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?

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

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

$\dfrac{\sqrt2-1}{2}$

$\dfrac{3-\sqrt3}{6}$

$\dfrac{\sqrt3-1}{2}$

Solution:

Let $p$ be the probability of heads. The chance of two heads and two tails in four flips is $\binom{4}{2}p^2(1-p)^2=6p^2(1-p)^2=\frac16.$

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.

2010 AMC 12A Problem 15

Solution:

Problem 15 in Other Years