Difference between revisions of "2008 AMC 10A Problems/Problem 22"

(solution)
(Replaced array with cases)
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==Solution==
 
==Solution==
We construct a tree showing all possible outcomes that Jacob may get after <math>3</math> flips.
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We construct a tree showing all possible outcomes that Jacob may get after <math>3</math> flips:
<!-- Yeah, kind've ugly latex - azjps -->
+
<cmath>
<cmath>6
+
6\quad\begin{cases}
\quad\left\{\begin{array}{ll}
 
  
\text{H}: 11 &\quad  
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\ \text{H}: 11 &\quad
\left\{\begin{array}{ll}
 
  
\text{H}: 21 &\quad
+
\begin{cases}
\left\{\begin{array}{ll}
 
\text{H}: \boxed{41}\\
 
\text{T}: 9.5
 
\end{array}\right.\\
 
  
\text{T}: 4.5 &\quad
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\ \text{H}: 21 &\quad
\left\{\begin{array}{ll}
+
\begin{cases}
\text{H}: \boxed{8}\\
+
\ \text{H}: \boxed{41}\\
\text{T}: 1.25
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\ \text{T}: 9.5
\end{array}\right.
+
\end{cases}\\
  
\end{array}\right.\\
+
\ \text{T}: 4.5 &\quad
 +
\begin{cases}
 +
\ \text{H}: \boxed{8}\\
 +
\ \text{T}: 1.25
 +
\end{cases}
  
\text{T}: 2 &\quad \left\{
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\end{cases}\\
\begin{array}{ll}
 
\text{H}: 3 &\quad\ \ \,
 
  
\left\{\begin{array}{ll}
+
\ \text{T}: 2 &\quad
\text{H}: \boxed{5}\\
 
\text{T}: 0.5
 
\end{array}\right.\\
 
  
\text{T}: 0 &\quad\ \ \,
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\begin{cases}
\left\{\begin{array}{ll}
+
 
\text{H}: \boxed{-1}\\
+
\ \text{H}: 3 &\qquad\!
\text{T}: \boxed{-1}
+
\begin{cases}
\end{array}\right.
+
\ \text{H}: \boxed{5}\\
\end{array}\right.
+
\ \text{T}: 0.5
\end{array}\right.
+
\end{cases}\\
 +
 
 +
\ \text{T}: 0 &\qquad\!
 +
\begin{cases}
 +
\ \text{H}: \boxed{-1}\\
 +
\ \text{T}: \boxed{-1}
 +
\end{cases}
 +
\end{cases}
 +
\end{cases}
 
</cmath>
 
</cmath>
There is a <math>\frac 58\ \mathrm{(D)}</math> chance that Jacob ends with an integer.  
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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>.
  
 
==See also==
 
==See also==

Revision as of 12:24, 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?

$\mathrm{(A)}\ \frac{1}{6}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{1}{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$

Solution

We construct a tree showing all possible outcomes that Jacob may get after $3$ flips: \[6\quad\begin{cases}  \ \text{H}: 11 &\quad  \begin{cases}  \ \text{H}: 21 &\quad \begin{cases} \ \text{H}: \boxed{41}\\ \ \text{T}: 9.5 \end{cases}\\  \ \text{T}: 4.5 &\quad \begin{cases} \ \text{H}: \boxed{8}\\ \ \text{T}: 1.25 \end{cases}  \end{cases}\\  \ \text{T}: 2 &\quad  \begin{cases}  \ \text{H}: 3 &\qquad\! \begin{cases} \ \text{H}: \boxed{5}\\ \ \text{T}: 0.5 \end{cases}\\  \ \text{T}: 0 &\qquad\! \begin{cases} \ \text{H}: \boxed{-1}\\ \ \text{T}: \boxed{-1} \end{cases} \end{cases} \end{cases}\] There is a $\frac{5}{8}$ chance that Jacob ends with an integer, so the answer is $\mathrm{(D)}$.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions