2008 AMC 10A Problems/Problem 22

Problem

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1. If it comes up tails, he takes half of the previous term and subtracts 1. What is the probability that the fourth term in Jacob's sequence is an integer?

$\mathrm{(A)}\ \frac{1}{6}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{1}{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$

Solution

We construct a tree showing all possible outcomes that Jacob may get after $3$ flips; we can do this because there are only 8 possibilities: \[6\quad\begin{cases}  \ \text{H}: 11 &\quad  \begin{cases}  \ \text{H}: 21 &\quad \begin{cases} \ \text{H}: \boxed{41}\\ \ \text{T}: 9.5 \end{cases}\\  \ \text{T}: 4.5 &\quad \begin{cases} \ \text{H}: \boxed{8}\\ \ \text{T}: 1.25 \end{cases}  \end{cases}\\  \ \text{T}: 2 &\quad  \begin{cases}  \ \text{H}: 3 &\qquad\! \begin{cases} \ \text{H}: \boxed{5}\\ \ \text{T}: 0.5 \end{cases}\\  \ \text{T}: 0 &\qquad\! \begin{cases} \ \text{H}: \boxed{-1}\\ \ \text{T}: \boxed{-1} \end{cases} \end{cases} \end{cases}\] There is a $\frac{5}{8}$ chance that Jacob ends with an integer, so the answer is $\mathrm{(D)}$.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS