Answer:
We have that 5 are man and 4 are woman.
1) the probabilty of selecting a man at random is equal to the number of men divided by the total number of aplicants, this would be:
p1 = 5/9 for the first one.
now, for the second, the number of men is 4, and the number of aplicants is 8 (because we already tooked one)
p2 = 4/8
and for the third one we have:
p3 = 3/7
then the total probability is p1*p2*p3 = (5*4*3)/(9*8*7) = 0.119
2) now, if all are woman we have the same thinking:
p1 = 4/9
p2 = 3/8
p3 = 2/7
P = p1*p2*p3 = (4*3*2)/(9*8*7) = 0.048
2) now we have two men and one woman.
if the selection is: man-man-woman.
p1 = 5/9
p2 = 4/8
p3 = 4/7
P1 = p1*p2*p3 = 0.159
Now, we also have the case where the selection is man-woman-man
p1 =5/9
p2 = 4/8
p3 = 4/7
P2 = p1*p2*p3 = 0.159
and the case where the selection is woman-man-man
p1 = 4/9
p2 = 5/8
p3 = 4/7
P3 = p1*p2*p3 = 0.159
you can see that the probabilitys are the same in every disposition, then the total probability here is P1 + P2 + P3 = 3*0.159 = 0.477
4) now the case where we have two women and one man, now we know that the order does not matter, so we suppose that the selection is woman-woman-man
p1 = 4/9
p2 = 3/8
p3 = 5/7
P1 = 0.119
and we have 3 permutations, so the total probability is:
P = 3*P1 = 0.357