Difference between revisions of "2016 AMC 12A Problems/Problem 19"
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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). | 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 | + | Then we sum to get <math>46</math>. There are a total of <math>2^8=256</math> possible sequences of 8 coin flips, so the probability is <math>\frac{46}{256}=\frac{23}{128}</math>. Summing, we get <math>23+128=\boxed{151}=\boxed{B}</math>. |
Revision as of 20:46, 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 sum to get 
. There are a total of 
possible sequences of 8 coin flips, so the probability is 
. Summing, we get 
.
