j_shreyans 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
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 the letters WWMM represents one way to get exactly 2 W's and 2 M's.
Thus, to account for ALL OF THE WAYS to get exactly 2 W's and 2 M's, the result above must be multiplied by the total number of ways to arrange the letters WWMM.
Number of ways to arrange 4 distinct elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the 2 identical W's and by another 2! to account for the 2 identical M's:
4!/(2!2!) = 6.
Thus, P(exactly 2 W) = 1/14 * 6 = 3/7.
The correct answer is
D.
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