Past Exam Questions

To find the probability distribution for the number of green pens Rod takes, we calculate the probabilities for 0, 1, 2, and 3 green pens.

Probability of 0 green pens:

\(P(0) = \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} = \frac{210}{720}\)

Probability of 1 green pen:

\(P(1) = \frac{3}{10} \times \frac{7}{9} \times \frac{6}{8} \times \binom{3}{1} = \frac{378}{720}\)

Probability of 2 green pens:

\(P(2) = \frac{3}{10} \times \frac{2}{9} \times \frac{7}{8} \times \binom{3}{2} = \frac{126}{720}\)

Probability of 3 green pens:

\(P(3) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720}\)

These probabilities sum to 1, confirming the distribution is correct.

(i) To find the mean of W, calculate the expected value:

Mean = \(\frac{1+1+1+2+3+3}{6} = \frac{11}{6} \approx 1.83\)

For the standard deviation, use:

SD = \(\sqrt{\frac{(1-1.83)^2 + (1-1.83)^2 + (1-1.83)^2 + (2-1.83)^2 + (3-1.83)^2 + (3-1.83)^2}{6}}\)

SD = \(\sqrt{\frac{29}{6}} \approx 0.898\)

(ii) The probability distribution table for X is calculated by considering all possible sums of two throws:

(iii) Given \(E(Y) = 8\), and \(Y\) follows a binomial distribution with \(p = \frac{1}{3}\), we have:

\(np = 8 \Rightarrow n = 24\)

Variance is given by \(Var(Y) = np(1-p) = 24 \times \frac{1}{3} \times \frac{2}{3} = \frac{16}{3} \approx 5.33\)

(i) To find the probability distribution of Y, consider the possible outcomes when two values of X are chosen:

• Y = 0: This occurs when both values are the same. The probabilities are:
  • P(2, 2) = 0.5 × 0.5 = 0.25
  • P(4, 4) = 0.4 × 0.4 = 0.16
  • P(6, 6) = 0.1 × 0.1 = 0.01
  Total probability for Y = 0 is 0.25 + 0.16 + 0.01 = 0.42.
• Y = 2: This occurs when the values are (2, 4), (4, 2), (4, 6), or (6, 4). The probabilities are:
  • P(2, 4) = 0.5 × 0.4 = 0.2
  • P(4, 2) = 0.4 × 0.5 = 0.2
  • P(4, 6) = 0.4 × 0.1 = 0.04
  • P(6, 4) = 0.1 × 0.4 = 0.04
  Total probability for Y = 2 is 0.2 + 0.2 + 0.04 + 0.04 = 0.48.
• Y = 4: This occurs when the values are (2, 6) or (6, 2). The probabilities are:
  • P(2, 6) = 0.5 × 0.1 = 0.05
  • P(6, 2) = 0.1 × 0.5 = 0.05
  Total probability for Y = 4 is 0.05 + 0.05 = 0.1.

The probability distribution table for Y is:

(ii) The expected value of Y is calculated as:

\(E(Y) = (0 imes 0.42) + (2 imes 0.48) + (4 imes 0.1) = 0 + 0.96 + 0.4 = 1.36\)

To find the probability distribution, we first determine \(k\) by ensuring the probabilities sum to 1:

\(k(-2)^2 + k(1)^2 + k(2)^2 + k(3)^2 = 1\)

\(4k + k + 4k + 9k = 18k = 1\)

\(k = \frac{1}{18}\)

Thus, the probability distribution table is:

To find \(E(X)\):

\(E(X) = \frac{4 \times (-2) + 1 \times 1 + 4 \times 2 + 9 \times 3}{18}\)

\(= \frac{-8 + 1 + 8 + 27}{18}\)

\(= \frac{28}{18} = \frac{14}{9}\)

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

\(\text{Var}(X) = \frac{4 \times (-2)^2 + 1 \times 1^2 + 4 \times 2^2 + 9 \times 3^2}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{16 + 1 + 16 + 81}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{114}{18} - \frac{196}{81}\)

\(= \frac{317}{81}\)

Let the probability of choosing a rope of length 3 m be denoted as \(P(3)\) and the probability of choosing a rope of length 5 m be \(P(5)\).

