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A committee of 10 members consists of men and women. If two members of the committee are to be selected at random, then what is the least number of women that the committee can have such that the probability that both members selected will be women is at least 0.2?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
03 Jan 2020, 22:11
Kudos
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Let the number of women = W
In selection of 2 members, the probability that both members selected will be women = WC2/10C2
W(W-1)/(10*9)>= 0.2
W(W-1)>=18
Clearly, W= 4 gives W(W-1)= 12
W= 5, gives W(W-1)= 20
hence, the min number of Women = 5
Answer C
In selection of 2 members, the probability that both members selected will be women = WC2/10C2
W(W-1)/(10*9)>= 0.2
W(W-1)>=18
Clearly, W= 4 gives W(W-1)= 12
W= 5, gives W(W-1)= 20
hence, the min number of Women = 5
Answer C
sjuniv32
A committee of 10 members consists of men and women. If two members of the committee are to be selected at random, then what is the least number of women that the committee can have such that the probability that both members selected will be women is at least 0.2?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
03 Jan 2020, 22:17
Kudos
Bookmarks
sjuniv32
A committee of 10 members consists of men and women. If two members of the committee are to be selected at random, then what is the least number of women that the committee can have such that the probability that both members selected will be women is at least 0.2?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Say there are x women, so ways to choose two women out of x = x(x-1), and ways to choose two out of 10 = 10*9=90
Hence probability that both are women = \(\frac{x(x-1)}{90}\geq{0.2}......x(x-1)\geq{18}\)
So, we look for product of TWO consecutive number such that there product is at least 18..
3*4=12<18....4*5=20>18...Hence x will be at east 5
C
