Question

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that 9

(i) both the balls are red 

(ii) one ball is red and the other is black 

(iii) one ball is white.

Answer 1

Given: urn containing 7 white, 5 black and 3 red

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

two balls are drawn at random, total possible outcomes are ¹⁵C₂ 

therefore n(S)= ¹⁸C₂ = 105 

(i) let E be the event that both balls are red 

n(E)= ³C₂ = 3

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{3}{105} = \frac{1}{35}\) 

(ii) let E be the event that one is red and other is black 

n(E)= ³C₁ ⁵C₁ = 15

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{15}{105} = \frac{1}{7}\) 

(iii) let E be the event that one ball is white 

n(E) = ⁸C₁ ⁷C₁ = 56

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{56}{105} = \frac{8}{15}\)