Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 10A Problems/Problem 22"
(→‎Solution)
(8 intermediate revisions by 6 users not shown)
Line 1: Line 1:
−
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?
+
==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]]?
  
−
<math>\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{5}{8} \qquad \textbf{(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 2 ===
+
==Solution==
−
{{solution}}
+
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>
 +
6\quad\begin{cases}
  
−
== See also ==
+
\ \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}
 +
</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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/2GLV1flwtUQ
 +
 
 +
~savannahsolver
 +
 
 +
==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]]
 +
{{MAA Notice}}

Revision as of 20:10, 4 June 2021

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 the fourth term in Jacob's sequence is an integer, so the answer is $\mathrm{(D)}$.

Video Solution

https://youtu.be/2GLV1flwtUQ

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_22&oldid=155269"