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Difference between revisions of "2010 AMC 12A Problems/Problem 15"

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== Problem 15 ==
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== Problem ==
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A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the probaiblity of an equal number of heads and tails is <math>\frac{1}{6}</math>. What is the probability that the coin lands on heads?
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A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the probability of an equal number of heads and tails is <math>\frac{1}{6}</math>. What is the probability that the coin lands on heads?
  
 
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math>
 
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math>
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The probability of flipping <math>2</math> heads and <math>2</math> tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.
 
The probability of flipping <math>2</math> heads and <math>2</math> tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.
  
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<cmath>\begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\
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6x^2(1-x)^2 &= \frac{1}{6}\\
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x^2(1-x)^2 &= \frac{1}{36}\\
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x(1-x) &= \pm\frac{1}{6}\end{align*}</cmath>
  
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<math>{4 \choose 2}x^2(1-x)^2 = \frac{1}{6}</math>
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As for the desired probability <math>x</math> both <math>x</math> and <math>1-x</math> are nonnegative, we only need to consider the positive root, hence
  
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<math>6x^2(1-x)^2 = \frac{1}{6}</math>
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<cmath>\begin{align*}x(1-x) &= \frac{1}{6}\\
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6x^2-6x+1&=0\end{align*}</cmath>
  
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<math>x^2(1-x)^2 = \frac{1}{36}</math>
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Applying the quadratic formula we get that the roots of this equation are <math>\frac{3\pm\sqrt{3}}{6}</math>. As the probability of heads is less than <math>\frac{1}{2}</math>, we get that the answer is <math>\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}</math>.
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<math>x(1-x) = \pm\frac{1}{6}</math>
 
  
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As for the desired probability <math>x</math> both <math>x</math> and <math>1-x</math> are nonnegative, we only need to consider the positive root, hence
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== See also ==
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{{AMC12 box|year=2010|num-b=14|num-a=16|ab=A}}
  
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<math>x(1-x) = \frac{1}{6}</math>
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}
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<math>6x^2-6x+1=0</math>
 
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Applying the quadratic formula we get that the roots of this equation are <math>\frac{3\pm\sqrt{3}}{6}</math>. As the probability of heads is less than <math>\frac{1}{2}</math>, we get that the answer is <math>\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}</math>.
 

Latest revision as of 20:47, 3 July 2013

Problem

A coin is altered so that the probability that it lands on heads is less than $\frac{1}{2}$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\frac{1}{6}$. What is the probability that the coin lands on heads?

$\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}$

Solution

Let $x$ be the probability of flipping heads. It follows that the probability of flipping tails is $1-x$.

The probability of flipping $2$ heads and $2$ tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.

\begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\ 6x^2(1-x)^2 &= \frac{1}{6}\\ x^2(1-x)^2 &= \frac{1}{36}\\ x(1-x) &= \pm\frac{1}{6}\end{align*}

As for the desired probability $x$ both $x$ and $1-x$ are nonnegative, we only need to consider the positive root, hence

\begin{align*}x(1-x) &= \frac{1}{6}\\ 6x^2-6x+1&=0\end{align*}

Applying the quadratic formula we get that the roots of this equation are $\frac{3\pm\sqrt{3}}{6}$. As the probability of heads is less than $\frac{1}{2}$, we get that the answer is $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$.

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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