + 9 \times 0.1 + 16 \times q\).

Substitute \(p = 0.2\) and \(q = 0.4\) to get \(4 \times 0.2 + 1 \times 0.2 + 9 \times 0.1 + 16 \times 0.4 = 8.3\).

Then \(\text{Var}(X) = 8.3 - 1.7^2 = 8.3 - 2.89 = 5.41\).

To find the values of p and q, we use the following conditions:

1. The sum of all probabilities must equal 1: \(0.15 + p + 0.4 + q = 1\).
2. The expected value is given by \(E(X) = 3.05\), which means \(1 \times 0.15 + 2 \times p + 3 \times 0.4 + 6 \times q = 3.05\).

From the first condition, we have:

\(p + q = 0.45\).

From the second condition, we have:

\(0.15 + 2p + 1.2 + 6q = 3.05\).

Simplifying the second equation:

\(2p + 6q = 3.05 - 0.15 - 1.2 = 1.7\).

Now, solve the simultaneous equations:

1. \(p + q = 0.45\)
2. \(2p + 6q = 1.7\)

From equation (1), express \(p\) in terms of \(q\):

\(p = 0.45 - q\).

Substitute into equation (2):

\(2(0.45 - q) + 6q = 1.7\).

\(0.9 - 2q + 6q = 1.7\).

\(0.9 + 4q = 1.7\).

\(4q = 1.7 - 0.9 = 0.8\).

\(q = 0.2\).

Substitute \(q = 0.2\) back into \(p = 0.45 - q\):

\(p = 0.45 - 0.2 = 0.25\).

Thus, the values are \(p = 0.25\) and \(q = 0.2\).

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:

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