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It is known that the computer disks produced by a company are defective with probability 0.02
independently of each other. Disks are sold in packs of 10.
A money back guarantee is offered if a pack contains more than 1 defective disk.
(a) What proportion of sales result in the customers getting their money back?
(b) In a stock of 100 packs, how many packs are expected to result in the customers getting their money back?
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(a) It is a Binomial Distribution probability problem.
To answer question (a), we should calculate the probability having more than 1 defective disk
in a pack of 10 disks.
It is the COMPLEMENT to the probability to have 0 (zero) or 1 (one) defective disk in the pack of 10 disks, which is
P = + = 0.8171 + 0.16675 = 0.98385.
So, in 1 - 0.98385 = 0.01615 = 1.615% cases the customers getting their money back. ANSWER
(b) In a stock of 100 packs, 1.615 packs are expected to result in the customers getting their money back. ANSWER
Solved and explained.