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

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SUMMARY

The discussion centers on calculating the probability of selecting at least one graduate from a group of 20 men, where 5 are graduates. The correct approach involves using combinations, specifically C(5,i) for graduates and C(15,3-i) for non-graduates, to find the probabilities for selecting 1, 2, or 3 graduates. The alternative method of only using C(5,i) without considering the non-graduate combinations leads to incorrect results. A more efficient method suggested is to calculate the probability of selecting no graduates and subtracting that from 1.

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s0ft
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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.
 
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hi s0ft! :smile:
s0ft said:
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?

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?

nothing that has anything to do with the given question :confused:

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

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