Difference between revisions of "2008 AMC 10A Problems/Problem 22"
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<math>\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}</math> | <math>\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}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We construct a tree showing all possible outcomes that Jacob may get after <math>3</math> flips; we can do this because there are only <math>8</math> possibilities: | We construct a tree showing all possible outcomes that Jacob may get after <math>3</math> flips; we can do this because there are only <math>8</math> possibilities: | ||
<cmath> | <cmath> | ||
| Line 46: | Line 46: | ||
</cmath> | </cmath> | ||
There is a <math>\frac{5}{8}</math> chance that the fourth term in Jacob's sequence is an integer, so the answer is <math>\mathrm{(D)}</math>. | There is a <math>\frac{5}{8}</math> chance that the fourth term in Jacob's sequence is an integer, so the answer is <math>\mathrm{(D)}</math>. | ||
| + | ==Solution 2== | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 16:19, 9 November 2022
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?
Solution 1
We construct a tree showing all possible outcomes that Jacob may get after 
flips; we can do this because there are only 
possibilities:
![\[6\quad\begin{cases} \ \text{H}: 11 &\quad \begin{cases} \ \text{H}: 21 &\quad \begin{cases} \ \text{H}: \boxed{41}\\ \ \text{T}: 9.5 \end{cases}\\ \ \text{T}: 4.5 &\quad \begin{cases} \ \text{H}: \boxed{8}\\ \ \text{T}: 1.25 \end{cases} \end{cases}\\ \ \text{T}: 2 &\quad \begin{cases} \ \text{H}: 3 &\qquad\! \begin{cases} \ \text{H}: \boxed{5}\\ \ \text{T}: 0.5 \end{cases}\\ \ \text{T}: 0 &\qquad\! \begin{cases} \ \text{H}: \boxed{-1}\\ \ \text{T}: \boxed{-1} \end{cases} \end{cases} \end{cases}\]](http://latex.artofproblemsolving.com/9/1/1/911a43419f7882cf84ca9d4df37cfe828a993b92.png)
There is a 
chance that the fourth term in Jacob's sequence is an integer, so the answer is 
.
Solution 2
Video Solution
~savannahsolver
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| 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. 
