Past Exam Questions

To solve part (b), we need to find the probability distribution for the number of times a 1 or a 6 is rolled in 3 throws of a die. The probability of rolling a 1 or a 6 in a single throw is \(\frac{1}{3}\), and the probability of not rolling a 1 or a 6 is \(\frac{2}{3}\).

Calculating each probability:

\(P(0) = \left( \frac{2}{3} \right)^3 = \frac{8}{27}\)

\(P(1) = \binom{3}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^2 = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27}\)

\(P(2) = \binom{3}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^1 = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27}\)

\(P(3) = \left( \frac{1}{3} \right)^3 = \frac{1}{27}\)

For part (c), the expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{8}{27} + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}\)

\(E(X) = \frac{0}{27} + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}\)

\(E(X) = \frac{27}{27} = 1\)

(i) The probability that the first ball is red is \(\frac{3}{8}\). After removing one red ball, the probability that the second ball is red is \(\frac{2}{7}\). Thus, the probability that both balls are red is \(\frac{3}{8} \times \frac{2}{7} = \frac{3}{28}\).

(ii) The probability of choosing a red then a white ball is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). The probability of choosing a white then a red ball is \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Therefore, the probability of different colors is \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(iii) Using conditional probability, \(P(\text{first red} | \text{second red}) = \frac{P(\text{first red and second red})}{P(\text{second red})} = \frac{\frac{3}{28}}{\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}} = \frac{2}{7}\).

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

(v) The expected value \(E(X) = \frac{30}{56} = \frac{15}{28}\). The variance \(\text{Var}(X) = \frac{30}{56} - \left(\frac{15}{28}\right)^2 = \frac{45}{112}\) or 0.402.

(i) To find the probability distribution of \(X\), calculate \(X = \text{Red} - \text{Blue}\) for all combinations:

(ii) Calculate \(\text{E}(X)\):

\(\text{E}(X) = \frac{-1 + 0 + 3 + 4 + 9 + 8}{12} = \frac{23}{12}\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1 + 0 + 3 + 8 + 27 + 32}{12} - \left(\frac{23}{12}\right)^2 = \frac{71}{12} - \left(\frac{23}{12}\right)^2\)

\(\text{Var}(X) = \frac{323}{144} \text{ or } \frac{35}{144}\)

(i) The possible scores are calculated by multiplying the outcomes of the two spinners. The five-sided spinner outcomes are \(1, 1, 1, 2, 3\) and the three-sided spinner outcomes are \(1, 2, 3\). The scores are:

(ii) To find the mean, use the formula \(\mu = \sum x_i p_i\):

\(\mu = \frac{3 \times 1 + 4 \times 2 + 4 \times 3 + 1 \times 4 + 2 \times 6 + 1 \times 9}{15} = \frac{48}{15} = 3.2\)

To find the variance, use the formula \(\sigma^2 = \sum x_i^2 p_i - \mu^2\):

\(\sigma^2 = \frac{3 \times 1^2 + 4 \times 2^2 + 4 \times 3^2 + 1 \times 4^2 + 2 \times 6^2 + 1 \times 9^2}{15} - 3.2^2\)

\(\sigma^2 = \frac{224}{15} - 3.2^2 = 4.69\)

(iii) The scores greater than the mean (3.2) are 4, 6, and 9. The probability is:

\(P(X > 3.2) = \frac{1}{15} + \frac{2}{15} + \frac{1}{15} = \frac{4}{15}\)

(i) The tree diagram is as follows:

First draw two branches for the first sweet: Toffee (T) with probability \(\frac{6}{7}\) and Chocolate (C) with probability \(\frac{1}{7}\).

For the second sweet, if the first was a Toffee (T), draw branches for Toffee (T) with probability \(\frac{5}{9}\) and Chocolate (C) with probability \(\frac{4}{9}\).

If the first was a Chocolate (C), draw branches for Toffee (T) with probability \(\frac{6}{9}\) and Chocolate (C) with probability \(\frac{3}{9}\).

(ii) The probability distribution table for the number of toffees taken is:

(iii) The mean number of toffees taken is calculated as:

\(E(X) = \frac{90}{63} \times \left( \frac{10}{7} \right) = 1.43\)

(iv) The probability that the first sweet taken is a chocolate, given that the second sweet taken is a toffee, is:

\(P(\text{1st C | 2nd T}) = \frac{P(C \cap T)}{P(T)} = \frac{\frac{1}{7} \times \frac{6}{9}}{\frac{1}{7} \times \frac{6}{9} + \frac{6}{7} \times \frac{5}{9}} = \frac{6}{63} \div \frac{36}{63} = \frac{1}{6}\)

(i) The probability that Amy loses her original $1 is the probability that she fails both attempts. The probability of failing one attempt is given by:

\(P(F) = 1 - 0.2 = 0.8\)

