2010 AMC 12A Problem 15

Question

A coin is altered so that the probability that it lands on heads is less than

\dfrac12,

and when the coin is flipped four times, the probability of an equal number of heads and tails is

\dfrac{1}{6}.

What is the probability that the coin lands on heads?

\dfrac{\sqrt{15}-3}{6}

\dfrac{6-\sqrt{6\sqrt6+2}}{12}

\dfrac{\sqrt2-1}{2}

\dfrac{3-\sqrt3}{6}

\dfrac{\sqrt3-1}{2}

The chance of two heads and two tails is

\binom{4}{2}p^2(1-p)^2

This reduces to

p(1-p)=\dfrac16;

solve the quadratic and take the root with

p\lt\dfrac12

Let

p

be the probability of heads. The chance of two heads and two tails in four flips is

\begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16. \end{aligned}

Thus

p^2(1-p)^2=\dfrac{1}{36},

so

p(1-p)=\dfrac16.

This gives

6p^2-6p+1=0,

so

p=\dfrac{3\pm\sqrt3}{6}.

Since

p\lt\dfrac12,

we take

p=\dfrac{3-\sqrt3}{6}.

Thus, D is the correct answer.

Problem 15 in other years