2008 AMC 10A Problems/Problem 22: Difference between revisions
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There is a <math>\frac{5}{8}</math> chance that Jacob ends with an integer, so the answer is <math>\mathrm{(D)}</math>. | There is a <math>\frac{5}{8}</math> chance that Jacob ends with an integer, so the answer is <math>\mathrm{(D)}</math>. | ||
==Video Solution== | |||
https://youtu.be/2GLV1flwtUQ | |||
~savannahsolver | |||
==See also== | ==See also== | ||
Revision as of 19:26, 5 October 2020
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
We construct a tree showing all possible outcomes that Jacob may get after 
flips; we can do this because there are only 8 possibilities:
![\[6\quad\begin{cases} \ \text{H}: 11 &\quad \begin{cases} \ \text{H}: 21 &\quad \begin{cases} \ \text{H}: \boxed{41}\\ \ \text{T}: 9.5 \end{cases}\\ \ \text{T}: 4.5 &\quad \begin{cases} \ \text{H}: \boxed{8}\\ \ \text{T}: 1.25 \end{cases} \end{cases}\\ \ \text{T}: 2 &\quad \begin{cases} \ \text{H}: 3 &\qquad\! \begin{cases} \ \text{H}: \boxed{5}\\ \ \text{T}: 0.5 \end{cases}\\ \ \text{T}: 0 &\qquad\! \begin{cases} \ \text{H}: \boxed{-1}\\ \ \text{T}: \boxed{-1} \end{cases} \end{cases} \end{cases}\]](http://latex.artofproblemsolving.com/9/1/1/911a43419f7882cf84ca9d4df37cfe828a993b92.png)
There is a 
chance that Jacob ends with an integer, so the answer is 
.
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 | ||
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