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: bag which contains 6 red, 8 blue and 4 white balls

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

two balls are drawn at random, therefore 

total possible outcomes are ¹⁸C₃ 

therefore n(S)=816 

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

E= {(W) (W) (R)} 

n(E)= ⁶C₁ ⁴C₂ = 36

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{36}{816}\) = \(\frac{3}{16}\) 

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

E= {(B) (B) (R)} 

n(E)= ⁸C₂ ⁶C₁ = 168

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{168}{816}\) = \(\frac{7}{34}\) 

(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)= ⁶C₁ ⁴C₁ ⁸C₁+ ⁶C₁ ⁴C₂ + ⁶C₁ ⁸C₂ = 396

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{396}{816}\) = \(\frac{33}{68}\)