Question 199498
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This is one where it is much simpler to calculate the probability of the event that is opposite to the desired event.  First of all, "Before the fifth house..." means that he will make a sale while visiting houses number 1 through 4.


If he has a 0.65 probability of making a sale, then there is a 1 - 0.65 = .35 chance that he will NOT make a sale at any given house.  Making a sale at at least one of the first four houses is 1 minus the probability that he makes no sales at all at the first four visits, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{ns} = (0.35)^4]


Therefore the probability that you are looking for is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{s} = 1 - P_{ns}]


You get to do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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