# Stochastic Variables

There are $7$ people in a room. $4$ male & $3$ female. $2$ persons are chosen randomly. The probability of "doing something right" if $2$ females are chosen is $0.6\,$. When 2 males are chosen the probability of "doing something right" is $0.1$ and when a male and a female are chosen the probability of "doing something right"is $0.3\,$.

A) What's the Expected Value of the probability on successful attempts

B) What's the Expected Value of the probability on successful attempts if $1$ male is chosen certainly.

*forgive me if I'm not clear enough cause this is a translated problem.

There are $\binom{7}{2}$ equally likely ways to choose $2$ people. There are $\binom{3}{2}$ ways to choose two females, and $\binom{3}{1}\binom{4}{1}$ ways to choose one female and one male, and $\binom{4}{2}$ ways to choose two males.
Thus the (simplified) probability of two females is $\frac{1}{7}$, the probability of one of each is $\frac{4}{7}$, and the probability of two males is $\frac{2}{7}$.
A) The probability that things turn out well is therefore $$(0.6)(1/7)+(0.3)(4/7)+(0.1)(2/7).$$
B) Here I will be making an interpretation of the question. I assume that we are asked to find the probability that things turn out well, given that at least one male is chosen. Let $p$ be the probability that one of each is chosen, given that at least one male is chosen. By the usual conditional probability calculation, we have $p=\frac{4/7}{6/7}$, that is, $2/3$. So the probability two males are chosen, given that at least one is chosen, is $1/3$. Thus the probability of success, given that at least one male was chosen, is $$(0.3)(2/3)+(0.1)(1/3).$$
For example, suppose that we are told that the people were chosen one at a time, and that the first person chosen is male. Then given this information, the probability that one is male and one is female is $3/6$, not $2/3$. The calculation will also be different if we are told that one specific male, say Adam, was chosen.