The probability distribution of no. of a

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Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

The probability distribution of no. of aces when two cards are drawn without replacement from a pack of 52 cards is :

Options:

$x_1$	0	1	2
$p_1$	$\frac{144}{221}$	$\frac{24}{221}$	$\frac{1}{221}$

$x_1$	0	1	2
$p_1$	$\frac{144}{169}$	$\frac{24}{144}$	$\frac{1}{144}$

$x_1$	0	1	2
$p_1$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

$x_1$	0	1	2
$p_1$	$\frac{188}{221}$	$\frac{24}{221}$	$\frac{1}{221}$

Correct Answer:

$x_1$	0	1	2
$p_1$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

Explanation:

The correct answer is Option (3) →

$x_1$	0	1	2
$p_1$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

$\text{P(X=0;Getting 0 Aces)}=\frac{{^{48}C}_1.{^{47}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{48×47}{52×51}$

$=\frac{188}{221}$

$\text{P(X=1;Getting 1 Aces)}=\frac{2×{^{48}C}_1.{^{4}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{2×48×4}{52×51}$

$=\frac{32}{221}$

$\text{P(X=2;Getting 2 Aces)}=\frac{{^4C}_1.{^{3}C}_1}{{^{52}C}_1.{^{51}C}_1}=\frac{4×3}{52×51}$

$=\frac{1}{221}$