Question

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:

Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is B

The given probabilities are \(k, 2k, 3k\) for \(X = 0, 1, 2\), respectively. Since the sum of probabilities must equal 1:

\[k + 2k + 3k = 1 \implies 6k = 1 \implies k = \frac{1}{6}.\]

For (A) \(k\): From above, \(k = \frac{1}{6}\). Match: (A) → (IV).

For (B) \(P(X<2)\): This is the sum of probabilities for \(X = 0\) and \(X = 1\):

\[P(X<2) = P(X = 0) + P(X = 1) = k + 2k = 3k.\]

Substituting \(k = \frac{1}{6}\):

\[P(X<2) = 3 \cdot \frac{1}{6} = \frac{1}{2}.\]

Match: (B) → (III).

For (C) \(E(X)\): The expected value is:

\[E(X) = \sum X \cdot P(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k = 0 + 2k + 6k = 8k.\]

Substituting \(k = \frac{1}{6}\):

\[E(X) = 8 \cdot \frac{1}{6} = \frac{4}{3}.\]

Match: (C) → (II).

For (D) \(P(1 \leq X \leq 2)\): This is the sum of probabilities for \(X = 1\) and \(X = 2\):

\[P(1 \leq X \leq 2) = P(X = 1) + P(X = 2) = 2k + 3k = 5k.\]

Substituting \(k = \frac{1}{6}\):

\[P(1 \leq X \leq 2) = 5 \cdot \frac{1}{6} = \frac{5}{6}.\]

Match: (D) → (I).

Final matching: (A) → (IV), (B) → (III), (C) → (II), (D) → (I).