If \( X \) is a random variable such that \( X \) can take values 0,1, 2 or 3. Then the expectation of \( X \) for the following data is :
| \( X \) | 0 | 1 | 2 | 3 |
| \( P(X) \) | \( k \) | \( k^2 \) | \( 1-5k^2 \) | \( k^2 \) |
(where \( k\gt 0 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{4}{3} \)
The probabilities must add up to 1:
\( \displaystyle k+k^2+(1-5k^2)+k^2=1 \)
\( \displaystyle k-3k^2=0\ \Rightarrow\ k(1-3k)=0 \)
Since k > 0, \( k=\frac{1}{3} \), and \( k^2=\frac{1}{9} \). The distribution is
\( \displaystyle P(0)=\frac{1}{3},\ P(1)=\frac{1}{9} \)
\( \displaystyle P(2)=1-\frac{5}{9}=\frac{4}{9},\ P(3)=\frac{1}{9} \)
\( \displaystyle E(X)=\sum x\,P(x) \)
\( \displaystyle =0+\frac{1}{9}+2\cdot\frac{4}{9}+3\cdot\frac{1}{9} \)
\( \displaystyle =\frac{1+8+3}{9} \)
\( \displaystyle =\frac{12}{9}=\frac{4}{3} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{3} \) — Incorrect. 1/3 is the value of k, not the expectation.
Option (2) 1 — Incorrect. 1 is the sum of the probabilities, not \( \sum x\,P(x) \).
Option (4) \( \displaystyle\frac{1}{9} \) — Incorrect. 1/9 is \( k^2 \), the probability P(X = 1) or P(X = 3), not the expectation.