A random variable X follow Poisson distribution such that P(X=1) = 2P(X=2) , then P(X=0) is

For any Poisson distribution, the probability formula is:

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

where $\lambda$ is the parameter of the distribution and $k$ is the value we're looking for.

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Given: $P(X=1) = 2P(X=2)$

For $P(X=1)$:

$P(X=1) = \frac{e^{-\lambda} \times \lambda^1}{1!}$

$P(X=1) = e^{-\lambda} \times \lambda$

For $P(X=2)$:

$P(X=2) = \frac{e^{-\lambda} \times \lambda^2}{2!}$

$P(X=2) = \frac{e^{-\lambda} \times \lambda^2}{2}$

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Using the condition $P(X=1) = 2P(X=2)$:

$e^{-\lambda} \times \lambda = 2 \times \left(\frac{e^{-\lambda} \times \lambda^2}{2}\right)$

$e^{-\lambda} \times \lambda = e^{-\lambda} \times \lambda^2$

Dividing both sides by $e^{-\lambda}$:

$\lambda = \lambda^2$

$\lambda^2 - \lambda = 0$

$\lambda(\lambda - 1) = 0$

This gives $\lambda = 0$ or $\lambda = 1$

Since $\lambda$ cannot be 0 for a valid Poisson distribution, $\lambda = 1$

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Finding $P(X=0)$ with $\lambda = 1$:

$P(X=0) = \frac{e^{-1} \times 1^0}{0!}$

$P(X=0) = \frac{e^{-1} \times 1}{1}$

$P(X=0) = \frac{1}{e}$

Therefore, $P(X=0) = \frac{1}{e}$