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Question from Jessica, a student:

Hi, ive tried this question and just wanted to check with you if my method is correct or not.

A small mobile phone retailer has found that one of their phones has a 12% probability of being faulty and a replacement having to be provided for the customer. They have just
received a trial order for 10 phones from their biggest customer who will take their business elsewhere if 20% or more items are faulty.

i) what is the probability that they will lose their biggest customer?

I used the binomial distribution formula using n=10, x=2, and p=0.12. i got the answer 0.233043 (23.3%). is this correct?



You found the probability that exactly 2 are defective. They will lose the customer if the number defective is 2 or 3 or 4 or ... or 10. It's easier to calculate the probability that they won't lose the customer.

Suppose P(X) is the probability that exactly X of the phones are defective. The probability that they will not lose the customer is then

P(0) + P(1).

The probability they will lose the customer is then

1 - [P(0) + P(1)].


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