Difference between revisions of "2008 AMC 10A Problems/Problem 22"
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==Problem== | ==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? | + | 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]]? |
<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== |
| − | {{ | + | We construct a tree showing all possible outcomes that Jacob may get after <math>3</math> flips. |
| + | <!-- Yeah, kind've ugly latex - azjps --> | ||
| + | <cmath>6 | ||
| + | \quad\left\{\begin{array}{ll} | ||
| + | |||
| + | \text{H}: 11 &\quad | ||
| + | \left\{\begin{array}{ll} | ||
| + | |||
| + | \text{H}: 21 &\quad | ||
| + | \left\{\begin{array}{ll} | ||
| + | \text{H}: \boxed{41}\\ | ||
| + | \text{T}: 9.5 | ||
| + | \end{array}\right.\\ | ||
| + | |||
| + | \text{T}: 4.5 &\quad | ||
| + | \left\{\begin{array}{ll} | ||
| + | \text{H}: \boxed{8}\\ | ||
| + | \text{T}: 1.25 | ||
| + | \end{array}\right. | ||
| + | |||
| + | \end{array}\right.\\ | ||
| + | |||
| + | \text{T}: 2 &\quad \left\{ | ||
| + | \begin{array}{ll} | ||
| + | \text{H}: 3 &\quad\ \ \, | ||
| + | |||
| + | \left\{\begin{array}{ll} | ||
| + | \text{H}: \boxed{5}\\ | ||
| + | \text{T}: 0.5 | ||
| + | \end{array}\right.\\ | ||
| + | |||
| + | \text{T}: 0 &\quad\ \ \, | ||
| + | \left\{\begin{array}{ll} | ||
| + | \text{H}: \boxed{-1}\\ | ||
| + | \text{T}: \boxed{-1} | ||
| + | \end{array}\right. | ||
| + | \end{array}\right. | ||
| + | \end{array}\right. | ||
| + | </cmath> | ||
| + | There is a <math>\frac 58\ \mathrm{(D)}</math> chance that Jacob ends with an integer. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2008|ab=A|num-b=21|num-a=23}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 10:55, 26 April 2008
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.
![\[6 \quad\left\{\begin{array}{ll} \text{H}: 11 &\quad \left\{\begin{array}{ll} \text{H}: 21 &\quad \left\{\begin{array}{ll} \text{H}: \boxed{41}\\ \text{T}: 9.5 \end{array}\right.\\ \text{T}: 4.5 &\quad \left\{\begin{array}{ll} \text{H}: \boxed{8}\\ \text{T}: 1.25 \end{array}\right. \end{array}\right.\\ \text{T}: 2 &\quad \left\{ \begin{array}{ll} \text{H}: 3 &\quad\ \ \, \left\{\begin{array}{ll} \text{H}: \boxed{5}\\ \text{T}: 0.5 \end{array}\right.\\ \text{T}: 0 &\quad\ \ \, \left\{\begin{array}{ll} \text{H}: \boxed{-1}\\ \text{T}: \boxed{-1} \end{array}\right. \end{array}\right. \end{array}\right.\]](http://latex.artofproblemsolving.com/8/9/5/8953b78071afd4b925e0e3f6a2a45bcdc7a17678.png)
There is a 
chance that Jacob ends with an integer.
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 | ||
