Bunuel wrote:

From a group of 10 people, a committee of four people is to randomly formed. Is the probability greater than 50% that the committee will contain more women than men?

More women than men in 4-person committee means one of the following 2 cases:

WWWW

WWWM

(1) The group from which the committee is formed contains more women than men.

If there are 10 women in the group then the probability of getting more women will obviously be greater than 50% (100%). I

If there are 6 women and 4 men, then \(P(W>M)=\frac{C^4_6}{C^4_{10}}+\frac{C^3_6*C^1_4}{C^4_{10}}=\frac{15}{210}+\frac{20*4}{210}=\frac{95}{210}<\frac{1}{2}\).

Not sufficient.

(2) The ratio of women to men in the group from which the committee is formed is greater than or equal to 3:2 --> \(\frac{W}{M}\geq{\frac{3}{2}}\). We can get two different answers if we consider 9 women and 1 men and 6 women and 4 men. Not sufficient.

(1)+(2) Consider the same examples as above. Not sufficient.

Answer: E.

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