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

Answer: (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

$P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$
$=\frac{\frac{1}{4}\cdot \frac{22}{425}}{\frac{1}{4}\cdot \frac{22}{425}+\frac{3}{4}\cdot \frac{26}{425}}$
$=\frac{22}{22+78}=\frac{11}{50}$