2024/25 AHSMC Part 1 Problem 6

Problem 6 of 16

Two boxes contain red and green balls. The first box has $7$ red balls and $5$ green balls. In the second box, there are $5$ red balls and an unknown number of green balls. If a ball is randomly selected from each box, the probability of selecting balls of the same colour is $\dfrac{4}{9}$ . How many green balls are in the second box?

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Solution

Let $n$ be the number of green balls in the second box. Then $\frac{4}{9} = \frac{7}{12}\cdot\frac{5}{n+5} + \frac{5}{12}\cdot\frac{n}{n+5} = \frac{35+5n}{12(n+5)}.$ Clearing denominators, $48(n+5) = 9(5n+35)$ , so $48n + 240 = 45n + 315$ , hence $n = 25$ . The answer is (B).

University of Alberta

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