Question

If \(X\) is a Poisson random variable with mean \(3\), then \[ P(|X-3|\lt 2) = \] is equal to:

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

Hint:

For a Poisson random variable with parameter \(\lambda\), \[ P(X=x)=e^{-\lambda}\frac{\lambda^x}{x!}. \] Always convert inequalities involving \(X\) into the corresponding integer values before computing probabilities.

Answer 1

The Correct Option is B

Step 1: Write the Poisson probability function.
Since \(X\) follows a Poisson distribution with mean \[ \lambda=3, \] its probability mass function is \[ P(X=x) = e^{-3}\frac{3^x}{x!}, \qquad x=0,1,2,\ldots \]

Step 2: Interpret the condition \(|X-3|\lt 2\).
We have \[ |X-3|\lt 2 \] \[ -2\lt X-3\lt 2 \] \[ 1\lt X\lt 5 \] Since \(X\) takes integer values, \[ X=2,3,4 \] Therefore, \[ P(|X-3|\lt 2) = P(X=2)+P(X=3)+P(X=4) \]

Step 3: Compute each probability.
For \(X=2\), \[ P(X=2) = e^{-3}\frac{3^2}{2!} = \frac{9}{2}e^{-3} \] For \(X=3\), \[ P(X=3) = e^{-3}\frac{3^3}{3!} = \frac{27}{6}e^{-3} = \frac{9}{2}e^{-3} \] For \(X=4\), \[ P(X=4) = e^{-3}\frac{3^4}{4!} = \frac{81}{24}e^{-3} = \frac{27}{8}e^{-3} \]

Step 4: Add the probabilities.
Hence, \[ P(|X-3|\lt 2) = e^{-3} \left( \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) \] \[ = e^{-3} \left( \frac{36+27}{8} \right) \] \[ = \frac{99}{8e^3} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{99}{8e^3}} \] Hence, the correct option is \[ \boxed{\left(2\right)\ \frac{99}{8e^3}} \]