Question

When you flip a biased coin the probability of getting a tail is 0.25.
How many times would you expect to get tails if you flip the coin 180 times?

Answer 1

Answer :-

Here's a way to solve it without using recursion.

Let Pk denote the probability of satisfying the above condition with k heads in n tosses.

P0=(n0)(.75)n This can be seen as placing 0 heads in a row of n tails.

P1=(n1)(0.75)(n−1)∗(0.25) This can be seen as placing 1 heads in a row of n-1 tails (so n positions for a head to be placed in).

P2=(n−12)(0.75)(n−2)∗(0.25)2 This can be seen as placing 2 heads in a row of n-2 tails such that no two heads are together (so n-1 positions available for heads to be placed in).

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Pk=(n−k+1k)(0.75)(n−k)∗(0.25)k

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It can be seen that for k>n/2,Pk=0 as it would not be feasible to place heads which are greater in number than tails + 1 in total.

The summation of all probabilities till Pk should be the answer.

I would only be glad if you pointed out some flaws in the above argument.