Difference between revisions of "2008 AMC 10A Problems/Problem 22"
I like pie (talk | contribs) (Replaced array with cases) |
(→Solution 1) |
||
| (9 intermediate revisions by 5 users not shown) | |||
| Line 4: | Line 4: | ||
<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 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> | ||
6\quad\begin{cases} | 6\quad\begin{cases} | ||
| Line 45: | Line 45: | ||
\end{cases} | \end{cases} | ||
</cmath> | </cmath> | ||
| − | There is a <math>\frac{5}{8}</math> chance that Jacob | + | 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== | ||
| + | We can see that as long as the last flip is heads, it will be an integer, so there is a <math>\dfrac{1}{2}</math> chance of this happening. | ||
| + | |||
| + | Doing a little casework, we see that the only possibility when the last flip is tails is when the third flip brings it to 0. There is a <math>\dfrac{1}{8}</math> chance of this happening. | ||
| + | |||
| + | Therefore, there is a <math>\dfrac{1}{2}+\dfrac{1}{8}=\boxed{(D)\dfrac{5}{8}}</math> possibility of ending with an integer. | ||
| + | |||
| + | ~nezha33 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/2GLV1flwtUQ | ||
| + | |||
| + | ~savannahsolver | ||
==See also== | ==See also== | ||
| Line 51: | Line 65: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:32, 11 August 2024
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
We can see that as long as the last flip is heads, it will be an integer, so there is a 
chance of this happening.
Doing a little casework, we see that the only possibility when the last flip is tails is when the third flip brings it to 0. There is a 
chance of this happening.
Therefore, there is a 
possibility of ending with an integer.
~nezha33
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. 