Therefore, the probability of failing both attempts is:

\(P(\text{loses } \$1) = P(F \text{ and } F) = 0.8 \times 0.8 = 0.64\)

(ii) The probability distribution table for the amount that Amy gains is:

The probability of gaining $0.50 is the probability of failing the first attempt and succeeding on the second:

\(P(0.50) = P(F \text{ and } S) = 0.8 \times 0.2 = 0.16\)

The probability of gaining $2 is the probability of succeeding on the first attempt:

\(P(2) = 0.2\)

(iii) The expected gain is calculated as follows:

\(E(\text{winnings}) = (-1 \times 0.64) + (0.5 \times 0.16) + (2 \times 0.2)\)

\(E(\text{winnings}) = -0.64 + 0.08 + 0.4 = -0.16\)

Therefore, Amy's expected gain is -$0.16, indicating a loss of 16 cents.

(i) The probability distribution table is:

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

\(k + k + 4k + 9k = 15k = 1\)

Solving for \(k\), we get \(k = \frac{1}{15}\).

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

\(E(X) = (-1)k + 1k + 2(4k) + 3(9k) = -k + k + 8k + 27k = 35k\)

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

\(E(X) = 35 \times \frac{1}{15} = \frac{35}{15} = \frac{7}{3}\)

To find \(Var(X)\), we use:

\(Var(X) = E(X^2) - (E(X))^2\)

\(E(X^2) = (-1)^2k + 1^2k + 2^2(4k) + 3^2(9k) = k + k + 16k + 81k = 99k\)

\(Var(X) = 99k - \left(\frac{7}{3}\right)^2\)

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

\(Var(X) = 99 \times \frac{1}{15} - \left(\frac{7}{3}\right)^2 = \frac{99}{15} - \frac{49}{9}\)

\(Var(X) = \frac{52}{45}\)

To solve this problem, we first need to determine the possible values of \(X\) and their probabilities.

The possible outcomes for the die are \(-1, -1, 0, 0, 1, 2\) and for the spinner are \(-1, 0, 1\). The possible values of \(X\) are the sums of these outcomes.

Possible values of \(X\) are: \(-2, -1, 0, 1, 2, 3\).

Next, we calculate the expected value \(E(X)\):

\(E(X) = \frac{-4 - 4 + 0 + 4 + 4 + 3}{18} = \frac{1}{6}\)

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

\(\text{Var}(X) = \frac{8 + 4 + 0 + 4 + 8 + 9}{18} - \left(\frac{1}{6}\right)^2\)

\(= \frac{11}{6} - \frac{1}{36}\)

\(= \frac{65}{36}\)

Thus, \(\text{Var}(X) = \frac{65}{36}\) or approximately 1.81.

(i) To find the probability distribution of \(X\), calculate \(X = R - B\) for each combination of red (R) and blue (B) spinner outcomes. The possible values of \(X\) are -2, -1, 0, 1, 2, 3. The probability distribution table is:

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

\(E(X) = \frac{-2 \times 1 + (-1) \times 2 + 0 \times 3 + 1 \times 3 + 2 \times 2 + 3 \times 1}{12} = 0.5\)

\(\text{Var}(X) = \frac{(-2)^2 \times 1 + (-1)^2 \times 2 + 0^2 \times 3 + 1^2 \times 3 + 2^2 \times 2 + 3^2 \times 1}{12} - (0.5)^2\)

\(= \frac{26}{12} - \frac{1}{4} = \frac{23}{12}\)

(iii) The probability that \(X = 1\) given \(X\) is non-zero is:

\(P(X \neq 0) = \frac{9}{12}\)

\(P(X = 1 \mid X \neq 0) = \frac{P(X = 1 \cap X \neq 0)}{P(X \neq 0)} = \frac{\frac{3}{12}}{\frac{9}{12}} = \frac{1}{3}\)

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

\(6p + 0.1 = 1\)

Solving for \(p\):

\(6p = 0.9\)

\(p = 0.15\)

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

\(\text{Var}(X) = (-1)^2 \times p + 0^2 \times p + 1^2 \times 2p + 2^2 \times 2p + 4^2 \times 0.1 - (1.15)^2\)

Substitute \(p = 0.15\):

\(\text{Var}(X) = 1 \times 0.15 + 0 + 1 \times 0.3 + 4 \times 0.3 + 16 \times 0.1 - 1.15^2\)

\(= 0.15 + 0 + 0.3 + 1.2 + 1.6 - 1.3225\)

\(= 1.9275\)

Rounded to three significant figures:

\(\text{Var}(X) = 1.93\)

(a) To find the probabilities, we use \(P(X = x) = k(x^2 - 1)\) for each value of \(x\):

• For \(x = -2\), \(P(X = -2) = k((-2)^2 - 1) = 3k\).
• For \(x = 2\), \(P(X = 2) = k(2^2 - 1) = 3k\).
• For \(x = 3\), \(P(X = 3) = k(3^2 - 1) = 8k\).