Given that there are four times as many ropes of length 3 m as there are ropes of length 5 m, we have:

\(P(3) = \frac{4}{5} \quad \text{and} \quad P(5) = \frac{1}{5}\)

The expectation \(E(X)\) is calculated as:

\(E(X) = 3 \times P(3) + 5 \times P(5) = 3 \times \frac{4}{5} + 5 \times \frac{1}{5} = \frac{12}{5} + \frac{5}{5} = \frac{17}{5} = 3.4\)

The variance \(\text{Var}(X)\) is calculated as:

First, find \(E(X^2)\):

\(E(X^2) = 3^2 \times P(3) + 5^2 \times P(5) = 9 \times \frac{4}{5} + 25 \times \frac{1}{5} = \frac{36}{5} + \frac{25}{5} = \frac{61}{5}\)

Then, use the formula for variance:

\(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{61}{5} - \left(\frac{17}{5}\right)^2 = \frac{61}{5} - \frac{289}{25}\)

Convert \(\frac{61}{5}\) to \(\frac{305}{25}\) and subtract:

\(\text{Var}(X) = \frac{305}{25} - \frac{289}{25} = \frac{16}{25} = 0.64\)

To solve this problem, we first need to determine the probability distribution of \(X\), the number of girls in the team.

We have 3 boys and 5 girls, and we are choosing a team of 4. The possible values for \(X\) are 1, 2, 3, and 4.

For \(X = 1\):

We choose 1 girl and 3 boys. The probability is given by:

\(P(X=1) = \frac{\binom{5}{1} \cdot \binom{3}{3}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

For \(X = 2\):

We choose 2 girls and 2 boys. The probability is given by:

\(P(X=2) = \frac{\binom{5}{2} \cdot \binom{3}{2}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 3\):

We choose 3 girls and 1 boy. The probability is given by:

\(P(X=3) = \frac{\binom{5}{3} \cdot \binom{3}{1}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 4\):

We choose 4 girls. The probability is given by:

\(P(X=4) = \frac{\binom{5}{4} \cdot \binom{3}{0}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

Thus, the probability distribution table is:

Given \(E(X) = \frac{5}{2}\), we calculate \(\text{Var}(X)\) using the formula:

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

First, calculate \(E(X^2)\):

\(E(X^2) = 1^2 \cdot \frac{1}{14} + 2^2 \cdot \frac{3}{7} + 3^2 \cdot \frac{3}{7} + 4^2 \cdot \frac{1}{14}\)

\(= \frac{1}{14} + \frac{12}{7} + \frac{27}{7} + \frac{16}{14}\)

\(= \frac{1}{14} + \frac{24}{14} + \frac{54}{14} + \frac{16}{14}\)

\(= \frac{95}{14}\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{95}{14} - \left(\frac{5}{2}\right)^2\)

\(= \frac{95}{14} - \frac{25}{4}\)

\(= \frac{95}{14} - \frac{87.5}{14}\)

\(= \frac{15}{28}\)

Thus, \(\text{Var}(X) = \frac{15}{28}\) (0.536).

(i) The score is 6 if the cards drawn are (3, 9) or (9, 3). The probability of each pair is \(\frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\). Therefore, the probability of a score of 6 is \(2 \times \frac{1}{25} = \frac{2}{25} = 0.08\).

(ii) The probability distribution table is:

(iii) The mean score is calculated as \(\sum x p = 0 \times 0.2 + 1 \times 0.24 + 2 \times 0.08 + 3 \times 0.08 + 4 \times 0.16 + 5 \times 0.16 + 6 \times 0.08 = 2.56\).

(iv) Judy wins if the score is 4, 5, or 6. The probability of these scores is \(0.16 + 0.16 + 0.08 = 0.4\). The probability of a draw (score 0) is 0.2. Therefore, the probability that Judy wins with the second choice is \(0.2 \times 0.4 = 0.08\).

