# Poisson distribution variance, probability, and mean.

Let $$X$$ be the poisson random variable such that $$\mathbb P(X=2)=9\mathbb P(X=4)+90\mathbb P(X=6)$$.

a) Find the mean and variance of $$X$$.

b) Find $$\mathbb P(X\geq1)$$.

c) Find $$\mathbb P(X\leq10)$$.

For a), I need to turn each probability into its respective $$p(x)$$ form and solve for $$\lambda$$. I got that$$\lambda$$ was $$1$$ so therefore, the mean and variance should be $$1$$.

Now, for the second question, I have that $$\mathbb P(X\geq 1)=1- \mathbb P(X<1)$$. Am I to assume that $$\lambda$$ is still $$1$$? If so, is $$1-\mathrm e^{-1}$$ the correct answer for b)?

For c), I'm just lost here... it was asked to solve this problem through R programming language. Correct me if I'm wrong, but is is not just simply $$\mathbb P(X \leq10)=1-\mathbb P(X>10)$$? If so, I'd think that solving for $$\mathbb P(X>10)$$ would be plugging in ppois(10,1) into R, but I'm just getting $$1$$ as the output... that means a $$100\%$$ change of $$\mathbb P(X>10)$$... sorry if this question isn't what this MSE is for but if you know what I'm talking about, that'd be really helpful.

• The is the second time today I've seen part (a) of this question posted here. Commented Oct 30, 2014 at 4:09

$$\Pr(X=2) = 9\Pr(X=4) + 90\Pr(X=6)$$ $$\frac{\lambda^2 e^{-\lambda}}{2!} = 9\frac{\lambda^4 e^{-\lambda}}{4!} + 90 \frac{\lambda^6 e^{-\lambda}}{6!}$$ Multiply both sides by $2e^\lambda/\lambda^2$: $$1 = \frac{9\lambda^2}{12} + \frac{90\lambda^4}{360} = \frac{3\lambda^2}{4} + \frac{\lambda^4}{4}$$ $$4 = 3\lambda^2+\lambda^4$$ $$(\lambda^2+4)(\lambda^2-1)=0$$ Since $\lambda$ must be real and positive, we have $\lambda=1$. From that, get the mean and variance and the probabilities from the usual formulas.