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!
Best Answer
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 $\frac1{10}$.