Assume drivers can be classified as either aggressive or non-aggressive. If the probability of getting into an accident is twice as high for aggressive drivers, and 10% of drivers are aggressive, what is the conditional probability that a driver is aggressive, given that they were in an accident?
I have: $S=\{A,N\}$
$P(S)=(\{A\}\cup\{N\})=1$
$A=.10$ and $A=2P(N)$, so $P(N)=0.05$
Then $P(A)+P(N)=1$, so $P(A)+0.05=1$
Meaning $P(A)=.95$?
The probability of $A$ given $B$ is defined as $$\frac{\textit{probability of A and B happening}}{\textit{probabilty of B happening}}$$.
Let's assume WLOG that the chance of ramming with a non-aggressive driver when met is $p$, meaning that the chance of hitting an aggressive one when met is $2p$. The chance of hitting an aggressive driver is $$10\% \cdot 2p = 20p\%$$ and the chance of hitting a non-aggressive driver is $$90\% \cdot p = 90p\%$$.
Plugging this into the conditional probability formula gives us $$\frac{20p\%}{20p\% + 90p\%} = \frac{2}{11}$$.
$\frac{2}{11}$ is also $18.\overline{18}\%$, by the way.
Why does it seem way less likely to have an aggressive driver in an accident than it should? The answer might be because although we focus on the twice chance of hitting, we forget about the sheer number of non-aggressive drivers.
This answer is also a little simplified as I don't mention both sides of the accident. If this is the intended answer, then the question is a little ambiguous.
