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A club has exactly 3 men and 7 women as members. If two members are se

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A club has exactly 3 men and 7 women as members. If two members are se  [#permalink]

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New post 22 Jul 2018, 20:32
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  25% (medium)

Question Stats:

84% (01:50) correct 16% (02:01) wrong based on 53 sessions

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A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)

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A club has exactly 3 men and 7 women as members. If two members are se  [#permalink]

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New post 22 Jul 2018, 20:40
Bunuel wrote:
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)



Best way is to find P of none and then subtract from 1
So 1st man can be selected with a P of 3/10
Next would be 2/9
Thus both as man = 3/10*2/9=6/90=1/15
So prob both are not man, which is same as at least one woman =1-1/15=14/15
A

Of course the lengthy process would be..
1) first is woman and second man = 7/10*3/9=7/30
2) first is man and second is woman=3/10*7/9=7/30
3) both are women= 7/10*6/9=14/30
Total = 7/30+7/30+14/30=(7+7+14)/30=28/30=14/15
A
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A club has exactly 3 men and 7 women as members. If two members are se   [#permalink] 22 Jul 2018, 20:40
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