Conditional probability of exponential random variable This question comes directly from a chapter in Gut's "Intermediate Probability" that focuses on conditional probability. I'm using this problem as more practice solving conditional probability questions.
The life $T$ (hours) of the bulb in an overhead project follows an exponential($10$)-distribution. During a normal week, it is used a Poisson($12$)-distributed number of lectures lasting exactly one hour each. Find the probability that a projector with a newly installed bulb functions throughout a normal week (without replacing the bulb).
The answer is given as $\exp \{-12(1-e^{-1/10})\} $, but I'm unsure how to arrive at it. My attempt was:
Let $T$ represent the lifetime of the bulb, and let $N$ represent the number of lectures given in a week.
$$\mathbb{P}(T > n | N=n) = \sum_{n=0}^\infty \Bigg \{ \int_{t=n}^\infty f_T(t) f_N(n) dt \Bigg \} $$
but this doesn't get me much, and I'm not even sure if I could solve the above equation.
I would appreciate any hints or suggestions for how to arrive at the given answer.
 A: The idea is that if the bulb lasts for an hour on a given day that it is used, then it survives to the next time it is used, and each time it is used, it behaves as if it were new, because its lifetime is exponentially distributed.  So we first need to calculate the probability of this survival:  this is simply $$p = \Pr[T > 1] = e^{-1/10}.$$  Now, if $N \sim \operatorname{Poisson}(\lambda = 12)$ is the random number of times it is used in a week, $X \mid N \sim \operatorname{Binomial}(N,1-p)$ counts the number of lectures that the bulb fails in a week.  We want the marginal (unconditional) probability $\Pr[X = 0]$.  To this end, we note $$\Pr[X = k] = \sum_{n=0}^\infty \Pr[X = k \mid N = n]\Pr[N = n]$$ by the law of total probability.  The conditional probability is binomial, and the second is Poisson, so $$\begin{align*} \Pr[X = 0] &= \sum_{n=0}^\infty \binom{n}{0} (1-p)^0 p^n e^{-12} \frac{12^n}{n!} \\ &= \sum_{n=0}^\infty e^{-12} \frac{(12p)^n}{n!} \\ &= e^{12p} e^{-12} \sum_{n=0}^\infty e^{-12p} \frac{(12p)^n}{n!} \\ &= e^{-12(1-p)} = e^{-12(1-e^{-1/10})}, \end{align*}$$ as claimed.
A: You need to find the probability that the life time $T$ of the bulb is more than the number of hours spent on lectures (which is equal to $N$, since each lecture is one hour), e.g. you need to find
$$P(T>N).$$
Here's one approach. If $N=n$ for some integer $n\geq 0$, then you also need $T>n$. I can thus rewrite the above probability as
$$P(T>N) = \sum_{n=0}^{\infty}P(T>n,N=n) = \sum_{n=0}^{\infty}P(T>n|N=n)P(N=n),$$
where the last equality uses the definition of conditional probability. There is another argument for this equality, but I don't know if you have covered it yet. I include it later in the post.
Continuing with the last sum, you know that $T$ and $N$ are independent, so $P(T>n|N=n) = P(T>n)$, which you know, since $T$ has an exponential distribution, and since $N$ is Poission, you also know the value of $P(N=n)$. Plugging it in, you get
$$\sum_{n=0}^{\infty} P(T>n|N=n)P(N=n) = \sum_{n=0}^{\infty} P(T>n)P(N=n)=\sum_{n=0}^{\infty}e^{-\frac{1}{10}n}e^{-12}\frac{12^n}{n!}$$
$$=e^{-12}\sum_{n=0}^{\infty}\frac{(e^{-\frac{1}{10}}12)^n}{n!}.$$
Recall that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ is true for any $x$, and thus the above expression can be rewritten as
$$e^{-12}\sum_{n=0}^{\infty}\frac{(e^{-\frac{1}{10}}12)^n}{n!} = e^{-12}e^{12\exp{(-1/10)}}=e^{-12(1-\exp(-1/10))},$$
which is what you are looking for.
Now, concerning my very first equality, you could also use the law of iterated expectations to conclude that
$$P(T>N) = \mathbb{E}[P(T>N|N)]=\sum_{n=0}^{\infty}P(T>n|N=n)P(N=n).$$
