Question

what is the probability that the committee of 7 persons from group of 6 woman and 5 men has at least 4 woman?

Answer 1

Answer)

As order here do not matters we can use combinations to find favorable and total outcomes

Ncr = n!/(r!*(n-r)!)

Where n! = n*n-1*n-2*n-3*n-4*n-5*....till 1

For example

5! = 5*4! = 5*4*3! = 5*4*3*2! = 5*4*3*2*1

We have 6 women and 5 men = 11

We need to choose 7 out of them

11c7 = 330

Favorable = at least 4 women

So here we have 4 cases

4, 5, 6, or 7 women

First case is that we have 4 women then we would have 3 men

6c4*5c3 = 150

Second case is that we have 5 women and 2 men

= 6c5*5c2

= 60

Third case is that we have 6 women and 1 men

= 6c6*5c1

= 5

Now we cannot have 7 women as total women is 6

So favorable = 5+60+150 = 215

Probability = favorable/total = 215/330 = 0.65151515151