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I (hypothetically) looked in a book of statistics which says that there is a 3% chance of getting in an auto accident per kilometer travelled. What is the formula for determining the chance of getting home safe given the number of kilometers you have to travel?

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  • $\begingroup$ So, if you travel 34 kilometers, you have over a 100% chance of getting in an accident? I must have had thousands of accidents by now.... $\endgroup$ – Gerry Myerson Nov 2 '17 at 2:24
  • $\begingroup$ That depends on the joint distribution of the events in the kilometers that you go. If you assume they're independent, then the number of auto accidents you get into is Binomial(n,0.03) distributed, and you can calculate that P(Binomial(n,0.03)=0)=$0.97^n$. $\endgroup$ – Ian Nov 2 '17 at 2:25
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    $\begingroup$ Call me paranoid then. So, if I travel 2 km, the chance is .97^2? $\endgroup$ – Paul FitzSimons Nov 2 '17 at 2:27
  • $\begingroup$ Deleted my previous comment - made a boo-boo in reading the question. Yes, the chance of getting home safe with a two km journey is indeed $0.97^2$. $\endgroup$ – Deepak Nov 2 '17 at 2:36
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    $\begingroup$ Your book of statistics deserves to be banned. It is like saying there is a 3% chance of falling down per kilometer you walk. Mathematics can't work well with rubbish statistics. Furthermore, even if you get the right statistics, you must not forget that it may be completely irrelevant to you. The likelihood of you getting into an accident depends a lot on some common sense factors such as whether you are careful and whether you are drunk. $\endgroup$ – user21820 Apr 6 '18 at 11:57
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Assuming independence of accident risk per km travelled, the probability of being accident-free for any one stretch of $1$ km is $1 - 0.03 = 0.97$.

The probability of being accident-free for $n$ such stretches is $0.97^n$.

That's what you've been asked to find. You can convert that to a percentage by multiplying by $100$.

If instead, you're asked about the probability of being involved in an accident over $n$ km, then this is $1-0.97^n$.

There are some big assumptions here: independence and even that you can divide distances in this sort of discrete fashion into $n$ "steps" meaningfully.

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  • $\begingroup$ Yes, I was thinking more of terms like doing an integral assuming you knew the average speed was, say, 3 m/s. And you were on the road for x number of seconds. With the same percentage and neglecting stops and speedups, what would be the integration here? Obviously, there has to be some continuous probability. $\endgroup$ – Paul FitzSimons Nov 2 '17 at 2:39
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If there is a probability $p$ of an accident on 1 km, then the probability of no accident on an integer number $k$ of kilometers is (assuming accidents/kilometers are independent) $$ (1-p)^k. $$ For instance, with $p=.03$ and 50 kms, $$ (1-.03)^{50}=21.8~\%. $$

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