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, then find the mean and variance of X. [CBSE 2015]

Answer 1

As, X represent the number of black balls drawn.

So, it can take values 0, 1 and 2. Yes, X is a random variable.

Now,

$$\begin{gathered} \mathrm{P}\left(X=0\right)=\mathrm{P}\left(RR\right)=\frac{5}{7}\times \frac{4}{6}=\frac{10}{21}, \\ \mathrm{P}\left(X=1\right)=\mathrm{P}\left(RB\mathrm{or}BR\right)=2\times \frac{5}{7}\times \frac{2}{6}=\frac{10}{21}, \\ \mathrm{P}\left(X=2\right)=\mathrm{P}\left(BB\right)=\frac{2}{7}\times \frac{1}{6}=\frac{1}{21} \\ \mathrm{Mean}=\sum p_{\mathit{i}}{x}_{\mathit{i}}=0\times \frac{10}{21}+1\times \frac{10}{21}+2\times \frac{1}{21} \\ =\frac{10}{21}+\frac{2}{21} \\ =\frac{12}{21} \\ =\frac{4}{7} \\ \mathrm{Also},\sum p_{\mathit{i}}{x_{\mathit{i}}}^2=0^2\times \frac{10}{21}+1^2\times \frac{10}{21}+2^2\times \frac{1}{21} \\ =\frac{10}{21}+\frac{4}{21} \\ =\frac{14}{21} \\ =\frac{2}{3} \\ \mathrm{So},\mathrm{variance}=\sum p_{\mathit{i}}{x_{\mathit{i}}}^2-{\left(\mathrm{Mean}\right)}^2 \\ =\frac{2}{3}-{\left(\frac{4}{7}\right)}^2 \\ =\frac{2}{3}-\frac{16}{49} \\ =\frac{98-48}{147} \\ =\frac{50}{147} \end{gathered}$$