A company prices its tornado insurance using the following assumptions: In any calendar year, there can be at most one tornado, In any calendar year, the probability of a tornado is 0.1

Question

Continued:

• The number of tornadoes in any calendar year is independent of the number of tornados in any other calendar year.

Using the company's assumptions, calculate the probability that there are fewer than 4 tornadoes in a 24-year period.

The correct answer is: 0.78573776095339

I honestly don't know where to start with solving this and I'm supposed to use my calculator.

Answer 1

We want to solve X, where P(X < = k) = P(X < = 3) and where X = # of tornados in 24 years and k = 3.

We further assume # of tornados in any given calendar year is independent of # of tornados in any other given calendar year, with there being no more than 1 tornado per year and probability of tornado in any given year is 10%.

The distribution that should be used to solve this problem is the Binomial and be aware there are 4 mandatory requirements for a variable to be distributed following Binomial:

1) Fixed number of trials [there are, in fact 24 trials; each year is a trial].

2) Each trial independent of others [as stated in the question, each year is independent of others].

3) There are only two outcomes [either there was a tornado or there wasn't one].

4) Probability of each outcome is constant from trial to trial [each year, probability of a tornado is 10%].

Binomial distribution has parameters n and p.

In Binomial distribution, n is the # of trials and p is probability of success in any given trial.

Therefore, X (i.e., # of tornados in 24 years) is distributed Binomially with n = 24 and p = 0.1.

As an aside, unintuitively, you'll need to think of a tornado's occurrence as being a success if you're going to correctly understand the problem.

Solve: P(X < = 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3).

Equation immediately above should make perfect sense - if it does not, stop right there and think about it before solving [equation above will make sense if you consider it for a minute or two].

Distribution function of P(X = k):

P(X=k) = p^(k)*(1-p)^(n-k)

Solve for k =0, k = 1, k = 2 and k = 3.

I'm sure you can do the combinatoric math to solve n-choose-k [where k=0,1,2,3] and can do the algebraic math to solve p^k*(1-p)^(n-k) , so I'll leave the rest to you.

Good luck.