Question

If \( X \) is a Poisson variate such that \[ 2P(X = 0) + P(X = 2) = 2P(X = 1), \text{ then } E(X) \] is equal to

• A. 1
• B. 1.5
• C. 2
• D. 1.75

Hint:

For Poisson distribution, the expected value is equal to the parameter \( \lambda \).

Answer 1

The Correct Option is B

Step 1: Using the Poisson distribution.
The Poisson distribution is given by \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \). Using the given condition, solve for \( \lambda \) and calculate \( E(X) = \lambda \).

Step 2: Conclusion.
Thus, the correct answer is option (B).

Final Answer: \[ \boxed{\text{(B) } 1.5} \]