MATHCOUNTS Exercises – 7

Solution Let \(P_{fair}\) and \(P_{crooked}\) denote the probability of the fair coin and the crooked coin landing on head. We have $$P_{fair} = \frac{1}{2}, \ \ \ P_{crooked} = \frac{2}{3}$$ The probability of selecting the fair coin or the crooked coin randomly is \(\dfrac{1}{2}\). Therefore the probability of both flips landing on head is $$\frac{1}{2}\cdot (P_{fair})^2+ \frac{1}{2}\cdot (P_{crooked})^2 = \frac{1}{2}\cdot(\frac{1}{2})^2+\frac{1}{2}\cdot(\frac{2}{3})^2=\frac{1}{8}+\frac{2}{9}=\frac{25}{72}$$ Therefore, the probability of the coin is the fair coin is $$ \frac{\dfrac{1}{8}}{\dfrac{25}{72}}=\frac{9}{25}$$ \(\blacksquare\)