# Probability Problem (maybe on Negative Binomial Distribution)

## 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?

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 = 7C2 . (0.1)2 . (0.9)5 = 0.124
Total Probability = 0.124 . 0.1 (for last hour) = 0.0124

I am trying the 2nd part.

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OK, here's the 2nd part -

P(working more than 8 hours) = P(selling less than 3 cars in 8 hours)
= P(selling 0 car in 8 hours) + P(selling 1 car in 8 hours) + P(selling 2 cars in 8 hours)
= 8C0 . (0.1)0 . (0.9)8 + 8C1 . (0.1)1 . (0.9)7 + 8C2 . (0.1)2 . (0.9)6
= 0.962

Ken G
Gold Member
It all looks right to me.