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

Given: A bag containing 6 red, 4 white and 8 blue balls.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Two balls are drawn at random.

Total possible outcomes are ¹⁸C₃

n (S) = 816

(i) Let E be the event of getting one red and two white balls

E = {(W) (W) (R)}

n (E) = ⁶C₁⁴C₂ = 36

P (E) = n (E) / n (S)

= 36 / 816

= 3/68

(ii) Let E be the event of getting two blue and one red

E = {(B) (B) (R)}

n (E) = ⁸C₂⁶C₁ = 168

P (E) = n (E) / n (S)

= 168 / 816

= 7/34

(iii) Let E be the event that one of the balls must be red

E = {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}

n (E) = ⁶C₁⁴C₁⁸C₁ + ⁶C₁⁴C₂ + ⁶C₁⁸C₂ = 396

P (E) = n (E) / n (S)

= 396 / 816

= 33/68