2019 AMC 10B Problems/Problem 21

Revision as of 17:23, 14 February 2019 by Will3145 (talk | contribs) (Solution)


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?

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


We first want to find out the sequences of coin flips that satisfy this equation. Since Debra sees two tails before two heads, her first flip can't be heads, as that would mean she would either end at tails or see two heads before she sees two tails. Therefore, her first flip must be tails. If you calculate the shortest way she can get two heads in a row and see two tails before she sees two heads, it would be T H T H H, which would be $\frac{1}{2^5}$, or $\frac{1}{32}$. Following this, she can prolong her coin flipping by adding an extra (T H), which is an extra $frac{1}{4}$ chance. Since she can do this indefinitely, this is an infinite geometric sequence, which means the answer is $\frac{\frac{1}{32}}{1-\frac{1}{4}}$ or $\boxed{B) \frac{1}{24}}$. -chen1046, LaTeX edit by will3145

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions

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