A committee of 9 persons is chosen from 10 men and 6 women?

Question

find the probability that the committee has
i)5 men and 4 women
ii) at least 1 woman

Answer 1

1)
number of ways of choosing 5 men from 10 is 10C5
number of ways of choosing 4 women from 6 is 6C4
number of ways of choosing 9 persons from 16 is 16C9

Prob that committee consists of 5 men and 4 women is...
( 10C5 * 6C4) / 16C9...

2)
To find the probability that the committee consists at least 1 women is equal to (1 - the probability that the committee has no woman)...

Probability that the committee has no woman is...
(10C9 * 6C0) / 16C9...
10C9 is number of ways 9 men from 10 can be selected...
6C0 is the number of ways 0 women from 6 can be selected..

subtract the above prob from total probability 1 and that will give you the probability that the committee consists at least 1 woman...

Answer 2

total number of people is 16
(a) (10 C 5 * 6 C 4) / (16 C 9)
= (252*15) / (11440)
=0.3304
therefor there is an approx 33% chance of this event occuring

(b) if there is at least one woman that means that there cant be zero women. the easiest way to do this is calculate the probability that there will be zero woman and then calculate P'(A) ... { P(A) + P'(A) = 1 }

( 10 C 9 * 6 C 0 ) / (16 C 9)
=10 / 11440
= 8.74x10^-4

therefore P'(A) = 1 - 8.74x10^-4
=0.9991
therefore there is a 99.91% chance that at least woman will be chosen

sry read it as nine first

Answer 3

To find answers for questions that ask for "at least 1 something", you can do by cases - what's it for exactly 1, exactly 2, .. exactly max number and add the prob. Generally it's simpler to find out what's the P(0), prob for 0, and do 1-P(0).

So you have 16C9 ways of choosing 9 people out of 16.
If you only have men, then this drops to 10C9 = 10 ways.
So the prob of getting no women is 10C9/16C9
and hence the prob of getting at least 1 woman is 1 minus this.

(nCm = n!/(m!*(n-m)!), eg 6C2 = 15

For 1) you find thenumber of ways of picking 5 men out of the10 - 10C5 and 4 women from 9, ie., 9C4.
If you multiply these, you get all the combinations of 5 men and 4 women. Again, divide by the total # of 9 person combinations and you get the prob. for i