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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.
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Solution
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
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