Past Exam Questions

(i) To find the probability distribution, consider all possible outcomes when two dice are thrown. There are 36 possible outcomes (6 sides on the first die and 6 sides on the second die).

- The score is 0 when both dice show the same number. There are 6 such outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, P(0) = \(\frac{6}{36}\).

- For scores 1 to 5, count the outcomes where the smaller number is 1, 2, 3, 4, or 5, respectively:

• Score 1: (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1) - 10 outcomes, so P(1) = \(\frac{10}{36}\).
• Score 2: (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2) - 8 outcomes, so P(2) = \(\frac{8}{36}\).
• Score 3: (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) - 6 outcomes, so P(3) = \(\frac{6}{36}\).
• Score 4: (4,5), (4,6), (5,4), (6,4) - 4 outcomes, so P(4) = \(\frac{4}{36}\).
• Score 5: (5,6), (6,5) - 2 outcomes, so P(5) = \(\frac{2}{36}\).

(ii) The expected score is calculated using the formula for expected value: \(E(X) = \sum x_i p_i\).

\(E(X) = 0 \times \frac{6}{36} + 1 \times \frac{10}{36} + 2 \times \frac{8}{36} + 3 \times \frac{6}{36} + 4 \times \frac{4}{36} + 5 \times \frac{2}{36}\)

\(E(X) = \frac{0 + 10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} \approx 1.94\)

(a) To find \(P(X = 2)\), we consider the scenarios where the highest number is 2. The possible outcomes are (2,2,2), (2,2,1), (2,1,2), (1,2,2), (2,1,1), (1,2,1), and (1,1,2). There are 7 such outcomes.

The total number of outcomes when spinning three 4-sided spinners is \(4 \times 4 \times 4 = 64\).

Thus, \(P(X = 2) = \frac{7}{64}\).

(b) To complete the probability distribution table, we need to find \(P(X = 1)\) and \(P(X = 4)\).

\(P(X = 1)\) occurs only when all spinners show 1, which is (1,1,1). Thus, \(P(X = 1) = \frac{1}{64}\).

For \(P(X = 4)\), we use the fact that the probabilities must sum to 1:

\(P(X = 4) = 1 - \left( \frac{1}{64} + \frac{7}{64} + \frac{19}{64} \right) = \frac{37}{64}\).

(i) The probability distribution table is:

Since the total probability must equal 1, we have:

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

\(10k = 1\)

Thus, \(k = \frac{1}{10}\).

(ii) The mean \(E(X)\) is calculated as:

\(E(X) = 1 \times \frac{1}{10} + 2 \times \frac{2}{10} + 3 \times \frac{3}{10} + 4 \times \frac{4}{10}\)

\(E(X) = \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{16}{10} = 3\)

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

\(\text{Var}(X) = \left(1^2 \times \frac{1}{10} + 2^2 \times \frac{2}{10} + 3^2 \times \frac{3}{10} + 4^2 \times \frac{4}{10}\right) - E(X)^2\)

\(\text{Var}(X) = \left(\frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10}\right) - 3^2\)

\(\text{Var}(X) = \frac{100}{10} - 9 = 1\)

Let X be the random variable representing the number of green sweets taken. The possible values of X are 0, 1, and 2.

Calculate the probabilities:

\(P(X = 0) = P(B, B) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}\)

\(P(X = 1) = P(G, B) + P(B, G) = \frac{2}{7} \times \frac{5}{6} + \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}\)

\(P(X = 2) = P(G, G) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21}\)

Calculate the expected value:

\(E(X) = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21} = \frac{4}{7}\)

Calculate the variance:

\(\text{Var}(X) = 0^2 \times \frac{10}{21} + 1^2 \times \frac{10}{21} + 2^2 \times \frac{1}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\)

To find the probability distribution for the number of white roses chosen, we calculate the probabilities for choosing 0, 1, or 2 white roses.

There are a total of 10 roses (5 yellow, 3 red, 2 white).

Probability of choosing 0 white roses:

The probability of choosing 3 non-white roses is:

\(P(0) = \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{42}{90}\)

Probability of choosing 1 white rose:

The probability of choosing 1 white and 2 non-white roses is:

\(P(1) = \left( \frac{2}{10} \times \frac{8}{9} \times \frac{7}{8} \right) \times 3 = \frac{42}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Probability of choosing 2 white roses:

The probability of choosing 2 white and 1 non-white rose is:

\(P(2) = \left( \frac{2}{10} \times \frac{1}{9} \times \frac{8}{8} \right) \times 3 = \frac{6}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Thus, the probability distribution table is:

(i) To complete the table, calculate the sum of the numbers from Spinner A and Spinner B for each combination. For example, if Spinner A shows 1 and Spinner B shows -3, then \(X = 1 + (-3) = -2\). Complete the table similarly for all combinations.

