Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

If the random variable x has the following distribution:

$X$

0

1

2

otherwise

$P(x)$

  $k$  

  $k$  

  $2k$  

0

Match List-I with List-II

List-I

List-II

(A) $k$

(I) $\frac{3}{4}$

(B) $P(x≥2)$

(II) $\frac{1}{4}$

(C) $P(X ≤2)$

(III) $\frac{1}{2}$

(D) $P(0 < x ≤ 2)$

(IV) 1

Choose the correct answer from the options given below.

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I) **

List-I

List-II

(A) $k$

(II) $\frac{1}{4}$

(B) $P(x≥2)$

(III) $\frac{1}{2}$

(C) $P(X ≤2)$

(IV) 1

(D) $P(0 < x ≤ 2)$

(I) $\frac{3}{4}$

Given distribution:

X : 0, 1, 2
P(X) : k, k, 2k

Total probability = 1:

$k + k + 2k = 4k = 1 \Rightarrow k = \frac{1}{4}$

Now compute:

$P(x \ge 2) = P(2) = 2k = \frac{1}{2}$

$P(X \le 2) = P(0) + P(1) + P(2) = 1$

$P(0 < x \le 2) = P(1) + P(2) = k + 2k = 3k = \frac{3}{4}$

Final answer: (A)–(II), (B)–(III), (C)–(IV), (D)–(I)