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Factories $A$ and $B$ produce computers. Factory $A$ produces $4$ times as many computers as factory $B$. The probability that an item produced by factory $A$ is defective is $0.014$, while the probability that an item produced by factory $B$ is defective is $0.048$.

A computer is selected at random and it is found to be defective. What is the probability it came from factory $A$?

I have figured that the fact that factory $A$ produced $4$ times as many computers does not factor into the question.

Let $P(A) = 0.014$ Let $P(B) = 0.048$

$P(A) = P(A ∩ B) + P(A ∩ Bc)$

I rearranged the equation to find the Probability of A and B compliment to find the answer.

$$P(A ∩ Bc) = P(A) - P(A ∩ B) = 0.014 - (0.014 \cdot 0.048) = 0.013328$$

What is the correct answer and solution?

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    Thank you Jeff for the edit. – 2017-02-01

2 Answers 2

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I have figured that the fact that factory $A$ produced $4$ times as many computers does not factor into the question.

No, that tells you the prior probability that a computer comes from factory $A$.   That is $P(A)=4 P(B)$

What you have otherwise been told is the rate of defects conditional on the source.

That is, $~P(D\mid A) = 0.014~$ and also, $~P(D\mid B) = 0.048$

What is the correct answer and solution?

It's a Bayes' Rule (and Law of Total Probability) questions.

You seek $P(A\mid D)$ the posterior probability that a computer comes from factory#A when given that it is defective.

$$\begin{align}P(A\mid D) ~&=~\dfrac{P(A)P(D\mid A)}{P(A)P(D\mid A)+P(B)P(D\mid B)} \\[1ex] &=~ \dfrac{4\times 0.014}{4\times 0.014~+~0.048} \\[1ex] &=~ \frac 7{13} \\[2ex] &\approx~0.538\end{align}$$

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    Hello, I appreciate the response. This seems like the right answer but for some reason I input the question and it is incorrect. – 2017-02-01
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If you select any (might be broken) computer the probability it comes from A is 0.8 or $\frac{4}{5}$ vice versa 0.2 or $\frac{1}{5}$ for B.

If A would't produce 4 times as much computers as B does then the probability of a broken computer coming from A is $\frac{0.014}{0.014+0.042}=\frac{1}{4}$

So the probability that a broken computer comes from A is $\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{5}$