Question

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribtution. [CBSE 2015]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{spade}\mathrm{in}\mathrm{a}\mathrm{draw}=\frac{13}{52}=\frac{1}{4}\mathrm{and} \\ q=1-p=1-\frac{1}{4}=\frac{3}{4} \\ \mathrm{Let}X\mathrm{denote}\mathrm{a}\mathrm{success}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{spade}\mathrm{in}\mathrm{a}\mathrm{throw}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=3\mathrm{and}p=\frac{1}{4} \\ \therefore \mathrm{P}\left(X=r\right)={}^3\mathrm{C}_r{p}^r{q}^{\left(3-r\right)}={}^3\mathrm{C}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{\left(3-r\right)}=\frac{{}^3\mathrm{C}_r{3}^{\left(3-r\right)}}{4^3}=\frac{27}{64}\left(\frac{{}^3\mathrm{C}_r}{3^r}\right),\mathrm{where},r=0,1,2,3 \\ \mathrm{So},\mathrm{the}\mathrm{probability}\mathrm{distibution}\mathrm{of}X\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{P}\left(X=r\right)=\frac{27}{64}\left(\frac{{}^3\mathrm{C}_r}{3^r}\right),\mathrm{where},r=0,1,2,3 \\ \mathrm{Now}, \\ \mathrm{Mean},\mathrm{E}\left(X\right)=np=3\times \frac{1}{4}=\frac{3}{4} \end{gathered}$$