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.
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