2010 AMC 12A Problems/Problem 15

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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2010 AMC 12A Problems/Problem 15

Problem 15

Solution