Question

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

Answer 1

Out of 18 balls, three balls can be drawn in ¹⁸C₃ ways.
∴ Total number of elementary events = ¹⁸C₃ = 816
(i)
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 = $\frac{36}{816}=\frac{3}{68}$

(ii)
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 = $\frac{168}{816}=\frac{7}{34}$

(iii)
There are six red balls out of which one red ball can be drawn in ⁶C₁ ways.
Two balls from the remaining 12 balls can be drawn in ¹²C₂ ways.
Therefore, one red two other coloured balls can be drawn in ⁶C₁× ¹²C₂ = 6 × 66 = 396 ways
∴ Favourable number of ways = 396
Hence, required probability = $\frac{396}{816}=\frac{33}{68}$