Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

Two numbers are selected without replacement at random, one at a time from the first six positive integers. Let x denotes the larger of the two numbers.

Match List-I with List-II

List-I

List-II

(A) $P(x = 2)$

(I) $\frac{4}{15}$

(B) $P(x = 3)$

(II) $\frac{1}{15}$

(C) $P(x = 4)$

(III) $\frac{2}{15}$

(D) $P(x = 5)$

(IV) $\frac{1}{5}$

Choose the correct answer from the options given below:

Options:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct Answer:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Explanation:

The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

List-I

List-II

(A) $P(x = 2)$

(II) $\frac{1}{15}$

(B) $P(x = 3)$

(III) $\frac{2}{15}$

(C) $P(x = 4)$

(IV) $\frac{1}{5}$

(D) $P(x = 5)$

(I) $\frac{4}{15}$

The first six positive integers are: 1, 2, 3, 4, 5, 6.

Total number of ways to choose 2 numbers (without replacement):

6C2 = 15

Let $x$ denote the larger of the two selected numbers.

(A) $P(x = 2)$:

Only one pair possible: (1,2) → larger = 2

So, $P(x=2) = \frac{1}{15}$

(B) $P(x = 3)$:

Valid pairs: (1,3), (2,3) → 2 pairs

$P(x=3) = \frac{2}{15}$

(C) $P(x = 4)$:

Valid pairs: (1,4), (2,4), (3,4) → 3 pairs

$P(x=4) = \frac{3}{15} = \frac{1}{5}$

(D) $P(x = 5)$:

Valid pairs: (1,5), (2,5), (3,5), (4,5) → 4 pairs

$P(x=5) = \frac{4}{15}$