I am not sure. Let's say that each person has h(is/er) own "killing rate" R, which tells the probability of killing somebody accidentally during a year, and characterising how good is (s)he at not killing people. We have some distribution P(R) in the population, which should be taken as the prior distribution for any person. Now, given that the person has killed (K) this year (and never before), h(is/er) posterior distribution is clearly P(R|K)=P(R)P(K|R)/P(K), where P(K|R)=R by definition and P(K) is the probability to kill with any rate, which is integral of P(K|R)P(R)dR, and that equals to R0, the population average killing rate. For the actual person's posterior average we get R=(int R^2 P(R)dR)/R0, which is (V0+R0^2)/R0, where V0 is the variance of the prior distribution, so the change from the prior R0 is of order V0/R0.

Now, we can plug in some real data, which I can't supply, but the point is that how strong evidence an accidental killing presents isn't clear and depends strongly on the width of the distribution P(R). If all people were almost equally good at not killing accidentally, then the fact that a person has killed says almost nothing except (s)he, and more so the victim, had simply bad luck.

I have assumed that R are small and we can disregard multiple killings in a year and similar effects.

Edit: math corrected

Well, I think your drunk-driving skill are only a factor to the extent that you're more likely not to get caught; but I doubt that if you had some kind of evidence of being an excellent drunk-driver, it would help you to get a lower sentence if you'd cause an accident anyhow.