Question

Three machines E₁, E₂ and E₃ in a certain factory producing electric bulbs, produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines E₁ and E₂are defective and that 5% of those produced by machine E₃ are defective. If one bulb is picked up at random from a day's production, calculate the probability that it is defective.

Answer 1

Given:
P(E₁) = 50% = 0.5
P(E₂) = 25% = 0.25
P(E₃) = 25% = 0.25
Let A be the event that the bulb picked is defective.
Then
P(A|E₁) = 4% = 0.04
P(A|E₂) = 4% = 0.04
P(A|E₃) = 5% = 0.05
The probability that a randomly picked bulb is defective can be found using the theorem of total probability.
That is, P(A) = P(E₁) × P(A|E₁) + P(E₂) × P(A|E₂) + P(E₃) × P(A|E₃)
               = 0.5 × 0.04 + 0.25 × 0.04 + 0.25 × 0.05
               = 0.02 + 0.01 + 0.0125
               = 0.0425