Question

If a random variable X follows poison distribution such that 3P(X=1)=P(X=2), then P(X=4) is,

• A. \(76e^{-4}\)
• B. \(18e^{-6}\)
• C. \(54e^{-6}\)
• D. \(36e^{-6}\)

Answer 1

The Correct Option is C

To solve the problem, we need to analyze the properties of the Poisson distribution. A random variable X follows a Poisson distribution with parameter λ, where the probability of X taking a value k is given by:

\(P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\)

Given that \(3P(X=1)=P(X=2)\), we have:

\[3 \cdot \frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}\]

Simplifying, we get:

\[3 \lambda = \frac{\lambda^2}{2}\]

Multiplying both sides by 2 to eliminate the fraction:

\[6 \lambda = \lambda^2\]

Rearranging gives:

\[\lambda^2 - 6 \lambda = 0\]

Factoring out λ:

\[\lambda(\lambda - 6) = 0\]

This implies \(\lambda = 0\) or \(\lambda = 6\). λ cannot be 0 in a Poisson distribution, so \(\lambda = 6\).

Now we need to find \(P(X=4)\):

\[P(X=4) = \frac{e^{-6} \cdot 6^4}{4!}\]

\[= \frac{e^{-6} \cdot 6^4}{24}\]

Calculating \(6^4\):

\[6^4 = 1296\]

Thus:

\[P(X=4) = \frac{1296 \cdot e^{-6}}{24}\]

\[= 54 e^{-6}\]

Therefore, the probability \(P(X=4)\) is \(54e^{-6}\).