2016 AMC 12A Problems/Problem 19

Revision as of 21:46, 5 February 2016 by FractalMathHistory (talk | contribs) (Solution)

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 $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$.

For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$.

For $5$ heads, we either start of with $4$ heads, which gives us $4\textbf{C}1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $HHHHHTTT$ and $HHHHTHTT$.

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{\textbf{(B) }151}.

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions

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