A study is being conducted in which the health of two independent groups of ten
policyholders is being monitored over a one-year period of time. Individual participants
in the study drop out before the end of the study with probability $0.2$ (independently of
the other participants).
Calculate the probability that at least nine participants complete the study in one of the
two groups, but not in both groups.
I set up a binomial model for each group where $X_i \sim B(10, 0.8)$ and then calculated $P[X_1 >= 9] + P[X_2 >= 9] – P[X_1 >=9] * P[X_2 >= 9]$, but it was incorrect. I thought this accounts for the given event, since I am subtracting off the probability of both groups having more than 9 completions. I'd appreciate it if someone tells me exactly why my method is wrong.
Best Answer
Let $~P~$ denote the probability that there are at least $~9~$ left, in a specific group.
Let $~Q = 1 - P.$
Then, the desired computation is
$$PQ + QP = 2PQ.$$
So, the problem reduces to computing $~P.$
$$P = ~\left[ ~\binom{10}{0}(0.2)^{0}(0.8)^{10 - 0}~\right] + ~\left[ ~\binom{10}{1}(0.2)^{1}(0.8)^{10 - 1}~\right].$$
$\underline{\text{Where you went wrong}}$
Your computation, using my variables, represents
$$2P - P^2.$$
What you actually want is
$$2PQ = 2P(1-P) = 2P - 2P^2.$$
Your answer, which is similar to an Inclusion-Exclusion computation counts the intersection twice, and then deducts it once. This isn't what you want, since you must exclude the intersection, not count it once.