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.
$$ \Pr(X=2) = 9\Pr(X=4) + 90\Pr(X=6) $$ $$ \frac{\lambda^2 e^{-\lambda}}{2!} = 9\frac{\lambda^4 e^{-\lambda}}{4!} + 90 \frac{\lambda^6 e^{-\lambda}}{6!} $$ 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.