(v) The probability that Judy wins with the \(n\)th choice is \((0.2)^{n-1} \times 0.4\).

(i) The probability distribution table for \(X\) is:

Since the sum of all probabilities must equal 1, we have:

\(k + 2k + 3k + 4k + 5k = 1\)

\(15k = 1\)

\(k = \frac{1}{15}\) (0.0667)

(ii) To find \(E(X)\), we calculate:

\(E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 3k + 4 \cdot 4k + 5 \cdot 5k\)

\(= k + 4k + 9k + 16k + 25k\)

\(= 55k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 55 \times \frac{1}{15} = \frac{55}{15} = \frac{11}{3}\) (3.67)

(i) To find \(P(X = 2)\), consider the combinations: \((0, 2)\) and \((2, 0)\).

\(P(0, 2) = \frac{6}{10} \times \frac{3}{7}\)

\(P(2, 0) = \frac{3}{10} \times \frac{4}{7}\)

\(P(X = 2) = \frac{6}{10} \times \frac{3}{7} + \frac{3}{10} \times \frac{4}{7} = \frac{30}{70} = \frac{3}{7}\)

(ii) The probability distribution of X is:

(iii) To find \(E(X)\) and \(\text{Var}(X)\):

\(E(X) = \sum x \cdot P(X = x) = \frac{13}{7}\)

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

\(\text{Var}(X) = \frac{120}{70} + \frac{208}{70} + \frac{108}{70} - \left(\frac{13}{7}\right)^2 = 2.78\)

(iv) Given \(X = 2\), find \(P(A = 2)\):

\(P(A = 2 \mid X = 2) = \frac{P(A = 2 \cap X = 2)}{P(X = 2)} = \frac{\frac{3}{10} \times \frac{4}{7}}{\frac{30}{70}} = 0.4\)

(i) The possible values of \(X\) are 0, 1, and 2. The probability distribution is:

(ii) To find \(E(X)\), use the formula:

\(E(X) = \sum x P(X = x) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = \frac{8}{7}\).

To find \(\text{Var}(X)\), use the formula:

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

First, calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times \frac{1}{7} + 1^2 \times \frac{4}{7} + 2^2 \times \frac{2}{7} = \frac{12}{7}\).

Then, \(\text{Var}(X) = \frac{12}{7} - \left(\frac{8}{7}\right)^2 = \frac{12}{7} - \frac{64}{49} = \frac{20}{49}\).

(i) To find \(P(X = 40)\), we need to find the value of \(r\) such that \(\frac{120}{r} = 40\). Solving for \(r\), we get:

\(40 = \frac{120}{r}\)

\(r = \frac{120}{40} = 3\)

Now, substitute \(r = 3\) into the probability formula:

\(P(X = 40) = \frac{3}{45} = \frac{1}{15}\)

(ii) Construct the probability distribution table for \(X\):

(iii) The modal value of \(X\) is the value with the highest probability. From the table, \(X = 40\) has the highest probability of \(\frac{3}{45}\).

(iv) To find the probability that \(X\) lies between 18 and 100, sum the probabilities for \(X = 20, 24, 30, 40, 60\):

\(P(18 < X < 100) = \frac{6}{45} + \frac{5}{45} + \frac{4}{45} + \frac{3}{45} + \frac{2}{45}\)

\(= \frac{20}{45} = \frac{4}{9}\)

(ii) To find the probability distribution of \(X\), count the occurrences of each sum in the table and divide by the total number of outcomes (36). The probabilities are:

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

\(E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + 6 \times \frac{5}{36} + 7 \times \frac{4}{36} + 8 \times \frac{4}{36} + 9 \times \frac{4}{36} + 10 \times \frac{4}{36} + 11 \times \frac{8}{36} + 12 \times \frac{4}{36}\)

\(= \frac{312}{36} = \frac{26}{3} \approx 8.67\)

(iv) The probability that \(X\) is greater than \(E(X)\) is:

\(P(X > 8.67) = P(X = 9, 10, 11, 12) = \frac{4}{36} + \frac{4}{36} + \frac{8}{36} + \frac{4}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556\)

(i) The tree diagram should have branches for each call attempt with probabilities 0.5 for Answered (A) and 0.5 for Unanswered (U). There should be 4 levels of branches, representing up to 4 attempts.

