Difference between revisions of "2016 AMC 12A Problems/Problem 19"
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==Problem== | ==Problem== | ||
| − | Jerry starts at 0 on the real number line. He tosses a fair coin 8 times. When he gets | + | Jerry starts at <math>0</math> on the real number line. He tosses a fair coin <math>8</math> times. When he gets heads, he moves <math>1</math> unit in the positive direction; when he gets tails, he moves <math>1</math> unit in the negative direction. The probability that he reaches <math>4</math> at some time during this process <math>\frac{a}{b},</math> where <math>a</math> and <math>b</math> are relatively prime positive integers. What is <math>a + b?</math> (For example, he succeeds if his sequence of tosses is <math>HTHHHHHH.</math>) |
| − | + | <math>\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313</math> | |
==Solution== | ==Solution== | ||
Revision as of 15:56, 4 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 6-8 heads, we are guaranteed to hit 4 heads, so the sum here is 
For 4 heads, you have to hit the 4 heads at the start so there's only one way
For 5 heads, we either start of with 4 heads, which gives us 4 ways to arrange the other flips, or we start off with five heads and one tail, which also has four ways (ignoring the overlap with the case of 4 heads to start).
Then we just sum to get 
.
