# Conditional Probability Given N=n

suppose that $N$ is a Poisson$(μ)$ random variable. Given $N=n$, random variables $X_1,X_2,X_3,\cdots,X_n$ are independent with uniform∼$(0,1)$ distribution. So there are a random number of $X$'s.

(a) Given $N=n$ what is the probability that all the $X$'s are less than $t$?

So I set up the problem as: $P(X<t\mid N=n)=\frac{P(X<t,N=n)}{P(N=n)}$

How do I compute this if it's correct, or what do I do next?

(b) What is the (unconditional) probability that all the $X$'s are less than $t$?

No idea how to start this one.

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If $0\le t \le 1$, and there are $n$ random variables $X_i$, the probability they are all $\lt t$ is, by independence, equal to $t^n$.
For the unconditional probability, note that $\Pr(N=n)=e^{-\mu} \frac{\mu^n}{n!}$. So the probability that all the $X_i$ are $\lt t$ is $$\sum_0^\infty e^{-\mu}\frac{\mu^n}{n!} t^n.$$ Since $\mu^n t^n=(\mu t)^n$, the above sum simplifies to $$e^{-\mu}e^{\mu t},$$ which can be "simplified" to $e^{-\mu(1-t)}$. For completeness, one should observe that the probability is $0$ if $t\lt 0$, and $1$ if $t\gt 1$.
Wouldn't $\sum_{n=1}^{\infty} e^{- \mu} \frac {\mu^{n}}{n!} {t^{n}}$ simplify to $e^{-\mu} {e^{\mu t}}$? –  WayMoreQuestionsThanAnswers May 2 '13 at 9:53
Actually, for $t \leq 0$, the probability is $e^{-\mu}$, because that is the probability that N will be zero, and the condition is trivially satisfied regardless of $t$. –  user3294068 Apr 23 '14 at 19:22