Question

A jar contains n chips. suppose a boy successively draws a chip from the jar, each time replacing the one drawn before drawing another. the process continues until the boy draws a chip that he has previously drawn. let x denote the number of draws, and compute its probability mass.

Answer 1

Well obviously you will never finish in one draw so f(1)=0 f(2) is the odds of getting it in 2 draws, this is 1/n because you have only 1 chip that you have previously drawn.

Now to finish on the kth draw you would have had to fail to finish on all previous draws and the odds of this is [(n-1)/n]*[(n-2)/n]*...*[(n-k+2)/n] and the odds of you getting on of the k-1 previous chips is (k-1)/n 

Multiplying these together we get,

f(k)=(k-1)*(n-1)!/[(n-k+1)!*n^(k-1)] 

And,

Holds for k=1 to k=n+1, for all k>n+1 f(k)=0 because if you end up at the (n+1) turn then you have to have drawn all n chips at some point previously thus on this turn you are guaranteed to draw one of them and thus can not proceed past this turn. 

So, to sum it all up if f(k) is the pmf then we have, 

If,

k>=1 and k<=n+1

Then,

f(k)=(k-1)*(n-1)!/(n-k+1)!*n^(k-1)