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
The problem deals with a Poisson distribution. We are given a relationship between the probabilities of two different outcomes, which allows us to find the parameter ($\lambda$) of the distribution. Once we have $\lambda$, we can calculate the probability of any other outcome.
Step 2: Key Formula or Approach
The probability mass function (PMF) for a Poisson distribution with parameter $\lambda$ is: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \] 1. Set up the given equation: $P(X=3) = P(X=5)$. 2. Substitute the PMF formula into the equation. 3. Solve for the parameter $\lambda$. 4. Use the found value of $\lambda$ to calculate $P(X=4)$.
Step 3: Detailed Explanation
1. Set up the equation: We are given $P(X=3) = P(X=5)$. Using the Poisson PMF: \[ \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!} \] 2. Solve for $\lambda$: Since $\lambda$ (the mean rate) must be positive, we know $\lambda \neq 0$, and $e^{-\lambda} \neq 0$. We can cancel these terms from both sides. \[ \frac{\lambda^3}{3!} = \frac{\lambda^5}{5!} \] \[ \frac{1}{3!} = \frac{\lambda^2}{5!} \] Rearrange to solve for $\lambda^2$: \[ \lambda^2 = \frac{5!}{3!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20 \] So, $\lambda = \sqrt{20}$. 3. Calculate P(X=4): Now we need to find $P(X=4)$ using $\lambda = \sqrt{20}$ and $\lambda^2=20$. \[ P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} \] \[ P(X=4) = \frac{e^{-\sqrt{20}} (\lambda^2)^2}{24} = \frac{e^{-\sqrt{20}} (20)^2}{24} \] \[ P(X=4) = \frac{e^{-\sqrt{20}} \cdot 400}{24} \] Simplify the fraction $\frac{400}{24}$: \[ \frac{400}{24} = \frac{100}{6} = \frac{50}{3} \] So, \[ P(X=4) = \frac{50}{3} e^{-\sqrt{20}} \] This matches the structure of the first option provided in the OCR, which is $\frac{50}{3e^{\sqrt{20}}}$. Step 4: Final Answer
The parameter of the Poisson distribution is $\lambda = \sqrt{20}$. The probability of $X=4$ is $P(X=4) = \frac{50}{3}e^{-\sqrt{20}}$.