Question

A bag contains 7 black and 4 red balls. If 3 balls are drawn at random find the probability that one is black and two are red.

Answer 1

3 balls can be drawn from 11 balls in ¹¹C₃

= `(11 × 10 × 9)/(1 × 2 × 3)`

= 11 × 15 ways

∴ n(S) = 11 × 15

Let B ≡ the event that one is black and two are red One black ball can be drawn from 7 black balls in ⁷C₁ ways and 2 red balls can be drawn from 4 red balls in ⁴C₂ ways.

∴ n(B) = ⁷C₁ × ⁴C₂ = 7 × 6

∴ n(B) = 7 × 6

∴ P(B) = `("n"("B"))/("n"("S"))`

= `(7  xx 6)/(11 xx 15)`

= `14/55`.