The probability distribution of no. of aces when two cards are drawn without replacement from a pack of 52 cards is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) →
| $x_1$ | 0 | 1 | 2 |
| $p_1$ | $\frac{188}{221}$ | $\frac{32}{221}$ | $\frac{1}{221}$ |
$\text{P(X=0;Getting 0 Aces)}=\frac{{^{48}C}_1.{^{47}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{48×47}{52×51}$
$=\frac{188}{221}$
$\text{P(X=1;Getting 1 Aces)}=\frac{2×{^{48}C}_1.{^{4}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{2×48×4}{52×51}$
$=\frac{32}{221}$
$\text{P(X=2;Getting 2 Aces)}=\frac{{^4C}_1.{^{3}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{4×3}{52×51}$
$=\frac{1}{221}$