Question

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that one ball is red and the other is black

Answer 1

Out of 15 balls, two balls can be drawn in ¹⁵C₂ ways.
∴  Total number of elementary events = ¹⁵C₂ = 105

Out of three red balls, one red ball can be drawn in ³C₁ ways; and out of five black balls, one black ball can be drawn in ⁵C₁ ways.
Therefore, one red and one black can be drawn in ³C₁× ⁵C₁ ways.
∴ Favourable number of ways = ³C₁× ⁵C₁ = 3× 5 = 15
Hence, required probability = \[\frac{15}{105} = \frac{1}{7}\]