Question

If a Poisson variate X satisfies the relation $P(X=3) = P(X=5)$, then $P(X=4) =$

• A. $\frac{50}{3e^{\sqrt{20}}}$
• B. $\frac{20000}{3e^{20}}$
• C. $\frac{125}{3e^{10}}$
• D. $\frac{25}{3e^{\sqrt{20}}}$

Hint:

In Poisson distribution problems where you are given an equality like $P(X=k_1) = P(X=k_2)$, the term $e^{-\lambda}$ will always cancel out. The problem then simplifies to an algebraic equation in $\lambda$ involving factorials. Be comfortable with simplifying factorial expressions like $\frac{n!}{k!} = n(n-1)\dots(k+1)$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept: For a Poisson distribution with parameter \( \lambda \), the probability of \( k \) occurrences is given by \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \).
Step 2: Solve for \( \lambda \): Given \( P(X=3) = P(X=5) \): \[ \frac{e^{-\lambda}\lambda^3}{3!} = \frac{e^{-\lambda}\lambda^5}{5!} \] Dividing both sides by \( e^{-\lambda}\lambda^3 \) (since \( \lambda>0 \)): \[ \frac{1}{6} = \frac{\lambda^2}{120} \] \[ \lambda^2 = \frac{120}{6} = 20 \implies \lambda = \sqrt{20} \]
Step 3: Calculate \( P(X=4) \): \[ P(X=4) = \frac{e^{-\lambda}\lambda^4}{4!} \] Substitute \( \lambda = \sqrt{20} \) and \( \lambda^4 = (\lambda^2)^2 = 20^2 = 400 \): \[ P(X=4) = \frac{e^{-\sqrt{20}}(400)}{24} \] Simplify the fraction \( \frac{400}{24} \): \[ \frac{400}{24} = \frac{100}{6} = \frac{50}{3} \] Thus, \[ P(X=4) = \frac{50}{3} e^{-\sqrt{20}} = \frac{50}{3e^{\sqrt{20}}} \]