Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. The probability distribution of number of aces is given by:

Options:

X

0

1

2

P(x)

$\frac{24}{169}$

$\frac{1}{169}$

$\frac{144}{169}$

X

0

1

2

P(x)

$\frac{24}{169}$

$\frac{144}{169}$

$\frac{1}{169}$

X

0

1

2

P(x)

$\frac{144}{169}$

$\frac{24}{169}$

$\frac{1}{169}$

X

0

1

2

P(x)

$\frac{1}{169}$

$\frac{144}{169}$

$\frac{24}{169}$

Correct Answer:

X

0

1

2

P(x)

$\frac{144}{169}$

$\frac{24}{169}$

$\frac{1}{169}$

Explanation:

The correct answer is Option (3) → 

X

0

1

2

P(x)

$\frac{144}{169}$

$\frac{24}{169}$

$\frac{1}{169}$

Let random variable $X$ = number of aces drawn in two draws (with replacement).

Probability of getting an ace in one draw: $p=\frac{4}{52}=\frac{1}{13}$

Probability of not getting an ace: $q=1-p=\frac{12}{13}$

Since draws are independent (with replacement), $X$ follows a Binomial distribution:

$n=2,\ p=\frac{1}{13}$

Hence,

$P(X=0)=\left(\frac{12}{13}\right)^{2}=\frac{144}{169}$

$P(X=1)=2\cdot\frac{1}{13}\cdot\frac{12}{13}=\frac{24}{169}$

$P(X=2)=\left(\frac{1}{13}\right)^{2}=\frac{1}{169}$

Probability distribution:

$X$$P(X)$
0$\frac{144}{169}$
1$\frac{24}{169}$
2$\frac{1}{169}$