2016 AMC 12A Problems/Problem 19

Problem

Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$ )

$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution

For 6-8 heads, we are guaranteed to hit 4 heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}$ For 4 heads, you have to hit the 4 heads at the start so there's only one way For 5 heads, we either start of with 4 heads, which gives us 4 ways to arrange the other flips, or we start off with five heads and one tail, which also has four ways (ignoring the overlap with the case of 4 heads to start).

Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of 8 coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{151}=\boxed{B}$ .

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2016 AMC 12A Problems/Problem 19

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