(New page: 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...) |
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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. | 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. | ||
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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*(1/100)*(1/99)*(1/98). I am not sure if this is right. Correct me if I am wrong. | ||
Wooi-Chen Ng | Wooi-Chen Ng | ||
Revision as of 18:20, 28 September 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*(1/100)*(1/99)*(1/98). I am not sure if this is right. Correct me if I am wrong.
Wooi-Chen Ng
