(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: