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

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<math>x(1-x) = \pm\frac{1}{6}</math>
 
<math>x(1-x) = \pm\frac{1}{6}</math>
  
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<math>6x^2-6x\pm1=0</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
  
 +
<math>x(1-x) = \frac{1}{6}</math>
  
−
Applying the quadratic formula we get <math>\frac{3\pm\sqrt{3}}{6}</math> or <math>\frac{3\pm\sqrt{15}}{6}</math>. The only answer that is less than <math>\frac{1}{2}</math> (since probability of heads is less than tails) and greater than <math>0</math> is <math>\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}</math>.
+
<math>6x^2-6x+1=0</math>
 +
 
 +
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>.

Revision as of 08:26, 16 February 2010

Problem 15

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 probaiblity 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.


${4 \choose 2}x^2(1-x)^2 = \frac{1}{6}$

$6x^2(1-x)^2 = \frac{1}{6}$

$x^2(1-x)^2 = \frac{1}{36}$

$x(1-x) = \pm\frac{1}{6}$

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

$x(1-x) = \frac{1}{6}$

$6x^2-6x+1=0$

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 $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$.

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