2016 AMC 12A Problem 19

Jerry starts at

0

on the real number line. He tosses a fair coin

8

times. When he gets heads, he moves

1

unit in the positive direction; when he gets tails, he moves

1

unit in the negative direction. The probability that he reaches

4

at some time during this process is

\dfrac{a}{b},

where

a

and

b

are relatively prime positive integers. What is

a+b?

(For example, he succeeds if his sequence of tosses is HTHHHHHH.)

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Solution

Count the sequences of

8

tosses whose running total reaches

4.

With at most

2

tails he certainly reaches

4,

contributing

\begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered}

sequences. With exactly

3

tails, he can first reach

4

on toss

4

or toss

6.

Reaching it on toss

4

requires four initial heads, after which the remaining head has

4

possible positions. Otherwise, tosses

5

and

6

are heads, one of the first four tosses is a tail, and the last two are tails, again giving

4

possibilities. Thus this case contributes

8.

With exactly

4

tails, only HHHHTTTT works, giving

1.

He cannot reach

4

with fewer than

4

heads. So there are

37+8+1=46

favorable sequences out of

2^8=256,

a probability of

\dfrac{46}{256}=\dfrac{23}{128}.

Then

a+b=23+128=151.

Thus, the correct answer is B.