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## Main Question or Discussion Point

The following problem is from "Probability and Statistics in Engineering - Hines, Montgomery"

A potential customer enters an automobile dealership every hour. The probability of a salesperson concluding a transaction is 0.10. She is determined to keep working until she has sold three cars. What is the probability that she will have to work exactly 8 hours? More than 8 hours?

She will have to work exactly 8 hours means - She will sell the last car at 8th hour and stop working.

The probability of selling a car at 8th hour (any hour) = 0.1. Thus we have got the last hour calculated.

We have the preceding 7 hours. Two cars will be sold at any two of the 7 hours and for remaining 5 hours no car will be sold.

Probability of that =

Total Probability = 0.124 . 0.1 (for last hour) = 0.0124

I am trying the 2nd part.

A potential customer enters an automobile dealership every hour. The probability of a salesperson concluding a transaction is 0.10. She is determined to keep working until she has sold three cars. What is the probability that she will have to work exactly 8 hours? More than 8 hours?

__Soln__: I have done it. Don't know whether the process is correct. Also I think the answer is low in value. Want to get your opinion.She will have to work exactly 8 hours means - She will sell the last car at 8th hour and stop working.

The probability of selling a car at 8th hour (any hour) = 0.1. Thus we have got the last hour calculated.

We have the preceding 7 hours. Two cars will be sold at any two of the 7 hours and for remaining 5 hours no car will be sold.

Probability of that =

^{7}C_{2}. (0.1)^{2}. (0.9)^{5}= 0.124Total Probability = 0.124 . 0.1 (for last hour) = 0.0124

I am trying the 2nd part.