(ii) To find the probability distribution of \(X\), count the frequency of each sum from the completed table and divide by the total number of outcomes (16). The probabilities are: \(-2: \frac{1}{16}, -1: \frac{2}{16}, 0: \frac{4}{16}, 1: \frac{3}{16}, 2: \frac{2}{16}, 3: \frac{3}{16}, 4: \frac{1}{16}\).

(iii) Calculate \(\text{Var}(X)\) using the formula \(\text{Var}(X) = E(X^2) - [E(X)]^2\). First, find \(E(X) = \sum x \cdot P(x) = 1\). Then, \(E(X^2) = \sum x^2 \cdot P(x) = \frac{62}{16}\). Thus, \(\text{Var}(X) = \frac{62}{16} - 1^2 = \frac{23}{8} = 2.875\).

(iv) To find the probability that \(X\) is even given \(X\) is positive, use conditional probability: \(P(\text{even} \mid \text{positive}) = \frac{P(\text{even and positive})}{P(\text{positive})}\). The even positive values are 2 and 4, with probabilities \(\frac{2}{16} + \frac{1}{16} = \frac{3}{16}\). The positive values are 1, 2, 3, and 4, with probabilities \(\frac{3}{16} + \frac{2}{16} + \frac{3}{16} + \frac{1}{16} = \frac{9}{16}\). Thus, \(P(\text{even} \mid \text{positive}) = \frac{3/16}{9/16} = \frac{5}{9}\).

(i) The tree diagram shows three stages, each with a probability of success \(S = 0.4\) and failure \(F = 0.6\). Each branch represents a login attempt.

(ii) The probability distribution is calculated as follows:

• \(P(X = 0) = 0.4\)
• \(P(X = 1) = 0.6 \times 0.4 = 0.24\)
• \(P(X = 2) = 0.6^2 \times 0.4 = 0.144\)
• \(P(X = 3) = 0.6^3 = 0.216\)

(iii) The expected number of unsuccessful attempts is calculated using \(E(X) = \sum x P(X = x)\):

\(E(X) = 0 \times 0.4 + 1 \times 0.24 + 2 \times 0.144 + 3 \times 0.216 = 1.176\)

(i) The probability that exactly one rabbit is white can be calculated by considering two cases: one white and one non-white rabbit.

Case 1: First rabbit is white, second is non-white: \(\frac{6}{9} \times \frac{3}{8}\).

Case 2: First rabbit is non-white, second is white: \(\frac{3}{9} \times \frac{6}{8}\).

Adding these probabilities gives: \(\frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8} = \frac{1}{2}\).

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

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

\(E(X) = 0 \times \frac{1}{12} + 1 \times \frac{1}{2} + 2 \times \frac{5}{12} = \frac{16}{12} = \frac{4}{3}\).

To find the probability distribution for \(S\), we list all possible combinations of 3 discs from the set \{1, 2, 4, 6, 7\}.

The possible combinations are: 124, 126, 127, 146, 147, 167, 246, 247, 267, 467.

For each combination, we determine the smallest number, \(S\).

Counting the occurrences of each smallest number:

• \(S = 1\): 124, 126, 127, 146, 147, 167 (6 combinations)
• \(S = 2\): 246, 247, 267 (3 combinations)
• \(S = 4\): 467 (1 combination)

The total number of combinations is 10.

The probability distribution is calculated as follows:

• \(P(S=1) = \frac{6}{10}\)
• \(P(S=2) = \frac{3}{10}\)
• \(P(S=4) = \frac{1}{10}\)

(i) To find \(P(X = 3)\), we consider the sequences where the process stops after 3 apples. The sequences are GRR and RGR. The probability for GRR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). The probability for RGR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). Therefore, \(P(X = 3) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).

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

(iii) To find the probability that all 3 peppers are orange given that at least 2 are orange, we use conditional probability:

\(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{P(3 \text{ orange })}{P(\text{at least 2 orange})}\).

\(P(3 \text{ orange }) = \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} = \frac{2}{7}\).

\(P(\text{at least 2 orange}) = P(2 \text{ orange, 1 yellow}) + P(3 \text{ orange})\).

\(P(2 \text{ orange, 1 yellow}) = \frac{5}{7} \times \frac{4}{6} \times \frac{2}{5} + \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} + \frac{2}{7} \times \frac{5}{6} \times \frac{4}{5} = \frac{6}{7}\).

