Question

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

Answer 1

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

Out of six red balls, one red ball can be drawn in ⁶C₁ ways.
Out of four white balls, two white balls can be drawn in ⁴C₂ ways .
Therefore, one red and two white balls can be drawn in ⁶C₁ × ⁴C₂ = 6 × 6 = 36 ways
∴ Favourable number of ways = 36
 Hence, required probability = \[\frac{36}{816} = \frac{3}{68}\]