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The probability of a randomly selected car crashing during a year is 0.040. If a family has four cars, at least one of the cars has a crash except in all cases except the case where none of the cars is involved in a car crash.
The probability that none of the cars is involved in a crash is `(1 - 0.04)^4 ~~ 0.8493`
This gives the probability that at least one car has a car crash as (1 - 0.8493) = 0.1506
Based on the probability given for one car crash there is no reason why the derived probability is incorrect.
There is a probability of 0.1506 that one of the cars has a crash.
The original question asked about the possibility of the derived probability being incorrect.
While there is no mathematical reason for the probability, here approximately 15% that at least one car is in a crash, to be incorrect there are some real world complications that would affect the probability.
For example, if I own 4 cars but one never leaves the garage. Or perhaps one is only driven on my property; never getting on a road, changing the probabilities. What if one of my cars is a derby car -- guaranteed to be in a crash this Saturday?
Thus an insurance agent setting rates by number of vehicles needs to also ask about the number of miles driven on each car among other questions, instead of charging as if all cars had an equal likelihood of being in an accident.
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