The random variable X has the following probability distribution
| X | 0 | 1 | 2 | 3 |
| P(X) | a | a | b | b |
such that \(E(x^2) = 2E(x)\), then the value of \(b\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{8}\)
The probabilities add up to 1:
\( \displaystyle 2a + 2b = 1 \Rightarrow a + b = \frac{1}{2} \)
The expectations are
\(E(X) = 0 \cdot a + 1 \cdot a + 2b + 3b = a + 5b\)
\(E(X^2) = 0 \cdot a + 1 \cdot a + 4b + 9b = a + 13b\)
Using \(E(X^2) = 2E(X)\):
\(a + 13b = 2a + 10b\)
\(a = 3b\)
Substituting in \(a + b = \frac{1}{2}\):
\( \displaystyle 4b = \frac{1}{2}, \quad b = \frac{1}{8} \)