Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are three caskets of treasure. The first casket contains 3 gold coins, the second casket contains 2 gold coins and 2 bronze coins, and the third casket contains 2 gold coins and 1 silver coin. You choose one casket at random and draw a coin from it. The probability that the coin you drew is gold has the value $\frac ab$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?

share|cite|improve this question
The original MathCounts-type problem must have had "has the value $a/b$". Value $ab$ does not make sense, since the probability is clearly $\lt 1$. – André Nicolas Nov 12 '12 at 15:59

If we draw from the first, our probability of gold is $\dfrac{3}{3}$. If we draw from the second, our probability of gold is $\dfrac{2}{4}$. And if we draw from the third, our probability of gold is $\dfrac{2}{3}$. So the required probability is $$\frac{1}{3}\cdot\frac{3}{3}+\frac{1}{3}\cdot \frac{2}{4}+\frac{1}{3}\cdot\frac{2}{3}.$$ This simplifies to $\dfrac{13}{18}$. The sum of numerator and denominator is $31$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.