Question

An urn contains $5$ red and $2$ black balls. Two balls are randomly drawn, without replacement. Let $X$ represent the number of black balls drawn. What are the possible values of $X$? Is $X$ a random variable? If yes, find the mean and variance of $X$.

Answer 1

Uru has $5$ red and $2$ black balls

No. of ways it can be drawn $=\{RR,RB,BR,BB\}$

$n\left(S\right)=4$

LEt $X$ represent black balls

Possible values of $X:X\left(RR\right)=0;X\left(RB\right)=1;X\left(BR\right)=1;X\left(BB\right)=2$

Possible values of $X$ are $0,1,2$

$P(X=0=\frac{1}{4},P(X=2)=\frac{2}{4},P(X=2)=\frac{1}{4}$

$\sum P_i{x}_i=P(X=0)+P(X=1)+P(X=2)=\frac{1}{4}+\frac{2}{4}+\frac{1}{\$}=1$

$X$ is a random variable

Now, $P(X=0=$P(No black)

$=\frac{{{}^{5}C}_2}{{{}^{7}C}_2}=\frac{20}{42}$

$P(X=1)=$P(1 black ball)$=\frac{{{}^{5}C}_1\times {{}^{2}C}_1}{{{}^{7}C}_2}=\frac{20}{42}$

$P(X=2)=P$(2 black balls)$=\frac{{{}^{2}C}_2}{{{}^{7}C}_2}=\frac{2}{42}$

$X=0;P(X=x)=\frac{20}{42}$

$X=1;P(X=x)=\frac{20}{42}$

$X=2;P(X=x)=\frac{2}{42}$