Probability exercise with accidents

Question

If we know that someone who drives has a 1/500 chance of getting into an accident, what is the probability of this person to get into an accident at least once in a year?

I really don't know how to solve this problem. I thought about multiplying it with 364 (the days) but it looks very wrong. Any help will be truly appreciated

Answer 1

The person either has an accident at least once in a year, or zero accidents that year. These two probabilities must sum to $1$.

The probability that zero accidents happen on any given day is $1 - \frac{1}{500} = \frac{499}{500}$. So to have zero accidents on every single day of the year, the events are all independent, which gives $(499/500)^{365}$.

So your probability is $1 - (499/500)^{365} \approx 0.52$.

Answer 2

Your question doesn't mention that how many times he drives the car and many other details . ( Like going from office and coming from office will count twice )

Let's say he goes n times office a year . Then it's probability in getting accident at least once a year $$=1-({499 \over 500})^n$$