Answer to Question #115445 in Statistics and Probability for Popcy

Question

A company prices its hurricane insurance using the following assumptions:
i In any calendar year, there can be at most one hurricane.
ii In any calendar year, the probability of a hurricane is 0.05 .
iii The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year.
Using the company’s assumptions, calculate the probability that there are fewer than 3 hurricanes in a 20-year period.

Answer 1

By the given assumptions the occurrence of hurricanes can be modeled as success/failure trials, with success meaning that a hurricane occurs in a given year and occurring with probability $p=0.05.$

Thus, the probability that there are fewer than 3 hurricanes in a 20-year period is equal to the probability of having less than 3 successes in 20 success/failure trials with $p=0.05.$

By the binomial distribution, this probability is 

$$P(X<3)=P(X=0)+P(X=1)+P(X=2)=$$

$$=\binom{20}{0}(0.05)^0(0.95)^{20-0}+\binom{20}{1}(0.05)^1(0.95)^{20-1}+$$

$$+\binom{20}{2}(0.05)^2(0.95)^{20-2}=$$

$$=(0.95)^{18}(0.95^2+20(0.05)(0.95)+190(0.05)^2)\approx0.9245$$

$0.92$