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:

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