Thus, \(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{\frac{2}{7}}{\frac{6}{7}} = \frac{1}{3}\).

Let \(X\) be the number of attempts Sharik makes until he gets the correct answer.

The possible values of \(X\) are 1, 2, and 3.

Probability that Sharik gets the correct answer on the first attempt (\(X = 1\)):

\(P(1) = \frac{1}{3}\)

Probability that Sharik gets the correct answer on the second attempt (\(X = 2\)):

First attempt is wrong and second attempt is correct:

\(P(2) = P(WC) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}\)

Probability that Sharik gets the correct answer on the third attempt (\(X = 3\)):

First two attempts are wrong and third attempt is correct:

\(P(3) = P(WWC) = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6}\)

Total probability for \(X = 2\) and \(X = 3\):

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

Probability distribution table:

Expected value \(E(X)\):

\(E(X) = 1 \times \frac{1}{3} + 2 \times \frac{1}{3} + 3 \times \frac{1}{3} = 2\)

(i) To find the probability of choosing exactly 2 hamsters, we use combinations. The total number of ways to choose 5 pets from 9 is \(\binom{9}{5}\). The number of ways to choose 2 hamsters from 3 is \(\binom{3}{2}\), and the number of ways to choose 3 rabbits from 6 is \(\binom{6}{3}\).

The probability is given by:

\(P(2) = \frac{\binom{3}{2} \times \binom{6}{3}}{\binom{9}{5}} = \frac{3 \times 20}{126} = \frac{60}{126} = \frac{10}{21}\)

(ii) The probability distribution table for X is constructed by calculating the probabilities for 0, 1, 2, and 3 hamsters:

• \(P(0) = \frac{\binom{6}{5} \times \binom{3}{0}}{\binom{9}{5}} = \frac{6}{126} = \frac{2}{42}\)
• \(P(1) = \frac{\binom{6}{4} \times \binom{3}{1}}{\binom{9}{5}} = \frac{45}{126} = \frac{15}{42}\)
• \(P(2) = \frac{\binom{6}{3} \times \binom{3}{2}}{\binom{9}{5}} = \frac{60}{126} = \frac{20}{42}\)
• \(P(3) = \frac{\binom{6}{2} \times \binom{3}{3}}{\binom{9}{5}} = \frac{15}{126} = \frac{5}{42}\)

The table is:

(a) To find \(P(B)\), consider the scenarios where at least one biased coin shows a head. The probability that a biased coin shows a tail is \(\frac{3}{4}\). The probability that both biased coins show tails is \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\). Therefore, \(P(B) = 1 - \frac{9}{16} = \frac{7}{16}\).

(b) To find \(P(A \mid B)\), use the formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). The event \(A \cap B\) occurs when all coins show heads or all show tails. The probability of all heads is \(\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}\). The probability of all tails is \(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}\). Thus, \(P(A \cap B) = \frac{1}{32}\). Therefore, \(P(A \mid B) = \frac{\frac{1}{32}}{\frac{7}{16}} = \frac{1}{14}\).

(c) The probability distribution for \(X\) is calculated as follows:

• \(P(X = 0) = \frac{9}{32}\)
• \(P(X = 1) = \frac{15}{32}\)
• \(P(X = 2) = \frac{7}{32}\)
• \(P(X = 3) = \frac{1}{32}\)

(i) To find the probability of obtaining exactly 1 head and 2 tails, consider the scenarios: HTT, THT, and TTH.

For HTT: \(P(H) = \frac{2}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(HTT) = \frac{2}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For THT: \(P(T) = \frac{1}{3}, P(H) = \frac{2}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(THT) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For TTH: \(P(T) = \frac{1}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(H) = \frac{1}{4}\) for Coin B.

\(P(TTH) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\).

Total probability: \(\frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{12}{36} + \frac{1}{36} = \frac{13}{36}\).

(ii) The probability distribution table is:

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

\(E(X) = 0 \times \frac{3}{36} + 1 \times \frac{13}{36} + 2 \times \frac{16}{36} + 3 \times \frac{4}{36}\).

\(E(X) = \frac{0}{36} + \frac{13}{36} + \frac{32}{36} + \frac{12}{36} = \frac{57}{36} = \frac{19}{12} \approx 1.58\).

(i) The probability of choosing exactly 2 paperback books is calculated using combinations. The number of ways to choose 2 paperback books from 6 is \(\binom{6}{2}\), and the number of ways to choose 2 hardback books from 2 is \(\binom{2}{2}\). The total number of ways to choose 4 books from 8 is \(\binom{8}{4}\).

