Question

If X is a Poisson random variate with mean 3, then \(P(|X-3|<2)\) =

• A. \( \frac{9}{2e^3} \)
• B. \( \frac{99}{8e^3} \)
• C. \( \frac{3}{2e^3} \)
• D. \( \frac{1}{3e^3} \)

Answer 1

The Correct Option is A

Given: \(X\) is a Poisson random variable with mean \( \lambda = 3 \).

The probability mass function (PMF) of a Poisson distribution is:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] So, for this case: \[ P(X = k) = \frac{e^{-3} \cdot 3^k}{k!} \]

We are asked to compute:

\[ P(|X - 3| < 2) \]

Step 1: Interpret the inequality

\[ |X - 3| < 2 \Rightarrow -2 < X - 3 < 2 \Rightarrow 1 < X < 5 \] Since \(X\) takes only integer values, this means: \[ X = 2, 3, 4 \]

Step 2: Calculate each probability

• \( P(X=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9}{2}e^{-3} \)
• \( P(X=3) = \frac{e^{-3} \cdot 3^3}{3!} = \frac{27}{6}e^{-3} = \frac{9}{2}e^{-3} \)
• \( P(X=4) = \frac{e^{-3} \cdot 3^4}{4!} = \frac{81}{24}e^{-3} = \frac{27}{8}e^{-3} \)

Step 3: Add the probabilities

\[ P(|X - 3| < 2) = \left( \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) e^{-3} = \left( 9 + \frac{27}{8} \right)e^{-3} = \frac{99}{8}e^{-3} \]

Final Answer:

\( \boxed{\frac{99}{8e^3}} \)

This matches the answer marked correct in the options (option b).