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However, out of 100 choose 3 ways of doing this, only one is what you want (the group with all three in it) | However, out of 100 choose 3 ways of doing this, only one is what you want (the group with all three in it) | ||
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| + | --[[User:Norlow|Norlow]] 00:33, 2 October 2008 (UTC) | ||
Latest revision as of 20:33, 1 October 2008
This question says that 100 people enter a contest and that different winners are selected at random for first, second and third prizes.
With 3 person winning a prize each. I am wondering if this if inclusion exclusion problem. Clearly, each person has a 1/100 chance of winning. So would this be using the formula of union of p(e1), p(e2) , p(e3)? I am quite confused to how to approach this problem.
Wooi-Chen Ng
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What I am thinking is, there are 6 combinations of 3 people each winning a prize. The first person has 1/100 chance, 2nd prize has 1/99 chance, third prize has 1/98 chance. Multiply them. 6*(3/100)*(2/99)*(1/98). I am not sure if this is right. Correct me if I am wrong.
Wooi-Chen Ng
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The problem is you are multiplying by 6 twice. There aren't 6 combinations of A winning first, B winning second, C winning third... only 1. There are 6 ways of A, B, C winning first, second, third in some order, and that's why you multiply by 6.
Think of it this way. You want to choose the 3 people ahead of time to win the 3 prizes, out of 100.
However, out of 100 choose 3 ways of doing this, only one is what you want (the group with all three in it)
--Norlow 00:33, 2 October 2008 (UTC)
