Difference between revisions of "2016 AMC 12A Problems/Problem 19"
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(2 tails in first four flips). In this case, the tails can be in <math>\binom{4}{2}=6</math> positions. | (2 tails in first four flips). In this case, the tails can be in <math>\binom{4}{2}=6</math> positions. | ||
− | 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. | + | 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>. |
==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 19:58, 30 December 2016
Contents
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 .
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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