# Probability that the driver is from G3

1. Jan 20, 2009

### 2RIP

There are three groups of drivers in a city: G1, G2, and G3.

G1 make up 30% of drivers, G2 make up 50% of drivers, and G3 make up the remaining 20% of drivers.

P(at least one accident for G1)=0.1 = P(A1)
P(at least one accident for G2)=0.3 = P(A2)
P(at least one accident for G3)=0.5 = P(A3)

If we randomly select a driver out of the groups and the driver has no accident, what is probability that the driver is from G3?

I tried $$P(G3 | (A1^{c} \cap A2^{c} \cap A3^{c}))$$. Is this is the right method of solving such a question? The probability I got from this happened to be zero which sort of confirms that it's incorrect. Any advice will be great.

Thanks a lot.

2. Jan 21, 2009

### regor60

Re: Probability

I like the brute force way since I don't know the notation. Assuming one hundred drivers in proportions given, 30 G1 drivers, etc. 1- prob of one accident is prob of no accidents, right ? So there are 27 no accident G1 drivers. Add up all the no accident drivers for G2 and G3, should be 72. What proportion of these 72 are G3 ? Well, there are 10 no accident G3 drivers, so 10/72 is probability.

3. Jan 22, 2009

### robert Ihnot

Re: Probability

It would be nice if the answer could be put in a form more in line with advanced probability thinking. So...?

For conditional probability: $$P(A\mid B)=\frac{P(A and B)}{P(B)}$$

So in this case A represents G3 and B represents no accident.

$$\frac{20*.5}{30*.9+50*.7+20*.5} = 10/27$$

I would not say that this "improves" the answer whatsoever, or even whether it applies.

Last edited: Jan 22, 2009