[Math] Conditional Probability of Good vs Bad Drivers

bayesianprobability

I have a problem I'm working on that I'm not quite getting. The problem starts as such:

Suppose that there are two types of drivers: good drivers and bad drivers. Let G be the event that a particular man is a good driver, A be the event that he gets into a car accident next year, and B be the event that he gets into a car accident the following year. Let P(G) = g and P(A|G) = P(B|G) = p1,P(A|G^c) = P(B|G^c) = p2, with p1 < p2. Suppose that given the information of whether or not the man is a good driver, A and B are independent. (for simplicity and to avoid being morbid, assume that the accidents being considered are minor and wouldn’t make the man unable to drive).

(a) Explain intuitively whether or not A and B are independent.

(b) Find P(G|A^c)

(c) Find P(B|A^c)

I am looking for hints on how to start or think about this problem, as I'm quite stuck at the moment. My work is as follows thus far:

A) If a driver gets into a car crash in the first year (meaning event A occurs, this means that it is more likely that the driver is a bad driver. Hence, it is more likely that event B occurs, meaning they are not independent.

B) I immediately expand our desired probability as follows:

$$P(G|A^c) = \frac{P(A^c|G)P(G)}{P(A^c)}$$

Using LOTP, we can expand the denominator as such:

$$P(G|A^c) = \frac{P(A^c|G)*g}{P(A^c|G)*g + P(A^c|G^c)P(G^c)}$$

Any hints?

Best Answer

You were on the right track, but turned left at Albuquerque.

You just want: $\def\P{\mathop{\mathsf P}}~\P(G\mid A^\complement)~{=\dfrac{\P(A^\complement\mid G)\P(G)}{\P(A^\complement)}\\ = \dfrac{\P(A^\complement\mid G)\P(G)}{\P(A^\complement\mid G)\P(G)+\P(A^\complement\mid G^\complement)\P(G^\complement)}}$

And you can finish it from here.


PS: Your argument on the interdependence of $A,B$ was good.

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