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Probability in Group

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by gmatNooB8787 » Thu May 03, 2012 2:35 pm
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?
1/14
1/7
2/7
3/7
1/2

solution : total people can be selected in 8C4 ways => 70
The 3 women can be selected in : 5C3 ways => 10
The 2 men can be selected in : 3C2 ways => 3
Hence total ways of selecting : 3*10 = 30

Hence Probability : 30/70 = 3/7.


Is there a different way to solving this problem ? Can anyone tell me what is the best way to solve this problem ?
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Source: — Problem Solving |

by Bill@VeritasPrep » Thu May 03, 2012 6:36 pm
Two spots for men: 3 choices for the first spot, 2 for the second. 3*2=6, but remember to divide by 2 (since AB and BA are both included) for a total of 3

Two spots for men: 5 choices for the first spot, 4 for the second. 5*4=20, divide by 2 for a total of 10.

3*10=30 groups of 2 men and 2 women

Four spots: 8 choices, 7 choices, 6 choices, 5 choices. Divide by 4!=24, which means you can do 8*7*6*5/24 = 48*7*5/24 = 2*7*5=70

30/70 = 3/7

Not necessarily any faster, though.
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