Binomial-like probability problem between two independent groups Question
A study is conducted to monitor the health of two independent groups in a year where there are 10 participants for each group. Each participant will quit from the study with probability 0.2, independent with the other. What is the probability at least 9 participants in one group and not both will follow the study until finish?
My thought the problem can be solved by using binomial distribution where $p=0.8$ and $q=1-p=0.2$. Let $X_1$ be the number of participants that will follow the study until finish in group 1 and $X_2$ be the number of participants that will follow the study until finish in group 2, then $$P[X_i\ge9]=\sum_{k=9}^{10}\binom{10}{k}0.8^k\cdot0.2^{10-k}\quad\text{and}\quad P[X_i<9]=\sum_{k=0}^{8}\binom{10}{k}0.8^k\cdot0.2^{10-k}$$
So, the probability of at least 9 participants in one group and not both will follow the study until finish is $$P[X_1\ge9]\cdot P[X_2<9]+P[X_2\ge9]\cdot P[X_1<9]$$
But I am not sure, is this correct? If not, what kind approach should I use to answer the problem? Any idea? Any help would be appreciated. Thanks in advance.
 A: More generally, let $X \sim \mathrm{Binomial}(n_x = 10, p_x = 0.2)$ count the number of drop outs in a single group of 10 participants.  Then let $Y \mid x \sim \mathrm{Binomial}(n_y = 2, p_y = \Pr[X \le x])$ count the number of successful groups, where the probability of success is defined as the probability that at most $x$ participants quit in the group.  We wish to determine $\Pr[Y = 1 \mid x = 1]$.  It is easy to compute $\Pr[X \le 1] = (0.8)^{10} + 10(0.2)(0.8)^9$, and then $\Pr[Y = 1 \mid x = 1] = \binom{2}{1}p_y (1-p_y).$
The flexibility of this model is evident:  for instance, we can see how to calculate the probability that in a study consisting of $n_y = 5$ groups of $n_x = 30$ participants each, at least $3$ are successful if "success" is defined as no more than $3$ participants in a group quit, where the individual probability of quitting is $p_x = 0.1$.
A: If I understand to your assignment, your aim is to compute the probability that at least 9 people in exactly one group will not quit. You ignored the fact that there are two groups and computed that at least 9 people in a group that you chose in advance will follow until finish.
The answer to your question is
$$P[X_1\ge9]\cdot P[X_2\le8]+P[X_2\ge9]\cdot P[X_1\le8]$$
where $P[X_i \geq 9]$ is computed exactly as you stated and $P[X_i \leq 8]= 1 - P[X \geq 9]$.
