A box contains four tickets, numbered 0,1,1, and 2. Let Sn be the sum of the numbers obtained from n draws at random with replacement from the box.(adsbygoogle = window.adsbygoogle || []).push({});

Find the exact formula for P(Sn=k) (k=0,1,2....)

I started by finding the probability of the first few sums:

P(Sn=0) = (1/4)^n

P(Sn=1) = n * (1/4)^(n-1) * (1/2)

P(Sn=2) = n * (1/4)^(n-1)*(1/4) + (n choose 2) * (1/4)^(n-2)(1/2)^2

Then I get stuck, since the combinations of draws that lead to the subsequent draws get much more complicated.

Is there a more simplified way to find the formula for P(Sn=k)?

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# Find the exact formula for P(Sn=k)

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