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 hears, 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 is $a/b$ , where $a$ and $b$ are relatively prime positive integers. What is $a+b$ ? (For example, he succeeds if his sequence of tosses is $HTHHHHHH$ .)

TODO: Answer choices

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 just sum to get $\boxed{46}$ .

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2016 AMC 12A Problems/Problem 19

Problem

Solution