Two cards are drawn with replacement fro

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

Two cards are drawn with replacement from a pack of 52 cards. The probability distribution of number of aces is :

Options:

$x_i$	0	1	2
$p_i$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

$x_i$	0	1	2
$p_i$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

$x_i$	0	1	2
$p_i$	$\frac{124}{169}$	$\frac{20}{169}$	$\frac{1}{169}$

$x_i$	0	1	2
$p_i$	$\frac{144}{169}$	$\frac{20}{169}$	$\frac{1}{169}$

Correct Answer:

$x_i$	0	1	2
$p_i$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Explanation:

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
$p_i$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$