Probability Question help

jrpersia

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Jun 4, 2010
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I recall seeing a post from some time ago (Set.'09) with a question about a Utility Company. It states:

A utility company in a large metropolitan area finds that 70% of its customers pay a given monthly bill in full.
a. Suppose two customers are chosen at random from the list of all customers. What is the probability that both customers will pay their monthly bill in full. I believe this one is simple...

P(A)P(B)=(0.7)(0.7)
=0.49

The question then goes on to ask...

b.What is the probability that at least one of them will pay in full.
P(A) = 0.7 Isn't this too simple?

I also then have a problem in this next question as Part C appears redundant. Any help or indication that I am on the right track would be greatly appreciated. I've included my work at the end of each sentence.

Refer to the above problem. A more detailed examination of the company records indicates that 95% of the customers who pay one monthly bill in full will also pay the next monthly bill in full; only 10% of those who pay less than the full amount one month will pay in full the next month.
a. Find the probability that a customer selected at random will pay two consecutive months in full. P(B given A) = (0.7)(0.95) = 0.665
b. Find the probability that a customer selected at random will pay neither of two consecutive months in full. P(not A and not B) = (0.3)(0.9) = .27
c. Find the probability that a customer chosen at random will pay exactly one month in full. P(A) = 0.7 (isn't that too simple?)

Thank you so much.
 
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