A bag contains 10 balls, out of which 2 are white, 5 are red and 3 are black balls. If three are drawn at random, then the probability that two balls are of same colour is

The total number of ways to draw three balls from ten is 10C3 = 120. Two balls being of the same color leaves the third ball as a different color, resulting in three valid cases: two white and one red, two white and one black, two red and one white, two red and one black, two black and one white, or two black and one red. Calculating the combinations for these mixed pairs gives (2C2 * 5C1) + (2C2 * 3C1) + (5C2 * 2C1) + (5C2 * 3C1) + (3C2 * 2C1) + (3C2 * 5C1) = 5 + 3 + 20 + 30 + 6 + 15 = 79. The required probability is therefore 79/120.

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