# Expected value problem

• Sep 3rd 2009, 11:09 AM
krje1980
Expected value problem
A warranty is written on a product worth \$10,000 so that the buyer is given \$8000 if it fails in the first year, \$6000 if it fails in the second, \$4000 if it fails in the third, \$2000 if it fails in the fourth, and zero after that. The probability of the product's failing in a year is 0.1; failures are independent of those of other years. What is the expected value of the warranty?

I'm a bit confused as to how to solve this problem. At first I thought I'd simply multiply the possible warrantpayments with 0.1 and add it together, but this answer is incorrect. Any help in solving this would be greatly appreciated!
• Sep 3rd 2009, 11:19 AM
Jameson
I'm not great at stats but it seems to me you shouldn't add the values together, but take an average of them. The probability of failure in each year is equal to any other year is my reasoning.
• Sep 4th 2009, 11:09 PM
krje1980
Where, according to the book the answer should be \$1809,80. I honestly don't know how to get this.
• Sep 5th 2009, 02:53 AM
mr fantastic
Quote:

Originally Posted by krje1980
where, according to the book the answer should be \$1809,80. I honestly don't know how to get this.

$(0.1)(8,000) + (0.9)(0.1)(6,000) + (0.9)^2(0.1)(4,000) + (0.9)^3(0.1)(2,000) = 1,809.80$
• Sep 6th 2009, 12:08 AM
krje1980
Great! Thanks a lot :)