Difference between revisions of "2016 AMC 12A Problems/Problem 19"
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For <math>4</math> heads, you have to hit the <math>4</math> heads at the start so there's only one way, <math>1</math>. | For <math>4</math> heads, you have to hit the <math>4</math> heads at the start so there's only one way, <math>1</math>. | ||
| − | For <math>5</math> heads, we either start | + | For <math>5</math> heads, we either start off with <math>4</math> heads, which gives us <math>4\textbf{C}1=4</math> ways to arrange the other flips, or we start off with five heads and one tail, which has <math>6</math> ways minus the <math>2</math> overlapping cases, <math>\text{HHHHHTTT}</math> and <math>\text{HHHHTHTT}</math>. Total ways: <math>8</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>. | 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>. | ||
Revision as of 16:39, 15 February 2016
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
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 
.
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. 
