A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | otherwise |
| P(X) | k | 2k | 3k | 0 |
Then:
(A) \(k=\displaystyle\frac{1}{6}\)
(B) \(P(X \lt 2)=\displaystyle\frac{1}{2}\)
(C) \(E(X)=\displaystyle\frac{3}{4}\)
(D) \(P(1 \lt X \lt 2)=\displaystyle\frac{5}{6}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
In a probability distribution the probabilities add up to 1:
\(k + 2k + 3k = 1\)
\( \displaystyle 6k = 1,\quad k = \frac{1}{6} \)
X < 2 means X = 0 or 1:
\( \displaystyle P(X \lt 2) = k + 2k = 3k = \frac{1}{2} \)
The expected value is the sum of x × P(x):
\(E(X) = 0(k) + 1(2k) + 2(3k) = 8k\)
\( \displaystyle = \frac{8}{6} = \frac{4}{3} \)
X takes only the values 0, 1, 2, so no value lies strictly between 1 and 2:
\(P(1 \lt X \lt 2) = 0\)
So \(k = \frac{1}{6}\), \(P(X \lt 2) = \frac{1}{2}\), \(E(X) = \frac{4}{3}\) and \(P(1 \lt X \lt 2) = 0\).
Checking the statements
(A) \(k=\frac{1}{6}\) — Correct. 6k = 1 gives k = 1/6.
(B) \(P(X \lt 2)=\frac{1}{2}\) — Correct. P(X = 0) + P(X = 1) = 3k = 1/2.
(C) \(E(X)=\frac{3}{4}\) — Incorrect. E(X) = 2k + 6k = 8k = 4/3.
(D) \(P(1 \lt X \lt 2)=\frac{5}{6}\) — Incorrect. No value of X lies strictly between 1 and 2, so this probability is 0.