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

Answer: (C)

(A), (B), (C) and (D)

Answer 2

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)