SOLUTION: 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

Algebra.Com
Question 1145776: 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 0.12.
• 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 17-year period.
Answer by VFBundy(438)   (Show Source): You can put this solution on YOUR website!
"Fewer than 4 tornadoes" is another way of saying 3,2,1, or 0 tornadoes.

Probability of exactly 3 tornadoes: = 0.1963
Probability of exactly 2 tornadoes: = 0.2878
Probability of exactly 1 tornado: = 0.2638
Probability of exactly 0 tornadoes: = 0.1138

Probability of "fewer than 4 tornadoes": 0.1963 + 0.2878 + 0.2638 + 0.1138 = 0.8617
RELATED QUESTIONS

A company prices its tornado insurance using the following assumptions: • In any... (answered by Boreal)
1. Assume that a procedure yields a binomial distribution with a trial repeated n=18... (answered by CPhill)
hi! my question is: The canada Calendar company reduces the prices of its calendars each (answered by Fombitz)
CASE STUDY: 1 The bulbs manufactured by a company gave a mean life of 3000 hours with... (answered by stanbon,shameer)
The R. R. Bowker Company collects information on the retail prices of books and publishes (answered by ewatrrr)
The probability that a certain state will be hit by a major tornado​ (category F4 or​ (answered by rothauserc)
Most insurance companies will replace a vehicle any time the estimated repair exceeds 80% (answered by edjones)
APEX Company produces and sells three products: X, Y and Z. In order to produce them, the (answered by solver91311)
What is the maximum number of Friday the 13ths that can occur in one calendar... (answered by Edwin McCravy)