An insurance company has three types of customers; high risk, medium risk and low risk. Twenty percent are high risk and 30% are medium risk. The probability that a customer has had at least one accident in the current year is .25 for high risk, .16 for medium risk and .10 for low risk.
What's the likelihood that a customer chosen at random has had at least one accident in the the current year?
I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Monday, August 27, 2007
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I came up with 14.8%. I took the weighted averages of the probabilities based on percentage of clientelle in that category.ReplyDelete
High: 20% X .20= .05
Med: 30% X .16= .048
Low: 50% X .10= .05
The sum of these probabilities is .148, or 14.8%.
Anon has the right idea, but he has a Typo in his solution:ReplyDelete
High: 20%X .25 = .05
Otherwise, good job.