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?

$\dfrac{1}{36}$

$\dfrac{1}{24}$

$\dfrac{1}{18}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

Solution:

If we flip heads first, then the only way to see the second tails before the next heads is with $HTT,$ which ends without two heads.

Therefore, we must start with tails. Then we need a heads to ensure that we don't end on two tails, and then another tails to ensure a second tails is seen.

This means we must start as $THT.$

Our sequence must also be alternating except the last two coins, or else there would be two consecutive flips that are the same causing it to stop. This means our sequence must be $THTH \cdots THH.$ This means that there is exactly one sequence of size $n$ for every odd $n$ at least $5,$ each with a probability of $\left(\frac 12\right) ^n.$ Thus, the total probability is $\frac 12^5 + \frac 12^7 + \cdots$ $= \frac 12^5\left( 1+ \frac 14 + \frac 14^2 \cdots \right)$ $= \dfrac 1{32} \cdot \dfrac{1}{1-\frac 14}$ $=\dfrac 1{32} \cdot \dfrac 43$ $= \dfrac 1{24}.$

Thus, the answer is B .

2019 AMC 10B Problem 21

Solution:

Problem 21 in Other Years