Question

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

Answer 1

Let $E_1$ be the event that the lost card is Diamond

Given that there are 13 diamonds in the deck $P\left(E_1\right)=\frac{13}{52}=\frac{1}{4}$

let $E_2$ be the even that the lost card is not a diamond $P\left(E_2\right)=\frac{3}{4}$

Let A be the event that the two cards drawn are both diamonds

$P\left(A/E_1\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_1}=\frac{No.ofwaysofdrawing2diamondscards}{Totalno.ofwaysofdrawing2cards}$

$=\frac{66}{1275}$

$P\left(A/E_2\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{78}{1275}$

Applying Bayes Theorem, where

$P\left(E_1/A\right)=\frac{P\left(E_1\right)\times \left(P\right(A/E_1\left)\right)}{P\left(E_1\right)\times P\left(A/E_1\right)+P\left(E_2\right)\times P\left(A/E_2\right)}$

$=\frac{\frac{66}{1275}\times \frac{1}{4}}{\frac{66}{1275}\times \frac{1}{4}+\frac{78}{1275}\times \frac{3}{4}}=\frac{11}{50}$