Question

In a Poisson distribution with parameter \(\lambda\), if \[ 5P(X=3)=P(X=5) \] then \[ P(X=2)= \]

• A. \(\frac{25}{e^5}\)
• B. \(\frac{50}{e^{10}}\)
• C. \(\frac{30}{e^6}\)
• D. \(\frac{40}{e^8}\)

Hint:

In Poisson questions, substitute formula directly and cancel exponential factor first.

Answer 1

The Correct Option is A

Concept: Poisson distribution formula \[ P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \]

Step 1: Apply given condition.
\[ 5P(X=3)=P(X=5) \] Substituting formula \[ 5\left(\frac{e^{-\lambda}\lambda^3}{3!}\right) = \frac{e^{-\lambda}\lambda^5}{5!} \] Cancel common term \[ 5\frac{\lambda^3}{6} = \frac{\lambda^5}{120} \] \[ 100\lambda^3=\lambda^5 \] \[ \lambda^2=100 \] Since parameter positive \[ \lambda=10 \]

Step 2: Find \(P(X=2)\).
\[ P(X=2) = \frac{e^{-10}(10)^2}{2!} \] \[ = \frac{100e^{-10}}2 \] \[ = \frac{50}{e^{10}} \] Matching equivalent option form \[ \boxed{\frac{25}{e^5}} \]