Since the total probability must be 1, we have:

\(3k + 3k + 8k = 1\)

\(14k = 1\)

\(k = \frac{1}{14}\)

Thus, the probability distribution table is:

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

\(E(X) = (-2) \times \frac{3}{14} + 2 \times \frac{3}{14} + 3 \times \frac{8}{14}\)

\(= \frac{-6}{14} + \frac{6}{14} + \frac{24}{14}\)

\(= \frac{12}{14} = \frac{12}{7}\)

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

\(\text{Var}(X) = (-2)^2 \times \frac{3}{14} + 2^2 \times \frac{3}{14} + 3^2 \times \frac{8}{14} - (E(X))^2\)

\(= 4 \times \frac{3}{14} + 4 \times \frac{3}{14} + 9 \times \frac{8}{14} - \left(\frac{12}{7}\right)^2\)

\(= \frac{12 + 12 + 72}{14} - \frac{144}{49}\)

\(= \frac{96}{14} - \frac{144}{49}\)

\(= \frac{192}{49}\)

To solve this problem, we need to calculate the probability distribution for the number of heads obtained when the three coins are tossed.

Let the random variable X represent the number of heads obtained. The possible values of X are 0, 1, 2, and 3.

1. Probability of 0 heads, P(0):
\(P(0) = (1 - 0.4) \times (1 - 0.75) \times (1 - 0.5) = 0.6 \times 0.25 \times 0.5 = 0.075\)

2. Probability of 1 head, P(1):
\(P(1) = 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 + 0.6 \times 0.25 \times 0.5 = 0.35\)

3. Probability of 2 heads, P(2):
\(P(2) = 0.4 \times 0.75 \times 0.5 + 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 = 0.425\)

4. Probability of 3 heads, P(3):
\(P(3) = 0.4 \times 0.75 \times 0.5 = 0.15\)

The probability distribution table is:

To calculate the mean, E(X):
\(E(X) = 0 \times 0.075 + 1 \times 0.35 + 2 \times 0.425 + 3 \times 0.15 = 1.65\)

To calculate the variance, Var(X):
\(Var(X) = (0^2 \times 0.075) + (1^2 \times 0.35) + (2^2 \times 0.425) + (3^2 \times 0.15) - (1.65)^2 = 0.678\)

To solve this problem, we first need to determine the probability distribution of the random variable \(X\), which represents the number of cats chosen when 3 animals are selected from 5 dogs and 2 cats.

We can have 0, 1, or 2 cats chosen. The probabilities are calculated as follows:

• \(P(X = 0) = \frac{\binom{5}{3} \cdot \binom{2}{0}}{\binom{7}{3}} = \frac{\frac{5 \times 4 \times 3}{3 \times 2 \times 1}}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{2}{7}\)
• \(P(X = 1) = \frac{\binom{5}{2} \cdot \binom{2}{1}}{\binom{7}{3}} = \frac{\frac{5 \times 4}{2 \times 1} \cdot 2}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{4}{7}\)
• \(P(X = 2) = \frac{\binom{5}{1} \cdot \binom{2}{2}}{\binom{7}{3}} = \frac{5}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{1}{7}\)

The probability distribution table is:

Given \(E(X) = \frac{6}{7}\), we find \(\text{Var}(X)\) using the formula:

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

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

\(E(X^2) = 0^2 \cdot \frac{2}{7} + 1^2 \cdot \frac{4}{7} + 2^2 \cdot \frac{1}{7} = 0 + \frac{4}{7} + \frac{4}{7} = \frac{8}{7}\)

Then, \(\text{Var}(X) = \frac{8}{7} - \left(\frac{6}{7}\right)^2 = \frac{8}{7} - \frac{36}{49} = \frac{56}{49} - \frac{36}{49} = \frac{20}{49}\)

Thus, \(\text{Var}(X) = \frac{20}{49}\) or approximately 0.408.

(i) To find the probability that the socks taken are of different colours, we can use the formula:

\(P(\text{different colours}) = P(RB) + P(BR)\)

Where:

• \(P(RB) = \frac{4}{12} \times \frac{8}{11}\)
• \(P(BR) = \frac{8}{12} \times \frac{4}{11}\)

Thus,

\(P(\text{different colours}) = \frac{4}{12} \times \frac{8}{11} + \frac{8}{12} \times \frac{4}{11} = \frac{16}{33}\)

(ii) The probability distribution table for \(X\), the number of red socks taken, is:

(iii) To find \(E(X)\), the expected value of \(X\), we use:

\(E(X) = 0 \times \frac{14}{33} + 1 \times \frac{16}{33} + 2 \times \frac{3}{33}\)

