2000 AIME I Problems/Problem 5

Problem

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?

Solution

Let $A$ be the first box. Let $B$ be the second box. Let $a$ be the number of marbles total in $A$ , and $b$ in $B$ .

Then, $a + b = 25$ , and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$ , 50 is the lowest, and $50|ab$ .

Then there are 2 pairs of $a$ and $b: (20,5),(15,10)$ , since $a$ and $b$ are identical in computing probability.

Case 1: The only combination that works is 18 black in first box, 3 black in second. Then, $P(\text{both white}) = \frac{2}{20} * \frac{2}{5} = \frac{1}{25},$ so $m + n = 26$ .

Case 2: The only combination that works is 9 black in both. Thus, $P(\text{both white}) = \frac{1}{10}*\frac{6}{15} = \frac{1}{25}$ . $m + n = 26$ .

Thus, $m + n = 26$ .

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2000 AIME I Problems/Problem 5

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