Question

A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that two are blue and one is red

Answer 1

Out of 18 balls, three balls can be drawn in ¹⁸C₃ ways.
∴ Total number of elementary events = ¹⁸C₃ = 816

Out of eight blue balls, two blue balls can be drawn in ⁸C₂ ways.
Out of six red balls, one red ball can be drawn  in ⁶C₁ ways .
Therefore, two blue and one red balls can be drawn in ⁸C₂ × ⁶C₁ = 28 × 6 = 168 ways
 ∴ Favourable number of ways = 168
Hence, required probability =

\[\frac{168}{816} = \frac{7}{34}\]