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#1
12-03-2005, 11:00 AM
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5 people choose a digit from 0-9
What is the probability that at least 2 people choose the same digit?
Can I use the backdoor approach like ... 1- [9/10*8/10*7/10*6/10] = 69.76% ... or is this the inclusion-exclusion problem and if so, would someone please give a simple explanation. Thanks for your answer. 
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#2
12-03-2005, 11:08 AM
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Re: 5 people choose a digit from 0-9
[ QUOTE ]
What is the probability that at least 2 people choose the same digit? Can I use the backdoor approach like ... 1- [9/10*8/10*7/10*6/10] = 69.76% ... or is this the inclusion-exclusion problem and if so, would someone please give a simple explanation. Thanks for your answer. [/ QUOTE ] You did it right. It's the birthday problem. 1 - P(365,n)/365^n, or in your case, 1 - P(10,5)/10^5 = 1 - P(9,4)/10^4, which is what you wrote. P(n,k) = n!/(n-k)! = C(n,k)*k!.
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