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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 $a+b$?

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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

1 Answer 1

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 $31$.

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