5 people choose a digit from 0-9

[SwissPoker]

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.

[BruceZ]

[ 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!.