vaccine success CDF A group of people is vaccinated twice (with two different vaccines). The probability of the first vaccine not working is 0.08, and the probability that the second vaccine doesn't work for the person for whom the first didn't work is 0.2. The probability of the second vaccine not working for the person for whom the first vaccine worked is 0.05. If we consider two people, find the CDF for the following random variables:
X - number of people for whom the first vaccine doesn't work
Y - number of people for whom the second vaccine doesn't work
This is seriously confusing, I don't even know where to start, or how to find these CDFs. Any help is appreciated
 A: X is easy, as the first one working or not does not depend on the second one at all. Y is a little harder: You can divide it up into two parts:


*

*The first one works, but the second one doesn't.

*Both don't work.


Now you can compute the probabilities of these two, add them up and you have Y.
A: For computing the CDFs, all you need to know are the probabilities $P(X=m)
$ and $P(Y=n)$ for nonnegative integers $m$ and $n$. I assume that the vaccination of one person is an event independent of the vaccination of another, and also hope you have been able to compute $P(X=m)$ straightforwardly.
For the second part, you may do the following: let's say we want to compute $P(Y=n)$. Then, among these $n$ people, for a certain number $m$ of them, where $0\leq m\leq n$, the first vaccine might not have worked. So, we have to first compute $P(Y=n|X=m)$, and thereafter, it is obvious by the law of total probability that
\begin{equation}
P(Y=n)=\sum\limits_{m=0}^{n} P(Y=n|X=m)\cdot P(X=m).
\end{equation}
Now, $P(Y=n|X=m)=\binom{n}{m}(0.2)^{m}(0.05)^{n-m}$ since there are $m$ people for which the first vaccine has not worked and the probability of the second vaccine not working for these people is $0.2$; similarly, there are $n-m$ people for whom the first vaccine has worked, and for these people, the second vaccine will not work with probability $0.05$. Finally, there are $\binom{n}{m}$ ways of choosing $m$ people out of $n$.
