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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*( | + | 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 | Wooi-Chen Ng | ||
Revision as of 17:23, 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*(3/100)*(2/99)*(1/98). I am not sure if this is right. Correct me if I am wrong.
Wooi-Chen Ng
