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 diamonds. Find the probability of the lost cards being a diamond.

Answer 1

Let $E_1$ the event that lost cards is a diamond $\Rightarrow n\left(E_1\right)$ = 13

$E_2$ : lost cards is not a diamond $\Rightarrow n\left(E_2\right)$ = 52-13=39 And, n(S)=52

Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive events. $\therefore P\left(E_1\right)=\frac{13}{52}=\frac{1}{4}$ and $P\left(E_2\right)=\frac{39}{52}=\frac{3}{4}$

Let E : the events that two cards drawn from the remaining pack are diamonds, When one diamond card is lost, there are 12 diamond cards out of 51 cards.

The cards can be drawn out of 12 diamond cards in ${}^{12}{C}_2$ ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ${}^{51}{C}_2$ ways. The probability of getting two cards, when one diamond card is lost, is given

by $P\left(\frac{E}{E_1}\right)\therefore P\left(\frac{E}{E_1}\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_2}$ = $\frac{\frac{12\times 11}{1\times 2}}{\frac{51\times 50}{1\times 2}}=\frac{12\times 11}{51\times 50}$

and $P\left(\frac{E}{E_2}\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{\frac{13\times 12}{1\times 2}}{\frac{51\times 50}{1\times 2}}=\frac{13\times 12}{51\times 50}$

By using Baye's theorem, we obtain

$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}=\frac{\frac{12\times 11}{51\times 50}\times \frac{1}{4}}{\frac{12\times 11}{51\times 50}\times \frac{1}{4}+\frac{13\times 12}{51\times 50}\times \frac{3}{4}}$

= $\frac{12\times 11}{12\times 11+13\times 12\times 3}=\frac{132}{132+468}=\frac{132}{600}=\frac{11}{50}$