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 18:58, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
