Question

An urn contain 4 white 2 green and 4 black balls. 3 balls are drawn at random. What is the probability that the drawn balls are 

(1) 3 white 

(2) 1 white and 2 green 

(3) one of each colour.

Answer 1

n(S) = (4 + 2 + 4)C₃ = ¹⁰C₃ = \(\frac{10\times9\times8}{3\times2\times1}\)= 120 

(1) Let A = 3 white balls ⇒ n(A) = ⁴C₃ = 1 

P(A) = \(\frac{n(A)}{n(A)}\) = \(\frac{1}{120}\)

(2) Let B: 1 white or 2 green 

n(B) = ⁴C₁ x ²C₂ = 4 x 1= 4

(3) Let C₁: one of each colour: 

n(B) = 4C₁ x 2C₁ x 4C₁ = 4x₂ x = 32.