(ii) The probability distribution table is completed as follows:

(iii) The expected number of unanswered calls \(E(X)\) is calculated as:

\(E(X) = \sum x \cdot P(X = x) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} + 4 \cdot \frac{1}{16}\)

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

\(E(X) = \frac{1}{4} + \frac{1}{4} + \frac{3}{16} + \frac{4}{16}\)

\(E(X) = \frac{1}{2} + \frac{7}{16} = \frac{15}{16}\)

Thus, \(E(X) = \frac{15}{16}\) or approximately 0.938 or 0.9375.

Let the events be defined as follows:

• Event R: A red paper clip is taken from box A.
• Event W: A white paper clip is taken from box A.

Calculate the probabilities for each scenario:

1. If a red paper clip is taken from box A and then a white paper clip from box B:

\(P(R, W) = \frac{5}{6} \times \frac{2}{10} = \frac{10}{60}\)

2. If a white paper clip is taken from box A and then a white paper clip from box B:

\(P(W, W) = \frac{1}{6} \times \frac{3}{10} = \frac{3}{60}\)

Now, calculate the probability distribution of X:

• \(x = 0\): No red paper clips are taken. This occurs when a white paper clip is taken from box A and then from box B. Thus, \(P(X=0) = \frac{3}{60}\).
• \(x = 1\): One red paper clip is taken. This can occur in two ways: either a red paper clip is taken from box A and a white paper clip from box B, or a white paper clip is taken from box A and a red paper clip from box B. The probability is \(P(X=1) = \frac{10}{60} + \frac{7}{60} = \frac{17}{60}\).
• \(x = 2\): Two red paper clips are taken. This occurs when a red paper clip is taken from both boxes. Thus, \(P(X=2) = \frac{40}{60}\).

The probability distribution table is:

(a) The probability distribution table for X is:

(b) To find E(X), calculate:

E(X) = \frac{1 \times 2 + 4 \times 3 + 10 \times 4 + 12 \times 5 + 9 \times 6}{36} = \frac{2 + 12 + 40 + 60 + 54}{36} = \frac{168}{36} = \frac{14}{3}

To find Var(X), calculate:

\(\text{Var}(X) = \frac{1 \times 2^2 + 4 \times 3^2 + 10 \times 4^2 + 12 \times 5^2 + 9 \times 6^2}{36} - \left(\frac{14}{3}\right)^2\)

\(= \frac{4 + 36 + 160 + 300 + 324}{36} - \left(\frac{14}{3}\right)^2\)

\(= \frac{824}{36} - \frac{196}{9} = \frac{10}{9}\)

To find the probability distribution for the number of green peppers taken, we consider the possible values of the random variable X, which are 0, 1, 2, and 3.

1. Probability of taking 0 green peppers:

The number of ways to choose 3 peppers from the 8 non-green peppers (3 red + 5 yellow) is \(\binom{8}{3} = 56\).

The total number of ways to choose 3 peppers from all 12 is \(\binom{12}{3} = 220\).

Thus, \(P(X=0) = \frac{56}{220} = \frac{14}{55}\).

2. Probability of taking 1 green pepper:

Choose 1 green pepper from 4 and 2 non-green peppers from 8: \(\binom{4}{1} \times \binom{8}{2} = 4 \times 28 = 112\).

Thus, \(P(X=1) = \frac{112}{220} = \frac{28}{55}\).

3. Probability of taking 2 green peppers:

Choose 2 green peppers from 4 and 1 non-green pepper from 8: \(\binom{4}{2} \times \binom{8}{1} = 6 \times 8 = 48\).

