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

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<math>\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313</math>
 
<math>\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313</math>
  
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==Solution==
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==Solution 1==
  
 
For <math>6</math> to <math>8</math> heads, we are guaranteed to hit <math>4</math> heads, so the sum here is <math>\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37</math>.
 
For <math>6</math> to <math>8</math> heads, we are guaranteed to hit <math>4</math> heads, so the sum here is <math>\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37</math>.
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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 <math>8</math> coin flips, so the probability is <math>\frac{46}{256}=\frac{23}{128}</math>. Summing, we get <math>23+128=\boxed{\textbf{(B) }151}</math>.
 
Then we sum to get <math>46</math>. There are a total of <math>2^8=256</math> possible sequences of <math>8</math> coin flips, so the probability is <math>\frac{46}{256}=\frac{23}{128}</math>. Summing, we get <math>23+128=\boxed{\textbf{(B) }151}</math>.
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==Solution 2==
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Reaching 4 will require 4,6, or 8 flips. Therefore we can split into 3 cases:
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(Case 1): The first four flips are heads. Then, the last four flips can be anything so <math>2^4=16</math> flips work.
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(Case 2): It takes 6 flips to reach 4. There must be one head in the first four flips so we don't repeat case 1. The tails can be in 4 positions. The next to flips must be heads. The last two flips can be anything so <math>2^2=4</math> flips work. <math>4*4=16</math>.
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(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.
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(1). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are <math>4*2=8</math> possibilities. 
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(2). In this case, the tails can be in<math>4C2=6</math> positions.
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Adding these cases up and taking the total out of <math>2^8=256</math> yields <math>\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}</math>. This means the answer is 23+128=151.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2016|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:53, 30 December 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 1

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 off 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, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$. Total ways: $8$.

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

Solution 2

Reaching 4 will require 4,6, or 8 flips. Therefore we can split into 3 cases:

(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ flips work.

(Case 2): It takes 6 flips to reach 4. There must be one head in the first four flips so we don't repeat case 1. The tails can be in 4 positions. The next to flips must be heads. The last two flips can be anything so $2^2=4$ flips work. $4*4=16$.

(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.

(1). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are $4*2=8$ possibilities.

(2). In this case, the tails can be in$4C2=6$ positions.

Adding these cases up and taking the total out of $2^8=256$ yields $\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}$. This means the answer is 23+128=151.

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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