imc

There are two ways the contestant can win. Case 1: The contestant guesses all three right. This can only happen $\frac{1}{3}\ast \frac{1}{3}\ast \frac{1}{3}=\frac{1}{27}$ of the time. Case 2: The contestant guesses only two right. We pick one of the questions to get wrong, $3$, and this can happen $\frac{1}{3}\ast \frac{1}{3}\ast \frac{2}{3}$ of the time. Thus, $\frac{2}{27}\ast 3$ = $\frac{6}{27}$. So, in total the two cases combined equals $\frac{1}{27}+\frac{6}{27}$ = $\boxed{\frac{7}{27}}$. More detailed explanation: For case 1, the contestant must guess all three correctly. The probability of guessing one problem right is $\frac{1}{3}$, so the probability of getting all three right is ${\left(\frac{1}{3}\right)}^3$. For case 2: we must choose one of the problems to answer correctly and two to answer incorrectly. The probabilities for guessing correctly and incorrectly are $\frac{1}{3}$ and $\frac{2}{3}$, respectively. So we have ${\left(\frac{1}{3}\right)}^2\cdot \frac{2}{3}\cdot 3$. The answer is the sum of probabilities of case 1 and 2, since there are no overcounts. $\frac{1}{27}+\frac{6}{27}=\frac{7}{27}$.