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Jury

by crackgmat007 » Mon May 25, 2009 11:56 am
If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?
a. 24/91
b. 45/91
c. 2/3
d. 67/91
e. 84/91

OA - D pls explain
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by mikeCoolBoy » Mon May 25, 2009 12:16 pm
I'm sure that there is a faster way to solve the problem but one solution is through combinations.

the pool consists of 15 people, 10 men (2/3) and 5 women (1/3).
we have to calculate the probability that the jury will comprise at least 2/3 of men, in order words at least 8 men. This includes a jury with 8 men, 9 men and 10 men.

first ways to select 12 people among 15 = C(15,12) = 15x14x13/3x2 = 5x7x13.

ways to select 8 men among 10 and 4 women among 5 = C(10,8) x (5,4) = (10 x 9)/2 x 5 = 9 x 5 x 5

ways to select 9 men among 10 and 3 women among 5 = C(10,9) x C(5,3) = 10 x 10

ways to select 10 men among 10 and 2 women among 5 = C(10,10) x C(5,2) = 1 x 10

probability = desire possibilities / total possibilities =

((9 x 5 x 5) + (10 x 10) + (10))/5x7x13 = ( 9x5 + 2x10 +5)/7x13 = 67/91
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by DarkKnight » Mon May 25, 2009 1:33 pm
[(10C8*5C4)/15C12 ] + [(10C9*5C3)/15C12] + [(10C10*5C2)/15C12] = 67/91.

Hence D.
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