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03 Apr 2010, 03:34
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A bus driver parks his bus at a bus depot having 20 parking slots during lunch break. There are a total of 20 buses parked in the depot including his bus and his bus is not parked at one of the ends. After returning from lunch he finds that there are only 12 buses parked in the lot including his own. If no other bus arrived at the depot during lunch break, then what is the probability that the two buses parked on either side of his bus have left?
03 Apr 2010, 06:43
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1. the number of buses that left the depot: 20 - 12 = 8
2. the number of combinations to take 8 buses out of 19 (20 - our bus): \(C^{19}_8\)
3. the number of combinations to take 2 buses parked on either side and other 6 buses out of 17 (20 - our bus - 2 buses parked on either side): \(C^{17}_6\)
4. the probability: \(p = \frac{C^{17}_6}{C^{19}_8} = \frac{17!8!*11!}{6!*11!*19!} = \frac{8*7}{19*18} = \frac{28}{171}\)
2. the number of combinations to take 8 buses out of 19 (20 - our bus): \(C^{19}_8\)
3. the number of combinations to take 2 buses parked on either side and other 6 buses out of 17 (20 - our bus - 2 buses parked on either side): \(C^{17}_6\)
4. the probability: \(p = \frac{C^{17}_6}{C^{19}_8} = \frac{17!8!*11!}{6!*11!*19!} = \frac{8*7}{19*18} = \frac{28}{171}\)
General Discussion
03 Apr 2010, 06:59
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walker
Walker please explain in detail why you divided both the combinations. I could not understand this
It this the only way or is there any other way to solve this problem
i got confused here and completely forgot abt total number of outcomes. Seems like spend too much time on maths on weekend (Excess of everything is bad 
) 
