Question

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 0.11.
• The number of tornadoes in any calendar year is independent of the number of tornados in any other calendar year.

Required:
Using the company's assumptions, calculate the probability that there are fewer than 3 tornadoes in a 14-year period.

Answer 1

The probability is
`P(X < 3) = P(X \le 3-1 = 2 ) = 0.8074`

Explanation:

From the question we are told that

The probability of a tornado is
`p = 0.11`

The sample size is
`n = 14`

Since the number of tornadoes in any calendar year is independent of the number of tornadoes in any other calendar year and there can be only outcome so we can evaluate probability using binomial distribution.

The probability of a tornado not occurring is mathematically evaluated

`q = 1 - p`

=>
`q = 1 - 0.11`

=>
`q = 0.89`

The probability that there are fewer than 3 tornadoes in a 14-year period is mathematically represented as

`P(X < 3) = P(X \le 3-1 = 2 ) = \left n } \atop {}} \right. C_2 * p ^2 q^(n- 2 ) + \left n } \atop {}} \right. C_1 * p ^1 q^(n- 1 ) + \left n } \atop {}} \right. C_0 * p ^r q^(n- 0 )`

`P(X < 3) = P(X \le 3-1 = 2 ) = \left 14 } \atop {}} \right. C_2 * (0.11) ^2 (0.89)^(14- 2 ) + \left 14 } \atop {}} \right. C_1 * (0.11) ^1 (0.89)^(14- 1 ) + \left 14 } \atop {}} \right. C_0 * (0.11) ^0 (0.89)^(14- 0 )`

`P(X < 3) = P(X \le 3-1 = 2 ) = 91 * 0.0121*0.247 + 14 * 0.11*0.2198 + 1 * 1 * 0.197`

`P(X < 3) = P(X \le 3-1 = 2 ) = 0.8074`