Question

A random variable \( X \) has the following probability distribution:

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

Then: 
(A) \( k = \frac{1}{6} \) 
(B) \( P(X < 2) = \frac{1}{2} \) 
(C) \( E(X) = \frac{3}{4} \) 
(D) \( P(1 < X \leq 2) = \frac{5}{6} \) 
Choose the correct answer from the options given below:

• A. (A) and (B) only
• B. (A), (B) and (C) only
• C. (A), (B), (C) and (D)
• D. (B), (C) and (D) only

Answer 1

The Correct Option is C

Step 1: Verify the value of \(k\): The sum of all probabilities must equal 1:

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

Hence, statement (A) is correct.

Step 2: Compute \(P(X < 2):\) Add the probabilities for \(X = 0\) and \(X = 1\):

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

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

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

Hence, statement (B) is correct.

Step 3: Calculate \(E(X):\) The expected value is given by:

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

Simplify:

\[ E(X) = 0 + 2k + 6k = 8k. \]

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

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

Hence, statement (C) is correct.

Step 4: Evaluate \(P(1 < X < 2):\) Only \(X = 2\) satisfies \(1 < X < 2\). Thus:

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

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

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

Hence, statement (D) is correct.

Correct Answer: 3. (A), (B), (C), and (D)