CUET Mathematics 2024 - If the random variable X has the following distribution: Match List-I with List-II: | PYQs + Solutions

CUET Mathematics 2024

Algebra

Probability

Easy

If the random variable $X$ has the following distribution:

$X$ 0 1 2 otherwise

$P(X)$ $k$ $2k$ $3k$ $0$

Match List-I with List-II:

List-I List-II

(A) $k$ (I) $\frac{5}{6}$

(B) $P(X < 2)$ (II) $\frac{4}{3}$

(C) $E(X)$ (III) $\frac{1}{2}$

(D) $P(1 \leq X \leq 2)$ (IV) $\frac{1}{6}$

$X$	0	1	2	otherwise
$P(X)$	$k$	$2k$	$3k$	$0$

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

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

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

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

✅ Correct Option: 2

First, we need to find the value of

k

For any probability distribution, the sum of all probabilities must equal 1:

P(X = 0) + P(X = 1) + P(X = 2) = 1

k + 2k + 3k = 1

6k = 1

k = \dfrac{1}{6}

Next, let's find

P(X < 2)

P(X < 2) = P(X = 0) + P(X = 1)

P(X < 2) = k + 2k = 3k = 3 \cdot \dfrac{1}{6} = \dfrac{1}{2}

For the expected value

E(X)

E(X) = \sum[x \cdot P(X = x)]

E(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k

E(X) = 2k + 6k = 8k = 8 \cdot \dfrac{1}{6} = \dfrac{4}{3}

For

P(1 \leq X \leq 2)

P(1 \leq X \leq 2) = P(X = 1) + P(X = 2)

P(1 \leq X \leq 2) = 2k + 3k = 5k = 5 \cdot \dfrac{1}{6} = \dfrac{5}{6}

Matching the lists:

(A)

k = \dfrac{1}{6}

matches with (IV)

(B)

P(X < 2) = \dfrac{1}{2}

matches with (III)

(C)

E(X) = \dfrac{4}{3}

matches with (II)

(D)

P(1 \leq X \leq 2) = \dfrac{5}{6}

matches with (I)