Each of two boxes contains both black and white marbles and the total number of marbles in two boxes is 25. One marble is taken out of each box. The probability that both marbles are black is $\frac{27}{50}$ . If the probability that both marbles are white is $\frac{m}{n}$ , where $m, n$ are relatively prime positive integers, then $m$ is

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Detailed Solution

Let

a, b

represent the number of marbles in each box, and without loss of generality let

a > b

. Then,

a + b = 25

, and since the

a b

may be reduced to form 50 on the denominator of

\frac{27}{50}, 50 | a b

. It follows that

5 | a, b

so there are 2 pairs of

a and b : (20, 5), (15, 10)

Case 1: Then the product of the number of black marbles in each box is 54, so the only combination that works is 18 black in first box, and 3 black in second. Then,

$P (both white) = \frac{2}{20} . \frac{2}{5} = \frac{1}{25}$

So, $m = 1$

Case 2: The only combination that works is 9 black in both. Thus

$P (both white) = \frac{1}{10} . \frac{6}{15} = \frac{1}{25}$

Thus, $m = 1$

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