Assume drivers can be classified as either aggressive or non-aggressive. If the probability of getting into an accident is twice as high...

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 $$P(A)=.95$$?

• This is very hard to follow. What is $A$? Is it an event? If so, what does it mean to say $A=.1$?
– lulu
Commented Sep 28, 2020 at 17:19
• I classified A as aggressive drivers and N as non-aggressive drivers. So in the question it says 10% of drivers are aggressive. That is why I had $A=.10$ Commented Sep 28, 2020 at 17:22
• I don't understand. When you say "$A=.1$ do you mean "the probability that a driver is aggressive is $.1$" But then you say $P(A)=.95$, so what does that mean?
– lulu
Commented Sep 28, 2020 at 17:23
– lulu
Commented Sep 28, 2020 at 17:25
• The line A=.10 and A=2P(N) is mixed up. You need to clarify what P is on that line. You seem to be mixing the fraction of drivers being aggressive with the probability of a driver in an accident had been aggressive. Commented Sep 28, 2020 at 17:30

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.

EDIT

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.

Another edit

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.