Question

In tossing a fair coin, what is

the probability that the second
"heads” occurs on the ninth toss?​

Answer 1

1/64

Explanation:

We want a second H in the 9th toss. So, we must get only 1 H in the first eight tosses.

So, in the first 8 tosses, we have 1 H and 7 T. Number of ways of getting 1 H and 7 T is 8!/7! = 8

Total possible outcomes = 2^9 = 512

Required probability = 8/512 = 1/64

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