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Jury

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Senior Manager
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26 May 2010, 01:04
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Can someone try to solve this?
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26 May 2010, 03:17

2/3 of 15 is 10Men
1/3 of 15 is 5Women

Jury consist of 10 Men and 5Women

We need to choose 12 from 15 - 15C12.

However of the 12, [highlight]"atleast "[/highlight] 2/3 should be men which means we should have atleast 8 or more men. This leads to the following selections

8m 4w
9m 3w
10m 2w

so the prob is (10c8*5c4 + 10c9*5c3 + 10c10*5c2)/15c12 ===> 67/91
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07 Jun 2010, 12:55
Why is (10C8 * 7C4)/15C12 incorrect?

10C8 should give us the no: of groups of 8 men and 7C4 should give us the groups of the remaining 4?
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08 Jun 2010, 07:39
dimitri92 wrote:
Can someone try to solve this?

2/3 men out of 15 = 10 men
1/3 women = 5

now if we need to have have at least 2/3 men in 12 member jury, that means we need to have at least 8 men,

Now the $$P (of having at least 2/3 men in 12 member jury)= 1 - P ( not having 2/3 men in the jury)$$

now, there is only one possiblity when we cant have 8 men in the 12 member jury i.e when all the women are selected. ( 5 women and 7 men)

So the $$probablity of having 5 women and 7 men = \frac{5C_5 * 10C_7}{15C_12}$$
$$15C_12$$ = total ways of choosing 12 members out of 15
$$P (of having at least 2/3 men in 12 member jury) = 1 - \frac{10*9*8*3*2}{3*2*15*14*13} = 1- \frac{24}{91}$$ $$=\frac{67}{91}$$
Re: Jury   [#permalink] 08 Jun 2010, 07:39
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