(a)OR(b)

A card from a well shuffled deck of $\displaystyle 52$ playing cards is lost. From the remaining cards of the pack, a card is drawn at random and is found to be a King. Find the probability of the lost card being a King.

Let $\displaystyle \mathrm{E}_{1}$ be the event of lost card is King, $\displaystyle \mathrm{E}_{2}$ be the event of lost card not a King and A be the event of drawing a King from remaining $\displaystyle 51$ cards. so, $\displaystyle \left.\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{13}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{12}{13}, \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\frac{3}{51}, \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\frac{4}{51}\right\}$ Now, Required probability is $\displaystyle \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)$, $\displaystyle \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{E}_{2}\right)}=\frac{\dfrac{1}{13} \times \dfrac{3}{51}}{\dfrac{1}{13} \times \dfrac{3}{51}+\dfrac{12}{13} \times \dfrac{4}{51}}=\frac{1}{17}$ Let $\displaystyle \mathrm{P}(1)=\mathrm{P}(3)=\mathrm{P}(5)=\mathrm{p}$, so $\displaystyle \mathrm{P}(2)=\mathrm{P}(4)=\mathrm{P}(6)=2 \mathrm{p}$ As, $\displaystyle \mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)+\mathrm{P}(5)+\mathrm{P}(6)=1 \Rightarrow 9 \mathrm{p}=1 \Rightarrow \mathrm{p}=\frac{1}{9} \mathrm{P}($ Getting $\displaystyle 6 )=\frac{2}{9}, \mathrm{P}($ Not getting six $\displaystyle )=\frac{7}{9}$ Let X represents the Number of sixes Possible values of X are $\displaystyle 0$, $\displaystyle 1$ or $\displaystyle 2$ Now, $\displaystyle \mathrm{P}(\mathrm{X}=0)=\frac{7}{9} \times \frac{7}{9}=\frac{49}{81}, \mathrm{P}(\mathrm{X}=1)=2 \times \frac{7}{9} \times \frac{2}{9}=\frac{28}{81}, \mathrm{P}(\mathrm{X}=2)=\frac{2}{9} \times \frac{2}{9}=\frac{4}{81}$ Required probability distribution of number of sixes is $\displaystyle \mathrm{P}(\mathrm{X})$ & $\displaystyle \frac{49}{81}$ & $\displaystyle \frac{28}{81}$ & $\displaystyle \frac{4}{81}$ Mean of $\displaystyle \mathrm{X}=\sum_{i=1}^{3} \mathrm{X}_{i} \mathrm{P}\left(\mathrm{X}_{i}\right)=0+\frac{28}{81}+\frac{8}{81}=\frac{4}{9}$