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Average Member

7 Posts

Posted - 10/06/2007 :  00:10:00  Show Profile
I am having a real hard time answering this question and I was wondering if someone can give me some pointers and how to solve this problem:

The probability that an Avon salesperson makes a sale to a prospective client on the first visit is 0.4. If the sale is not made on the first visit, the probability that it will be made on the second visit is 0.65. A prospective client is never visited more than twice if no sale is made. What is the probability that the salesperson will make a sale to a prospective client?

At first I thought you just multiply 0.4 by 0.65 but the answer seems way too low. Also, I thought it would be 0.4+0.65/(1-0.4)+(1-0.65) might be the answer but it was over 1 so I know that's not right. So if someone can help me out, that would be really appreciated =)

Thank you
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Advanced Member

1001 Posts

Posted - 10/06/2007 :  01:22:36  Show Profile
0.4*0.65 would be the probability of making a sale BOTH times, except that the second trip wouldn't be made.

You can calculate it directly.

A sale on the second try REQUIRES failure on the first. Pr(Failure on the first) = 1 - 0.4 = 0.6

0.4 + 0.6*0.65

You can calculate it from the back door.
Pr(making a sale) = 1 - Pr(NOT making a sale)

1 - 0.6*0.35

You can construct a nice tree:

Sale1 0.4

NoSale1 0.6
-----Sale2 0.65
-----NoSale2 0.35

Total Probability

Sale1 0.4
NoSale1Sale2 0.6*0.65
NoSale1NoSale2 0.6*0.35
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