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

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 .

2019 AMC 10B Problem 21

Answer: B

Difficulty rating: 1660

Solution:

Problem 21 in Other Years