Question

Lightbulbs are manufactured by three factories, I, II, and III. 10% of the lightbulbs were made in factory I. 40% of the lightbulbs were made in factory II. 50% of the lightbulbs were made in factory III.

Suppose that 3% of the lightbulbs produced by factory I are defective. 1% of the lightbulbs produced by factory II are defective. 2% of the lightbulbs produced by factory III are defective.

If a randomly selected lightbulb is defective, what is the probability that the lightbulb came from factory II? Use Bayes’ rule.

Answer 1

Solution:

Let us define some events.

A : A bulb is defective.

E_{^{​​​​​}​​​​​​1} : Bulb is made in factory 1.

E_{​​​​​2} : Bulb is made in factory 2.

E_{​​​​​​3} : Bulb is made in factory 3.

We have following information:

Probability that a bulb is made in factory 1 is,

P(E_{​​​​​​1}) = 10/100 = 0.1

Probability that a bulb is made in factory 2 is,

P(E_{​​​​​2}) = 40/100 = 0.4

Probability that a bulb is made in factory 3 is,

P(E_{​​​​​3}) = 50/100 = 0.5

P(A|E_{​​​​​​1}) = 3/100 = 0.03, P(A|E_{​​​​​2}) = 1/100 = 0.01 and

P(A|E_{​​​​​3}) = 2/100 = 0.02,

We have to find P(E_{​​​​​​2}|A).

According to Bayes rule, if E_{​​​​​​1}, E_{​​​​​​2},........, E_{​​​​​​n} are mutually disjoint events with P(E_{​​​​​​i}) ≠ 0, (i = 1,2,.........,n), then for any arbitrary event A which is a subset of such that P(A) > 0, we have

If a randomly selected lightbulb is defective, the probability that the lightbulb came from factory 2 is 0.2353.