Question

Let X be a Poisson variate with $P(X = 3) = P(X = 4)$. Then $P(X \neq 0)$ is

• A. $e^{-4}$
• B. $1 - e^{-4}$
• C. $1 - e^{-3}$
• D. $e^{-3}$

Hint:

For any Poisson distribution, if $P(X = k) = P(X = k+1)$, then the parameter $\lambda$ is exactly equal to $k+1$. Here, $k=3 \implies \lambda = 4$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A Poisson distribution is a discrete probability distribution parameterized by its mean rate parameter $\lambda$.
Key Formula or Approach:
The probability mass function of a Poisson distribution is:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

Step 2: Detailed Explanation:
We are given the condition:
\[ P(X = 3) = P(X = 4) \]
Substitute $k = 3$ and $k = 4$ into the probability mass function:
\[ \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^4}{4!} \]
Since $e^{-\lambda} \neq 0$ and assuming $\lambda \neq 0$, we can divide both sides by $e^{-\lambda} \lambda^3$:
\[ \frac{1}{3!} = \frac{\lambda}{4!} \]
Expand the factorials ($3! = 6$ and $4! = 24$):
\[ \frac{1}{6} = \frac{\lambda}{24} \implies \lambda = \frac{24}{6} = 4 \]
Now, calculate the probability $P(X \neq 0)$:
\[ P(X \neq 0) = 1 - P(X = 0) \]
Compute $P(X = 0)$ using $\lambda = 4$:
\[ P(X = 0) = \frac{e^{-4} 4^0}{0!} = e^{-4} \]
Substitute this value back:
\[ P(X \neq 0) = 1 - e^{-4} \]
Therefore, the probability is $1 - e^{-4}$.

Step 3: Final Answer
The correct option is (B).