Thus, the probability is:

\(P(\text{exactly 2}) = \frac{\binom{6}{2} \cdot \binom{2}{2}}{\binom{8}{4}} = \frac{15}{70} = \frac{3}{14}\)

(ii) The probability distribution table for X is:

(iii) Given \(E(X) = 3\), we find \(\text{Var}(X)\) using the formula \(\text{Var}(X) = \sum x^2 p - (E(X))^2\).

\(\text{Var}(X) = \frac{12}{14} + \frac{72}{14} + \frac{48}{14} - 3^2\)

\(= \frac{3}{7}\)

To find the probability distribution of \(X\), we consider the possible values of \(X\): 0, 1, or 2.

Case 1: \(X = 0\)

This means no 5s are chosen. There are 6 non-5 digits (1s and 3s) and 3 5s. The probability of choosing two non-5s is:

\(P(0) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12}\)

Case 2: \(X = 1\)

This means one 5 is chosen. From part (ii), \(P(1) = 0.5\).

Case 3: \(X = 2\)

This means two 5s are chosen. From part (i), \(P(2) = \frac{6}{72} = \frac{1}{12}\).

The probability distribution table is:

(i) To find the probability that the sum of the numbers on the three cards is 11, consider the possible combinations: (3, 4, 4), (4, 3, 4), or (4, 4, 3). The probability for each combination is calculated as follows:

\(\text{Prob} = \left( \frac{4}{10} \times \frac{6}{9} \times \frac{5}{8} \right) \times 3C1 = \frac{360}{720} = \frac{1}{2}\)

Alternatively, using combinations:

\(\frac{\binom{6}{2} \times \binom{4}{1}}{\binom{10}{3}} = \frac{1}{2}\)

(ii) The probability distribution table for the sum of the numbers on the three cards is:

Calculations for each sum:

• \(P(3, 3, 3) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720}\)
• \(P(3, 3, 4) = \frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} \times 3C1 = \frac{216}{720}\)
• \(P(4, 4, 4) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720}\)

(i) If the coin shows a head, the smallest sum of two dice throws is 2 (1+1). Therefore, X = 1 can only occur if the coin shows a tail and the first die shows 1. The probability of a tail is \(\frac{1}{2}\) and the probability of rolling a 1 is \(\frac{1}{4}\). Thus, \(P(X = 1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

(ii) For X = 3, if the coin shows a head, possible outcomes are (1,2) and (2,1), each with probability \(\frac{1}{16}\). If the coin shows a tail, the first die must show 3, with probability \(\frac{1}{8}\). Therefore, \(P(X = 3) = \frac{2}{32} + \frac{1}{8} = \frac{3}{16}\).

(iii) Completing the table:

(iv) Events Q and R are exclusive because if a tail is thrown, X cannot be 7. Thus, \(P(Q \cap R) = 0\).

The possible values of the random variable X are 0, 1, and 2.

To find the probability distribution, consider the following cases:

1. P(X = 0): No chocolates are taken. This occurs if a toffee or boiled sweet is taken from Susan's bag and then a toffee or boiled sweet is taken from Ahmad's bag.

\(P(T, B) + P(T, T) = \frac{5}{12} \times \frac{2}{10} + \frac{5}{12} \times \frac{5}{10} = \frac{7}{24} \approx 0.292\)

\(2. P(X = 2): Two chocolates are taken. This occurs if a chocolate is taken from Susan's bag and then a chocolate is taken from Ahmad's bag.\)

\(P(C, C) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120} = \frac{7}{30} \approx 0.233\)

\(3. P(X = 1): One chocolate is taken. This is the complement of the other two probabilities.\)

\(P(X = 1) = 1 - \frac{7}{24} - \frac{7}{30} = \frac{19}{40} \approx 0.475\)

Thus, the probability distribution is:

(i) To find \(P(X = 9)\), we consider the combinations of scores that sum to 9: (1, 4, 4), (2, 3, 4), and (3, 3, 3). The probabilities are calculated as follows:

\(P(1, 4, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 3 = \frac{3}{64}\)

\(P(2, 3, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 6 = \frac{6}{64}\)

\(P(3, 3, 3) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\)

Adding these probabilities gives \(P(X = 9) = \frac{3}{64} + \frac{6}{64} + \frac{1}{64} = \frac{10}{64}\).

(ii) The completed probability distribution table is:

(iii) To determine if \(R\) and \(S\) are independent, we calculate:

\(P(S) = \frac{6}{64}\)

\(P(R \cap S) = \frac{3}{64}\)

Since \(P(R \cap S) \neq P(R) \times P(S)\), events \(R\) and \(S\) are not independent.