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)

Hint:

When solving problems involving probabilities, always ensure the sum of probabilities equals 1, as this is a fundamental rule. For expected value calculations, use the formula \( E(X) = \sum X \cdot P(X) \). If you're given a probability distribution, carefully evaluate the sums and the expected values, and remember to substitute correctly when necessary. Finally, check your work for consistency and ensure you match the terms correctly.

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).

Answer 2

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

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

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

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

Match: (D) → (I).

Final matching:

• (A) → (IV)
• (B) → (III)
• (C) → (II)
• (D) → (I)