Question

Three machines E₁, E₂, E₃ in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of machines E₁ and E₂ are defective, and that 5% of those produced on E₃ are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.

Answer 1

Let D be the event that the picked up tube is defective

Let A₁, A₂ and A₃ be the events that the tube is produced on machines E₁, E₂ and E₃, respectively.

P(D) = P(A₁) P(D|A₁) + P(A₂) P(D|A₂) + P(A₃) P(D|A₃) ......(1)

P(A₁) = `50/100 = 1/2`

P(A₂) = `1/4`

P(A₃) = `1/4`

Also P(D|A₁) = P(D|A₂) 

= `4/100`

= `1/25`

P(D|A₃) = `5/100`

= `1/20`

Putting these values in (1), we get

P(D) = `1/2 xx 1/25 + 1/4 xx 1/25 + 1/4 xx 1/20`

= `1/50 + 1/100 + 1/80`

= `17/400`

= 0.0425