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