Difference between revisions of "2016 AMC 12A Problems/Problem 19"
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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 <math>23+128=\boxed{\textbf{(B) }151}</math>. | 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 <math>23+128=\boxed{\textbf{(B) }151}</math>. | ||
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+ | ==Solution 3== | ||
+ | |||
+ | (Inspired by https://www.youtube.com/watch?v=EZYzjmd_f2g by Walt S) | ||
+ | |||
+ | Draw a triangle of dots as seen in the video. Now, we'll do something different than he did. Draw a dotted line between dots that correspond to 3 on the number line and dots that correspond to 4 on the number line. This is the threshold that must be passed to have a successful sequence. We consider the three entry points to the immediate right of this line. Doing a tiny bit of fun bashing as we slide down this triangle, we find there is 1 way to cross the line for the first time through the uppermost entry point, 4 ways to cross the line for the first time through the middle entry point, and 14 ways to cross the line for the first time through the lowermost entry point. | ||
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+ | Superimpose Pascal's triangle over the dots. Notice that the number of ways to get to a dot is equal to its corresponding number on Pascal's triangle. Rows on Pascal's triangle sum to powers of 2. There are <math>2^4=16</math> ways to get to the 4th row (remember that we start with the 0th row), and we found that 1 of them makes us cross the threshold for the first time. There are <math>2^6=64</math> ways to get to the 6th row, and we found that 4 of them make us cross the threshold for the first time. There are <math>2^8=256</math> ways to get to the 8th row, and we found that 14 of them make us cross the threshold for the first time. | ||
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+ | Adding up these probabilities <cmath>\frac{1}{16}+\frac{4}{64}+\frac{14}{256}=\frac{23}{128}</cmath> and summing <math>a</math> and <math>b</math>, <cmath>a+b=23+128=\boxed{\textbf{(B)}\ 151}</cmath> we get our answer. | ||
==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 12:19, 24 July 2017
Problem
Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this process where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is )
Solution 1
For to heads, we are guaranteed to hit heads, so the sum here is .
For heads, you have to hit the heads at the start so there's only one way, .
For heads, we either start off with heads, which gives us ways to arrange the other flips, or we start off with five heads and one tail, which has ways minus the overlapping cases, and . Total ways: .
Then we sum to get . There are a total of possible sequences of coin flips, so the probability is . Summing, we get .
Solution 2
Reaching 4 will require either 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 possibilities work.
(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so flips work. .
(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 tail in first four flips). 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 possibilities.
(2 tails in first four flips). In this case, the tails can be in positions.
Adding these cases up and taking the total out of yields . This means the answer is .
Solution 3
(Inspired by https://www.youtube.com/watch?v=EZYzjmd_f2g by Walt S)
Draw a triangle of dots as seen in the video. Now, we'll do something different than he did. Draw a dotted line between dots that correspond to 3 on the number line and dots that correspond to 4 on the number line. This is the threshold that must be passed to have a successful sequence. We consider the three entry points to the immediate right of this line. Doing a tiny bit of fun bashing as we slide down this triangle, we find there is 1 way to cross the line for the first time through the uppermost entry point, 4 ways to cross the line for the first time through the middle entry point, and 14 ways to cross the line for the first time through the lowermost entry point.
Superimpose Pascal's triangle over the dots. Notice that the number of ways to get to a dot is equal to its corresponding number on Pascal's triangle. Rows on Pascal's triangle sum to powers of 2. There are ways to get to the 4th row (remember that we start with the 0th row), and we found that 1 of them makes us cross the threshold for the first time. There are ways to get to the 6th row, and we found that 4 of them make us cross the threshold for the first time. There are ways to get to the 8th row, and we found that 14 of them make us cross the threshold for the first time.
Adding up these probabilities and summing and , we get our answer.
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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