What is the probability that a component is selected given the selected component is faulty? Good day all,
I'm new to probability theory and am currently working on a problem and was looking for some feedback on my work.
The question is this: One of two components is selected at random and tested. Component 1 is faulty with probability 1/5 and Component 2 is faulty with probability 1/10. What is the probability that Component 2 was the one selected given that the selected component is faulty?
So firstly, I tried working out the probability of selecting a faulty component.
Let $F$ be the event that the selected component is faulty. Then let $A$ be the event that Component 1 is faulty and let $B$ be the event that Component 2 is faulty.
$$P(F)=P(A\cup B)$$
$$=P(A)+P(B)-P(A\cap B)$$
$$\frac{1}{2}\cdot\frac{1}{5}+\frac{1}{2}\cdot\frac{1}{10}-0$$
$$=\frac{3}{20}$$
Firstly, I've no idea whether that is correct or not, but if it is, would I then proceed to calculate the conditional probability like this:
$$P(B|F)=P(B\cap F)/P(F)$$
$$=\frac{1}{20}\cdot\frac{20}{3}$$
$$=\frac{1}{3}$$
I feel like I've really missed something. I'm not sure exactly how to calculate $P(B\cap F)$ but my intuition says that $P(B\cap F)=P(B)$ in this instance. I'm also unsure of whether my working even makes sense. Any helpful feedback would be appreciated. Thanks in advance for your patience and time!
 A: Your description of $A$ and $B$ isn’t quite right: you want $A$ to be the event that Component $1$ is selected (which you don’t actually need), and $B$ to be the event that Component $2$ is selected.
Your calculation of $P(F)$ is fine. $P(B\cap F)$ is the probability that you picked Component $2$ and Component $2$ is faulty. The events $B$ and $F$ are independent, so this $P(B)\cdot P(F)$. We’re told that the component was picked ‘at random’, which is a sloppy way of saying that each component had probability $\frac12$ of being picked. Thus, $P(B)\cdot P(F)=\frac12\cdot\frac1{10}=\frac1{20}$. Thus, your computation used the correct value of $P(B\cap F)$ (and came to the correct final result)but contrary to what you then said, this is not $P(B)$: $P(B)$ is $\frac12$.
You could also use Bayes’ theorem for this problem:
$$P(B\mid F)=\frac{P(F\mid B)\cdot P(B)}{P(F)}\;.$$
We already know that $P(F)=\frac3{20}$ and $P(B)=\frac12$, and $P(F\mid B)$ is simply the probability that Component $2$ is faulty given that it was picked. Whether or not Component $2$ is faulty does not depend on whether it was picked, so this is simply $\frac1{10}$.
