An unbiased coin is tossed until a head appears and then tossed until a tail appears. If the tosses are independent, what is the probability that a total of exactly n tosses will be required?

Passafaromx

Passafaromx

Answered question

2022-08-12

Geometric probabilities solution verification
An unbiased coin is tossed until a head appears and then tossed until a tail appears. If the tosses are independent, what is the probability that a total of exactly n tosses will be required?
My attempt:
P(n tosses required to produce one head and one tail) = P ( x tosses needed for first head ) × P ( y tosses needed for first tail ) where   x + y = n .
So, the probability becomes ( 1 2 ) x 1 ( 1 2 ) ( 1 2 ) y 1 ( 1 2 ) = ( 1 2 ) n .
This is not the correct answer, however. The correct answer is ( 1 2 ) n ( n 1 ). Can someone pleas explain what I did incorrectly and where the n 1 factor is coming from?

Answer & Explanation

Riya Cline

Riya Cline

Beginner2022-08-13Added 17 answers

Step 1
Your thinking is almost there - except in your argument, x could be anything from 1 to n 1 (since getting a head on xth toss and then getting a tail on the y = n x th toss gives you a head and tail on exactly the nth toss). So if we sum your expression over all possible combinations of x and y = n x, we have n 1 times the expression you got.
Step 2
More generally, this is an example of (a variant of) the Negative Binomial distribution, which counts the number of successes needed for r events to happen.

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