Bayes Theorem Problem An insurance company classifies customers as accident-prone or not accident-prone. An accident-prone customer has a 0.3 probability of submitting a claim each year. A
customer who is not accident-prone has a 0.1 probability of submitting a claim each year.
One fifth of the customers are accident-prone. What is the probability that a randomly
chosen customer will submit a claim in the next year?
My solution: Let $A=$ {person is accident prone}, $B=$ {person is not accident prone}, and $C=$ {person will submit a claim}. We know$$P(C|A)=0.3, P(C|B)=0.1$$
My guess is that we have to solve for P(C), but I'm not sure if that is right.
 A: Call the non-accident prone people $\bar A$ (read as "not A" or "A bar") then
$$
P(C|A)=0.3\\
P(C|\bar A)=0.1\\
P(A)=0.2\\
P(\bar A)=1-P(A)=0.8
$$
since the person is chosen randomly, and finally

 $$P(C)=P(C|A)P(A)+P(C|\bar A)P(\bar A).\\$$

A: The problem does not require Bayes Theorem as the title suggests. It would require, for instance, if the question were "$C$, a claim, was observed, what is the probability of being submited by an accident prone person $A$?". Translating into probabilities the question is
$$\text{Pr}(A|C)=?$$
Now you might benefit from using Bayes theorem:
$$\text{Pr}(A|C)=\dfrac{\text{Pr}(C|A)\text{Pr}(A)}{\text{Pr}(C)}$$
As you have $\text{Pr}(C|A)=0.3$ and $\text{Pr}(A)=0.2$ your problem would be finding out $\text{Pr}(C)$.
By the "Law of Total Probabilities":
$$\text{Pr}(C)=\text{Pr}(C|A)\Pr(A)+\text{Pr}(C|B)\Pr(B)$$
as $A$ and $B$ are 2 events that are mutually exclusive and collectively exhaustive. Making the substitutions you will find
$$\text{Pr}(C)=0.3\times 0.2 + 0.1\times 0.8=0.14\ \  \text{and}\ \ \text{Pr}(A|C)=\dfrac{0.06}{0.14}=3/7.$$