Thus, \(P(X=2) = \frac{48}{220} = \frac{12}{55}\).

4. Probability of taking 3 green peppers:

Choose all 3 peppers from the 4 green peppers: \(\binom{4}{3} = 4\).

Thus, \(P(X=3) = \frac{4}{220} = \frac{1}{55}\).

The probability distribution table is:

(i) Mario pays $1 for each throw. After missing twice, he hits on the third throw and receives $3. His total cost is $3, so his profit is \(3 - 3 = 0\). However, the mark-scheme states $2, indicating a profit calculation of \(3 - 1 = 2\) for the third throw.

(ii) The probability that Mario’s profit is $0 is the probability that he misses all five throws or hits on the fourth or fifth throw. This is calculated as:

\(P(MMMMH) + P(MMMMM) = 0.8^3 \times 0.2 + 0.8^4 \times 0.2 = 0.184\)

(iii) The probability distribution table for Mario’s profit is:

(iv) The expected profit \(E(X)\) is calculated as:

\(E(X) = 4 \times 0.2 + 2 \times 0.288 + 0 \times 0.184 - 1 \times 0.328 = 0.8 + 0.576 - 0.328 = 1.05\)

(i) The probability of a single die landing on a green face is \(\frac{1}{4} = 0.25\). The probability of a die not landing on a green face is \(1 - 0.25 = 0.75\).

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

For 4 dice landing on a green face, \(n = 5\), \(k = 4\), \(p = 0.25\):

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

\(= 5 \times (0.25)^4 \times 0.75\)

\(= 5 \times 0.00390625 \times 0.75\)

\(= 0.0146484375\)

Rounding to 4 decimal places gives \(0.0146\).

(ii) The probability distribution of \(X\) is calculated using the binomial formula for each value of \(X\) from 0 to 5:

To find the probability distribution of L, we consider all possible combinations of drawing 3 balls from the set {1, 2, 3, 4, 5} and determine the largest number in each combination.

The possible values for L are 3, 4, and 5. The probabilities are calculated as follows:

For the expectation, we use the formula:

\(E(L) = \sum l \cdot P(L = l) = 3 \times 0.1 + 4 \times 0.3 + 5 \times 0.6 = 4.5\)

For the variance, we use the formula:

\(\text{Var}(L) = \sum l^2 \cdot P(L = l) - (E(L))^2 = 3^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.6 - 4.5^2 = 0.45\)

(i) To find the probability distribution of \(X\), we consider all possible outcomes of throwing two dice. There are a total of 36 outcomes.

- For \(x = 1\), the outcomes are \((1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1)\). There are 11 outcomes, so \(P(X = 1) = \frac{11}{36}\).

- For \(x = 2\), the outcomes are \((2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2)\). There are 9 outcomes, so \(P(X = 2) = \frac{9}{36}\).

- For \(x = 3\), the outcomes are \((3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)\). There are 7 outcomes, so \(P(X = 3) = \frac{7}{36}\).

- For \(x = 4\), the outcomes are \((4,4), (4,5), (4,6), (5,4), (6,4)\). There are 5 outcomes, so \(P(X = 4) = \frac{5}{36}\).

- For \(x = 5\), the outcomes are \((5,5), (5,6), (6,5)\). There are 3 outcomes, so \(P(X = 5) = \frac{3}{36}\).

- For \(x = 6\), the outcome is \((6,6)\). There is 1 outcome, so \(P(X = 6) = \frac{1}{36}\).

(ii) To find \(\mathbb{E}(X)\), we calculate:

\(\mathbb{E}(X) = 1 \times \frac{11}{36} + 2 \times \frac{9}{36} + 3 \times \frac{7}{36} + 4 \times \frac{5}{36} + 5 \times \frac{3}{36} + 6 \times \frac{1}{36}\)

\(= \frac{11}{36} + \frac{18}{36} + \frac{21}{36} + \frac{20}{36} + \frac{15}{36} + \frac{6}{36}\)

\(= \frac{91}{36}\)