An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn, one by one, at random without replacement. Find the probability distribution of the number of white balls drawn. Also, find the mean and the variance of the number of white balls drawn.

Mean = $\frac{3}{5}$ and Variance = $\frac{28}{75}$

The correct answer is Option (1) → Mean = $\frac{3}{5}$ and Variance = $\frac{28}{75}$

Three balls drawn, one by one, without replacement is same as drawing three balls simultaneous.

Total number of balls = 5 + 2 + 3 = 10, 2 balls are white and 8 balls are non-white.

Let X denote the number of white balls drawn in a random draw of three balls, then X can take values 0, 1, 2.

$\text{P(0)= P (3 non-white balls)}=\frac{{^8C}_3}{{^{10}C}_3}=\frac{8.7.6}{1.2.3}×\frac{1.2.3}{10.9.8}=\frac{7}{15}$,

$\text{P(1) = P(1 white, 2 non-white)} =\frac{{^2C}_1×{^8C}_2}{{^{10}C}_3}=2×\frac{8.7}{1.2}×\frac{1.2.3}{10.9.8}=\frac{7}{15}$,

$\text{P(2) = P(2 white, 1 non-white)} =\frac{{^2C}_2×{^8C}_1}{{^{10}C}_3}=2×8×\frac{1.2.3}{10.9.8}=\frac{1}{15}$,

∴ The probability distribution of white balls is $\begin{pmatrix}0&1&2\\\frac{7}{15}&\frac{7}{15}&\frac{1}{15}\end{pmatrix}$.

Mean = $μ=∑p_ix_i = \frac{7}{15}×0+\frac{7}{15}×1+\frac{1}{15}×2=\frac{9}{15}=\frac{3}{5}$

$∑p_i{x_i}^2=\frac{7}{15}×0^2+\frac{7}{15}×1^2+\frac{1}{15}×2^2=0+\frac{7}{15}+\frac{4}{15}=\frac{11}{15}$

Variance = $∑p_i{x_i}^2-μ^2=\frac{11}{15}-(\frac{3}{5})^2=\frac{11}{15}-\frac{9}{25}=\frac{55-27}{75}=\frac{28}{75}$