A box contains 10 red balls, 9 green balls, and some blue balls. Find the probability of getting three balls of different colours when 3 balls are drawn at random, if the total number of balls in the box is 36.

1/15
2/7
9/14
3/14

Correct answer: 3/14

Solution

Since total balls are 36, blue balls = \(36-10-9=17\). The number of ways to get one red, one green, and one blue is \(10\times 9\times 17\). Dividing by total ways \(\binom{36}{3}\) gives \(\frac{3}{14}\).

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