Question regarding basic probability. I'm looking for confirmation that what I for did this question is correct:
"If two people are randomly chosen from
a group of eight women and six men, what is the probability
that (a) both are women (b) one is a man and the other is a
woman?"
a) The probability that both are women: $\frac{8}{14}\times\frac{7}{13}\approx.31$
b) The probability that one is a man and the other is a woman: $\frac{6}{14}\times\frac{8}{13}\approx.26$
 A: The first one is correct.
However, to see the problem with part $b$ try to answer this question:
What is the probability that the first person is men and the second
one is a woman ?
A: There are two possibilities of getting one of each gender:
First "man" then "woman" or first "woman" then "man"
A: Unfortunately, no. Your method works in part (a), but fails in part (b).
For a more generally workable method, we could proceed as follows. There are $\binom{14}2=91$ ways to pick $2$ people from a group of $14$. For part (a), there are $\binom82=28$ ways to pick two people from a group of $8$, so the answer is $\frac{28}{91}=\frac4{13},$ as you calculated.
For part (b), though, there are $8$ ways to choose one woman out of $8$ and $6$ ways to choose one man out of $6$, so there are $48$ ways to choose one man and one woman. Thus, the probability is $\frac{48}{91},$ instead.
Basically, what you found is the probability that a man was picked first and then a woman. However, for our purposes, we could have picked a woman first, instead, so we need to multiply this probability by $2$ to get the correct answer.
