Q.There are 2 black and 2 white balls in a box. Two balls are drawn without replacement from it. Obtain probability distribution of the number of white balls in the selected balls. Hence find its mean and variance.

Total ways $\binom{4}{2}=6$; $P(0)=\frac16,P(1)=\frac46,P(2)=\frac16$; $E(X)=1$, $E(X^2)=\frac43$, $\text{Var}=\frac43-1=\frac13$.

A box has 2 white and 2 black balls; 2 balls are drawn without replacement. Total ways $=\binom{4}{2}=6$. Let $X=$ number of white balls, $X=0,1,2$.

$P(X=0)=\frac{\binom{2}{0}\binom{2}{2}}{6}=\frac{1}{6},\quad P(X=1)=\frac{\binom{2}{1}\binom{2}{1}}{6}=\frac{4}{6},\quad P(X=2)=\frac{\binom{2}{2}\binom{2}{0}}{6}=\frac{1}{6}.$

Mean:

$E(X)=\sum xP(x)=0\cdot\tfrac16+1\cdot\tfrac46+2\cdot\tfrac16=\frac{0+4+2}{6}=\frac{6}{6}=1.$

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