Two cards are drawn with replacement from a pack of 52 cards. The probability distribution of number of aces is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1)
No of aces = 4 → total cards = 52
no. of drawn = 2 → Remaining/non aces = 48
case (1) → 0 aces ⇒ $\frac{{^{48}C}_1×{^{48}C}_1}{{^{52}C}_1×{^{52}C}_1}=\frac{144}{169}$
case (2) → 1 ace ⇒ $\frac{2!{^{48}C}_1×{^{4}C}_1}{{^{52}C}_1×{^{52}C}_1}=\frac{24}{169}$
case (3) → 2 aces ⇒ $\frac{{^{4}C}_1×{^{4}C}_1}{{^{52}C}_1×{^{52}C}_1}=\frac{1}{169}$
So table
| $x_i$ | 0 | 1 | 2 |
| $p_i$ | $\frac{144}{169}$ | $\frac{24}{169}$ | $\frac{1}{169}$ |