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.