Combinatorics/Probability Insurance accident An insurance company classifies people as normal or accident prone. Suppose that the probability that a normal person has an accident in a specified year is 0.2 and that for an accident prone person this probability is 0.6. Further suppose that 18% of the policyholders are accident prone. A policyholder had no accidents in a specified year. What is the probability that he or she is accident prone?
What I did:
$P(\text{Normal&NoAccident}) = 0.82 \times 0.80 = 0.6560$
$P(\text{Accident Prone & No Accident} ) = 0.18 \times 0.94 = 0.1692$
$P(\text{No Accident}) = 0.6560 + 0.1692 = 0.8252$
$P(\text{Accident Prone} | \text{No Accident}) = P(\text{Accident Prone & No Accient}) / P(\text{No Accident}) = 0.2050$
I feel like this is too simple for a probability class. Is there something I am missing?
 A: Some of your efforts are not correct because you confused conditional events and intersection of events. To avoid such mistakes in similar exercises, choose some notation for the events, write down the probabilities you are given and the probabilities that you are asked to find. Here is a suggestion:


*

*Let $N$ denote the event that a person is normal, with $P(N)=1-0.18=0.82$,

*Let $N^c$ denote the event that a person is (not normal) accident prone, with $P(N^c)=0.18$,

*Let $A$ denote the event that a policyholder has an accident, with $$P(A\mid N)=0.2$$ and $$P(A \mid N^c)=0.6$$ 


That is all what you are given and you are asked to find $P(N^c\mid A^c)$. By the Bayes Rule (BR) and the Law of Total Probability (LTB) (for the denominator), you have that 
\begin{align}P(N^c\mid A^c)&\overset{(BR)}=\frac{P(A^c\mid N^c)P(N^c)}{P(A^c)}=\frac{\left(1-P(A\mid N^c)\right)\cdot 0.18}{1-P(A)}\\[0.3cm]&\overset{(LTP)}=\frac{(1-0.6)\cdot 0.18}{1-\left(P(A\mid N)P(N)+P(A\mid N^c)P(N^c)\right]}\\[0.3cm]&=\frac{(1-0.6)\cdot 0.18}{1-\left[0.2\cdot0.82+0.6\cdot0.18\right]}=\frac{0.072}{0.728}=0.0989\end{align}
