Question

of the light bulbs made by acme, 7% are defective. only 2% of the light bulbs made by baker are defective. an office buys over 1000 light bulbs per year. they order 75% of their bulbs from acme and 25% from baker. a light bulb is chosen at random and found to be defective. what is the probability that it was produced by baker? a \\( \frac { 2 } { 23 } \\) b \\( \frac { 7 } { 25 } \\) c \\( \frac { 2 } { 7 } \\) d \\( \frac { 1 } { 3 } \\)

Explanation:

Step1: Calculate the probability of defective bulbs from Acme

Let \(P(A)\) be the probability of choosing a bulb from Acme (\(P(A)=0.75\)), and \(P(D|A)\) be the probability of a defective bulb given it's from Acme (\(P(D|A) = 0.07\)). Then \(P(A\cap D)=P(A)\times P(D|A)=0.75\times0.07=\frac{3}{4}\times\frac{7}{100}=\frac{21}{400}\)

Step2: Calculate the probability of defective bulbs from Baker

Let \(P(B)\) be the probability of choosing a bulb from Baker (\(P(B) = 0.25\)), and \(P(D|B)\) be the probability of a defective bulb given it's from Baker (\(P(D|B)=0.02\)). Then \(P(B\cap D)=P(B)\times P(D|B)=0.25\times0.02=\frac{1}{4}\times\frac{2}{100}=\frac{2}{400}\)

Step3: Use Bayes' theorem

By Bayes' theorem \(P(B|D)=\frac{P(B\cap D)}{P(A\cap D)+P(B\cap D)}\). Substitute the values: \(P(B|D)=\frac{\frac{2}{400}}{\frac{21 + 2}{400}}=\frac{2}{23}\)

Answer:

A. \(\frac{2}{23}\)