Question

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.)

$69$

$151$

$257$

$293$

$313$

Solution:

Count the sequences of $8$ tosses whose running total reaches $4.$ With at most $2$ tails he certainly reaches $4,$ contributing $\binom80+\binom81+\binom82=1+8+28=37$ sequences.

With exactly $3$ tails he reaches $4$ only if he does so before the second tail, which allows at most one tail in the first $5$ tosses; this gives $4+4=8$ sequences. 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.

2016 AMC 12A Problem 19

Solution:

Problem 19 in Other Years