Answer to Question #136595 in Statistics and Probability for Anarfi Peter

Question

If X has a Poisson distribution with parameter λ, and if
P(X=0)=0.2.
Evaluate
P(X>2).

Answer 1

Poisson distribution has following PDF:

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

We can use given value $P(X=0)=0.2$ to find the value of $\lambda$.

$P(X=k) = \frac{1}{1} e^{-\lambda} = 0.2$

$\lambda = - \ln{0.2=1.609}$, and, since the coefficient $e^{-\lambda}$ is the same for every X=k, we can use directly $e^{-\lambda}=0.2$.

$P(X>2) = 1 - P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)$

$P(X=0)=0.2$

$P(X=1) = \frac{1.609}{1} 0.2 =0.322$

$P(X=2) = \frac{1.609^2}{2 \cdot 1} 0.2=0.259$

So, $P(X>2) = 1 -0.2-0.322-0.259 = 0.219$

Answer: $P(X>2)=0.219$