2019 AMC 10B Problems/Problem 21

Problem

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?

$\textbf{(A) } \frac{1}{36} \qquad \textbf{(B) } \frac{1}{24} \qquad \textbf{(C) } \frac{1}{18} \qquad \textbf{(D) } \frac{1}{12} \qquad \textbf{(E) } \frac{1}{6}$

Solution

We firstly want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the seecond head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is $THTHH$ , which has a probability of $\frac{1}{2^5} = \frac{1}{32}$ . Furthermore, she can prolong her coin flipping by adding an extra $TH$ , which itself has a probability of $\frac{1}{2^2} = \frac{1}{4}$ . Since she can do this indefinitely, this gives an infinite geometric series, which means the answer (by the geometric series sum formula) is $\frac{\frac{1}{32}}{1-\frac{1}{4}} = \boxed{\textbf{(B) }\frac{1}{24}}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2019 AMC 10B Problems/Problem 21

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