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A bus driver parks his at a bus depot having 20 PARKING SLOTS during lunch breaks
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 find that there are only 12 buses parked in the lot including his own .What is the probability that the 2 buses parked on either of his bus have left ?
plz explain yours work ..... thanks,
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yaieks... no answer set either... certainly one where some numbers would have helped.
Here's my attempt.
Total Outcomes: Buses that leave can be selected in 19C8 (our hero's bus is still there, so 20-1) ways
Favorable outcomes: My approach here would be... since the requirement is for two buses on either side to have left.. so the three buses were lined up together... so the other 6 buses that left could have left in (20-3)C(8-2) ways or 17C6 ways.
So Probability that the two buses next to our hero's bus left = Probability that 6 other buses apart from these three have left from the remaining buses = 17C6/19C8
Is this correct?
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.