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Difference between revisions of "2016 AMC 12A Problems/Problem 19"

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Then we sum to get <math>46</math>. There are a total of <math>2^8=256</math> possible sequences of 8 coin flips, so the probability is <math>\frac{46}{256}=\frac{23}{128}</math>. Summing, we get <math>23+128=\boxed{151}=\boxed{B}</math>.
 
Then we sum to get <math>46</math>. There are a total of <math>2^8=256</math> possible sequences of 8 coin flips, so the probability is <math>\frac{46}{256}=\frac{23}{128}</math>. Summing, we get <math>23+128=\boxed{151}=\boxed{B}</math>.
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==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 11:44, 5 February 2016

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 $23+128=\boxed{151}=\boxed{B}$.


See Also

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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