# Difference between C(5,i)/C(20,3) and (C(5,i)*C(20-5,3-i))/C(20,3)

1. Apr 17, 2013

### s0ft

The question goes this way:
Five men in a group of 20 are graduates. If three men are chosen out of 20 at random, what is the probability of at least one being graduates?
This can be done by adding the probabilities for exactly 1 graduate being included in the group, exactly 2 and exactly 3 like this :
P = P(1) + P(2) + P(3)
= C(5,1)*C(15,2)/C(20,3) + C(5,2)*C(15,1)/C(20,3) + C(5,3)*C(15,0)/C(20,3)
Now, the numerators of these terms give the total number of favourable cases for P(1), P(2) and P(3). But if I only put the primary criterion i.e. number of graduates to be chosen from total i.e. C(5,1), C(5,2) and C(5,3) and remove the multiplicative terms for non graduates combinations i.e. C(15,2), C(15,1) and C(15,0) what would the probability now give?
At first, I couldn't figure that I had to account for all possible favourable cases for each term and I did exactly what I wrote above. I put only the C(5,1) and so on without any thoughts for the second part, C(15,2)... After I found something was wrong, I managed to get to the right solution. But the question in the above paragraph arose.

2. Apr 17, 2013

### tiny-tim

hi s0ft!
nothing that has anything to do with the given question

(btw, it would have been a lot quicker to find the probability of no graduates, and then subtract from 1 )