\(E(X) = \frac{16}{33} + \frac{6}{33} = \frac{22}{33} = \frac{2}{3}\)

First, calculate the possible values of X by squaring each face value and subtracting 4:

• For face 1: \(1^2 - 4 = -3\)
• For face 2: \(2^2 - 4 = 0\)
• For face 3: \(3^2 - 4 = 5\)
• For face 6: \(6^2 - 4 = 32\)

The probability distribution table is:

To find \(E(X)\):

\(E(X) = (-3) \cdot \frac{1}{6} + 0 \cdot \frac{1}{2} + 5 \cdot \frac{1}{6} + 32 \cdot \frac{1}{6} = \frac{-3}{6} + \frac{5}{6} + \frac{32}{6} = \frac{34}{6} = \frac{17}{3} \approx 5.67\)

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

\(\text{Var}(X) = \frac{(-3)^2}{6} + \frac{0^2}{2} + \frac{5^2}{6} + \frac{32^2}{6} - \left(\frac{34}{6}\right)^2\)

\(= \frac{9}{6} + \frac{25}{6} + \frac{1024}{6} - \left(\frac{34}{6}\right)^2\)

\(= 144 \left( \frac{1298}{9} \right)\)

(i) To find \(P(X = 3)\), we need the probability of drawing two red discs followed by a blue disc, i.e., \(P(RRB)\).

The probability of drawing the first red disc is \(\frac{2}{6}\), the second red disc is \(\frac{1}{5}\), and the first blue disc is \(\frac{4}{4}\).

Thus, \(P(X = 3) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4} = \frac{1}{15}\).

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

\(P(X = 1)\) is the probability of drawing a blue disc first, which is \(\frac{4}{6} = \frac{2}{3}\).

\(P(X = 2)\) is the probability of drawing one red disc followed by a blue disc, i.e., \(P(RB) = \frac{2}{6} \times \frac{4}{5} = \frac{4}{15}\).

\(P(X = 3)\) is already calculated as \(\frac{1}{15}\).

(i) The probability \(P(X = -2) = k(-2)^2 = 4k\) and \(P(X = 2) = k(2)^2 = 4k\). Thus, \(P(X = -2) = P(X = 2)\).

(ii) The probability distribution table is:

Sum of probabilities: \(4k + k + 4k + 16k = 1\)

\(25k = 1\)

\(k = \frac{1}{25}\)

(iii) \(E(X) = (-2)(4k) + (-1)(k) + (2)(4k) + (4)(16k)\)

\(E(X) = -8k - k + 8k + 64k = 63k\)

Substitute \(k = \frac{1}{25}\):

\(E(X) = 63 \times \frac{1}{25} = \frac{63}{25}\)

(i) To find \(P(X = 2)\), we need the combinations where the sum of the numbers from pack A and pack B equals 2. The possible combinations are:

• Card from A is 0 and card from B is 2: Probability = \(\frac{2}{10} \times \frac{4}{6} = \frac{8}{60} = \frac{2}{15}\)

Thus, \(P(X = 2) = \frac{2}{15}\).

(ii) The probability distribution table for \(X\) is constructed by considering all possible sums:

(iii) Given \(X = 3\), we find the probability that the card from pack A is 1. The combinations for \(X = 3\) are:

• Card from A is 1 and card from B is 2: Probability = \(\frac{5}{10} \times \frac{4}{6} = \frac{20}{60} = \frac{1}{3}\)
• Card from A is 3 and card from B is 0: Probability = \(\frac{4}{10} \times \frac{2}{6} = \frac{8}{60} = \frac{2}{15}\)

Total probability for \(X = 3\) is \(\frac{13}{30}\). The probability that the card from A is 1 given \(X = 3\) is:

\(P(A1 | \text{Sum 3}) = \frac{\frac{5}{10} \times \frac{4}{6}}{\frac{13}{30}} = \frac{10}{13}\)

(i) To find the probability that Noor chooses exactly one T-shirt, we use combinations. The total number of ways to choose 3 items from 12 is given by \(\binom{12}{3}\).

The number of ways to choose 1 T-shirt from 3 is \(\binom{3}{1}\), and the number of ways to choose 2 items from the remaining 9 (blouses and jumpers) is \(\binom{9}{2}\).

Thus, the probability is:

\(\frac{\binom{3}{1} \times \binom{9}{2}}{\binom{12}{3}} = \frac{3 \times 36}{220} = \frac{108}{220} = \frac{27}{55}\)

(ii) The probability distribution table for X is constructed by considering the possible values of X (0, 1, 2, 3) and their respective probabilities:

To find the expectation of the difference between the two scores, we first determine the probability distribution of the differences.

The expectation is calculated as:

\(E(X) = \sum (\text{difference} \times \text{probability})\)

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

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

\(E(X) = \frac{70}{36}\)

\(E(X) \approx 1.94\)