2019 AMC 10B Problem 21

Question

Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?

\dfrac{1}{36}

\dfrac{1}{24}

\dfrac{1}{18}

\dfrac{1}{12}

\dfrac{1}{6}

Answer: B

Concepts:basic probability geometric sequence

Difficulty rating: 1660

Small hint:

The sequence must alternate until the final matching pair

Big hint:

Sum the probabilities of the possible odd lengths

Solution:

Before the final repeated flip, the sequence must alternate. If it starts with

H

, the second head necessarily occurs before the second tail, so a successful sequence must start with

T

. To see a second tail before ending with

HH

, it must begin

THT

.

Thus the successful sequences are

THTHH,THTHTHH,\ldots

: exactly one sequence of each odd length at least

5

. Their total probability is

\begin{gathered} \frac1{2^5}+\frac1{2^7}+\cdots\\ =\frac1{32}\cdot\frac1{1-\frac14}\\ =\frac1{24}. \end{gathered}

Thus, the answer is B .