If a fair coin is tossed thrice find the probability that there are
at most one tail

Explanation

Let `a_(n)` be the number of strings of H and T of length n with no two adjacent H's.
Then `a_(1) = 2` (H or T) and `a_(2) = 3` (HT, TH, T T)
Now consider `a_(n + 2)`.
Also, `a_(n + 2) = a_(n + 1) + a_(n)` (`because` The string must begin with T or HT)
So, `a_(3) = 5, a_(4) = 8, a_(5) = 8 + 5 = 13`.
Hence, the required probability = `(13)/(2^(5)) = (13)/(32)`