Binomial Distribution with Conditional Probability I am currently going through some sample problems in a book I have and I am unsure on how to approach a solution.
I am given the following.

The probability that a person will be helped by a certain medicine is .90. A doctor will be seeing 12 patients on this day.

Based on that data I can conclude that p=.9 and q=.1
The following question is based on the given data.

Find P(A | B) in which P(The medicine helped the first person | The medicine helped exactly 11 people)

The first step I did was
$$
P(B)=\binom{12}{11}.9^{11}(.1)^{1}\approx 0.3766
$$
By using the binomial distribution formula, I find that P(B) is approximately 0.3766. At this point I am stuck. I believe I am to use Bayes' Theorem at this point, but I am not sure how to find the answer with it.
 A: The question has nothing to do with the binomial distribution. It is enough to know that all people look alike under the binomial distribution. If all you know is that exactly 11 out of 12 people were helped, then the set of people helped is a random subset of 11 people. The probability that any given person is in that set is $11/12$.
This is probably not the way you were intended to solve the exercise, which is a pity. What they wanted you to do was to calculate $P(B)$ and $P(A \cap B)$. You already calculated $P(B) = \binom{12}{11} p^{11} q = 12 p^{11}q$. As for $P(A \cap B)$, the first person is helped with probability $p$, and as for the other 11, we know that exactly 10 were helped, so $P(A \cap B) = p \cdot \binom{11}{10} p^{10} q = \binom{11}{10} p^{11} q = 11 p^{11} q$. Dividing, we obtain $P(A|B) = P(A \cap B)/P(B) = 11/12$.
The situation becomes more interesting if you know that at least 11 people were helped. If $p_{11},p_{12}$ are the probabilities that exactly 11 or 12 were helped (which you can calculate using the formula for binomial distribution) then conditioned on the information, you know that with probability $p_{11}/(p_{11}+p_{12})$ exactly 11 people were helped, and with probability $p_{12}/(p_{11}+p_{12})$ exactly 12, so the probability that a particular person were helped is $(11/12) p_{11}/(p_{11}+p_{12}) + (12/12) p_{12}/(p_{11}+p_{12})$.
