Probability Problem (maybe on Negative Binomial Distribution)

[Debdut]

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.

 

[Debdut]

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]

It all looks right to me.