From a group of 10 people, a committee of four people is to

Bunuel,

I am not clear that we you added 6C4 + 6C3x 4C1 in statement 1

P(W>M) = P(WWWW) + P(WWWM).

If there are 6 women and 4 men, then, then \(P(WWWW)=\frac{C^4_6}{C^4_{10}}\) and \(P(WWWM)=\frac{C^3_6*C^1_4}{C^4_{10}}\).

Hope it's clear.

Hi Bunuel,

I have one question, are n't both statements saying same thing ?

Statement 1: more women than men => w >= 6 and m <= 4 =>
if w = 6, m = 4, ratio w/m = 3/2
if w = 7, m = 3, ratio w/m = 7/3
ratio w/m >= 3/2

Statement 2: is exactly saying w/m >= 3/2

but answer is definitely (E) as we will get diff probabilities with cases w = 6, m = 4 and w = 7, m = 3 .....

Please clarify

Thanks

Quick and easy solution without any calculation.

If you have 3 women in the committee of 4 people, ratio of women to men is 3:1.

So, in order to have the probability of more women than men in the committee greater than 0.5, total number of women must be greater than 3/4, i.e. 7.5 out of 10. So, we need at least 8 women in the total population to get the probability higher than 0.5.


Statement-1:
The number of women is higher than number men in the total population.
No. of women > 5.
No. of women may or may not be greater than or equal to 8.

So, statement 1 is not sufficient.

Statement-2:
The number of women is 6 or higher in the total population.
No. of women may or may not be greater than or equal to 8.

So, statement 2 is not sufficient.

Combining both statements:
Even after combining both statements, no. of women may or may not be greater than or equal to 8.

So, ans is E.

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