Question

Two cards are drawn successively with replacement from a well shuffled deck of $52$ cards. Find the probability distribution of the number of aces.

• A. $\frac{1}{169}$
• B. $\frac{1}{221}$
• C. $\frac{1}{265}$
• D. $\frac{4}{663}$

Answer 1

The correct option is A $\frac{1}{169}$

Total no. of cards = 52

Total no. of aces = 4

$P\left(ace\right)=\frac{4}{52}=\frac{1}{13}$

$P(non-ace)=\frac{48}{52}=\frac{12}{13}$

Let X be a random variable, such that

X = no. of aces obtained in 2 draws.

Case I:- X = 0 (No aces are obtained)

$P(X=0)=P\left(\text{non-ace and non-ace}\right)$

$=P(non-ace)\times P(non-ace)$

$=\frac{12}{13}\times \frac{12}{13}$

$=\frac{144}{169}$

Case II:- X = 1 (1 ace is obtained)

$P(X=1)=P\left(\text{(ace and non-ace) or (non-ace and ace)}\right)$

$=\left(P\right(non-ace)\times P(non-ace\left)\right)+\left(P\right(ace)\times P(non-ace\left)\right)$

$=(\frac{12}{13}\times \frac{1}{13})+(\frac{1}{13}\times \frac{12}{13})$

$=\frac{24}{169}$

Case III:- X = 2 (two aces are obtained)

$P(X=2)=P\left(\text{ace and ace}\right)$

$=P\left(ace\right)\times P\left(ace\right)$

$=\frac{1}{13}\times \frac{1}{13}$

$=\frac{1}{169}$