2016 AMC 12B Problem 19

Question

Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?

\dfrac18

\dfrac17

\dfrac16

\dfrac14

\dfrac13

Answer: B

Concepts:geometric distribution independent events

Difficulty rating: 1910

Small hint:

A single player gets his first head on flip

n

with probability

\left(\tfrac12\right)^n

Big hint:

All three stop on flip

n

with probability

\left(\tfrac18\right)^n;

sum this geometric series over

n\ge1

Solution:

A player’s first head comes on flip

n

with probability

\left(\tfrac12\right)^n.

All three stopping on the same flip

n

has probability

\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n.

Summing over

n\ge1,

\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17.

Thus, the correct answer is B.

2016 AMC 12B Problem 19

Answer: B

Difficulty rating: 1910

Small hint:

Big hint:

Solution:

Problem 19 in other years