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 diamonds. Find the probability of the missing card to be a diamond.

• A. $\frac{39}{50}$
• B. $\frac{11}{50}$
• C. $\frac{23}{25}$
• D. $\frac{3}{26}$

Answer 1

The correct option is B $\frac{11}{50}$

Let $E_1,E_2,E_3,E_4$ A be the events as defined below:
$E_1$ = the missing card is diamond.
$E_2$ = the missing card is heart.
$E_3$ = the missing card is spade.
$E_4$ = the missing card is club.
A = drawing two diamonds cards from the remaining cards
Then $P\left(E_1\right)=\frac{13}{52}=\frac{1}{4},P\left(E_2\right)=\frac{13}{52}=\frac{1}{4}$
$P\left(E_3\right)=\frac{13}{52}=\frac{1}{4}$
and $P\left(E_4\right)=\frac{13}{52}=\frac{1}{4}$
$P\left(\frac{A}{E_1}\right)$ = probability of drawing two diamonds cards given that on diamond card is missing = $\frac{{}^{12}{C}_2}{{}^{51}{C}_2}$
$P\left(\frac{A}{E_2}\right)$ = probability of drawing two diamonds cards given that one heart card is missing $=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}$
Similarly, $P\left(\frac{A}{E_3}\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}$
and $P\left(\frac{E_1}{A}\right)=\frac{{}^{13}{C}_2}{{}^{31}{C}_2}$
By, Baye's rule
Required probability = $P\left(\frac{E_1}{A}\right)$
$=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)+P\left(E_4\right)P\left(\frac{A}{E_4}\right)}$
$=\frac{11}{50}$