Auniversity received 9 applications for three postdoctorate fellowships. five of the applicants are men and four are women. find these probabilities: 1) all 3 selected are men, 2) all 3 selected are women, 3) two men and one woman are selected, 4) two women and one man are selected.

There are ways to select 3 from 9.

There are ways to select 3 men from 5 men and ways to select 3 women from 4 women.

If all three selected are men, then no woman was selected, and we have

ways we can do that.

1)

2) If all 3 are women then no man was selected, and we have

ways we can do that.

3) If 2 men and 1 woman were selected, the we have ways we can do that.

4) If 2 women and 1 man is selected, then we have  

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

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: