HeyArnold wrote:A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

A) 1/14
B) 1/7
C) 2/7
D) 3/7
E) 1/2

[spoiler]OA: D, my logic was to say P(Woman) and P(Woman) and P(man) and P(man) = 5/8 x 4/7 x 3/6 x 2/5 = 1/14, clearly a trap by this question, but what is my error? [/spoiler]

P(exactly n times) = P(one way) * total possible ways.

Let W = woman and M = man.

P(one way):
P(1st person is W) = 5/8. (8 people, 5 of them W.)
P(2nd person is W) = 4/7. (7 people left, 4 of them W.)
P(3rd person is M) = 3/6. (6 people left, 3 of them M.)
P(4th person is M) = 2/5. (5 people left, 2 of them M.)
Since we want all of these events to happen together, we multiply the fractions:
P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.

Total possible ways:
Any arrangement of WWMM will yield exactly 2 W and 2 M.
Thus, the result above must be multiplied by the number of ways to arrange the 4 elements WWMM.
Number ways to arrange WWMM = 4!/(2!2!) = 6.

Thus, P(exactly 2 W) = 1/14 * 6 = 3/7.

The correct answer is D.

HeyArnold, you correctly determined the probability that the first 2 people chosen are women that the last 2 people chosen are men. But this is only ONE WAY to choose exactly 2 women. You need to account for ALL the ways in which exactly 2 women could be chosen, as shown in my solution above.

**** yeah! I got a probability question right!

Total possibilities of selecting four person teams out of a group of eight available people: 8C4 = 70

Total possibilities of selecting one of three men: 3C1 = 3

Total possibilies of selecting two women out of five available: 5C2 = 10

Now we will have Men*Women/Total

Men*Women: 3*10=30

30/70 = 3/7

Answer: D

@ GMAT GURU NY - Can you please explain the statement

Total possible ways:
Any arrangement of WWMM will yield exactly 2 W and 2 M.
Thus, the result above must be multiplied by the number of ways to arrange the 4 elements WWMM.
Number ways to arrange WWMM = 4!/(2!2!) = 6.

Another simple solution could be:
P(2W out of 5W AND 2M out of 3M)/P(4 members out of 5W+3M)
=5C2*3C2/8C4
=10*3/70
=3/7

Hence ans is D

factor26 wrote:@ GMAT GURU NY - Can you please explain the statement

Total possible ways:
Any arrangement of WWMM will yield exactly 2 W and 2 M.
Thus, the result above must be multiplied by the number of ways to arrange the 4 elements WWMM.
Number ways to arrange WWMM = 4!/(2!2!) = 6.

P(WWMM) = 1/14.

But WWMM is only ONE way to select exactly 2 women.
Now we have to account for ALL of the ways to select exactly 2 women.
Any arrangement of the letters WWMM will yield exactly 2 women and 2 men.
Thus, the result above (1/14) must be multiplied by the number of ways to arrange the letters WWMM.

The number of ways to arrange 4 elements = 4! = 4*3*2*1 = 24.

But WWMM includes IDENTICAL elements -- WW and MM -- which decrease the number of UNIQUE arrangements.
Why is there a decrease in the number of unique arrangements?
Because when the identical elements swap positions, the arrangement DOESN'T CHANGE.

To account for the duplicate arrangements, the result above (4! = 24) must be divided by the number of ways to arrange each set of duplicates.
The number of ways to arrange the two W's = 2! = 2*1 = 2.
The number of ways to arrange the two M's = 2! = 2*1 = 2.
Thus, the total number of ways to arrange WWMM = 4!/(2!2!)= 6.

Putting it all together:
P(exactly 2 women) = 1/14 * 6 = 3/7.

3M 5W
need 2m 2w

3C2*5C2/8C4=3/7

2nd approach (lengthy one)-

((3*2/2)*(5*4)/2)/(8*7*6*5/4!)=3/7

In this question, one need not worry about how you are arranging the committee but how you are selecting the committee. Exactly 2 women can be selected in 5C2 ways and then you have to have 2 men in the committee who can be selected in 3C2 ways and the no of ways the committee can be selected in 5C2*3C2
The no of ways a committee can be selected in 8C4. P = 5C2*3C2/8C4 = 3/7