Question

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced in to urn, otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the Third ball is black is _______

(a) (23/30)

(b) (17/20) 

(c) (19/20)

(d) None   

Answer 1

The correct option (a) (23 /30)   

Explanation:

P(E) = P(E₁) ∙ P(E/E₁) + P(E₂) ∙ P(E/E₂) + P(E₃) ∙ P(E/E₃) + P(E₄) ∙ P(E/E₄) ...(1)

Let       E₁ = both 1^st and 2^nd balls are white.

E₂ = 1^st is white and 2^nd is black

E₃ = 1^st is black and 2^nd is white

E₄ = both balls are black.

∴ P(E₁) = (2/4) × (1/3) = (1/6), P(E₂) = (2/4) × (2/3) = (1/3), 

P(E₃) = (2/4) × (2/5) = (1/5) and P(E₄) = (2/4) × (3/5) = (3/10)

∴ from (1)

P(E)  = (1/6)(1) + (1/3)(3/4) + (1/5)(3/4) + (3/10)(2/3)

= (1/6) + (1/4) + (3/20) + (2/10)

= (10/24) + (7/20)

= [(50 + 42)/120] 

= (92/120)

= (23/30).