Question

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both clubs. Find the probability of the lost card being a club.

• A. $\frac{11}{50}$
• B. $\frac{17}{50}$
• C. $\frac{13}{50}$
• D. $\frac{19}{50}$

Answer 1

The Correct Option is A

Let $E_1$, $E_2$ and $A$ be the events defined as follows : $E_1 =$ lost card is of club, $E_2 =$ lost card is not of club and $A =$ two cards drawn are both of clubs Then $P\left(E_{1}\right) = \frac{13}{52} =\frac{1}{4}$ and $P\left(E_{2}\right) = \frac{39}{52} = \frac{3}{4}$ When one card is lost, number of remaining cards in the pack $= 51$. When $E_1$ has occurred i.e. a card of club is lost, then the probability of drawing $2$ cards of club from the remaining pack i.e. $P(A|E_1) = \frac{^{12}C_{2}}{^{51}C_{2}} = \frac{66}{1275} = \frac{22}{425}$ When $E_2$ has occurred i.e. when a card of clubs is not lost, then the probability of drawing $2$ cards of club from the remaining pack i.e. $ P(A|E_2) = \frac{^{13}C_{2}}{^{51}C_{2}} = \frac{78}{1275} = \frac{26}{425}$ We want to find $P(E_1|A)$. By Bayes' theorem $P\left(E_{1}|A\right) = \frac{P\left(E_{1}\right)P\left(A |E_{1}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$ $=\frac{\frac{1}{4}\cdot\frac{22}{425}}{\frac{1}{4}\cdot\frac{22}{425}+\frac{3}{4}\cdot\frac{26}{425}}$ $= \frac{22}{22+78} = \frac{11}{50}$