Question:
Let \( X \) be a random variable having the Poisson distribution with mean \( 1 \). Let \( g: \mathbb{N} \cup \{0\} \to \mathbb{R} \) be defined by
\[g(x) = \begin{cases} 1 & \text{if } x \in \{0, 2\} \\ 0 & \text{if } x \notin \{0, 2\} \end{cases}\]
Then \( E(g(X)) \) is equal to:
Let \( X \) be a random variable having the Poisson distribution with mean \( 1 \). Let \( g: \mathbb{N} \cup \{0\} \to \mathbb{R} \) be defined by
\[g(x) = \begin{cases} 1 & \text{if } x \in \{0, 2\} \\ 0 & \text{if } x \notin \{0, 2\} \end{cases}\]
Then \( E(g(X)) \) is equal to:
\[g(x) = \begin{cases} 1 & \text{if } x \in \{0, 2\} \\ 0 & \text{if } x \notin \{0, 2\} \end{cases}\]
Then \( E(g(X)) \) is equal to:
Updated On: Oct 1, 2024
- \( e^{-1} \)
- \( 2e^{-1} \)
- \( \frac{5}{2} e^{-1} \)
- \( \frac{3}{2} e^{-1} \)
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The Correct Option is D
Solution and Explanation
The correct option is (D): \( \frac{3}{2} e^{-1} \)
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