SOLUTION: Mike took the tires off of two cars. The tires were all identical. He didn't keep them separated, so he got them mixed up. One of the cars is supposed to have the same tires put ba

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Question 1078958: Mike took the tires off of two cars. The tires were all identical. He didn't keep them separated, so he got them mixed up. One of the cars is supposed to have the same tires put back on the same wheels he took them from. What is the probability that Mike will pick the right four tires from the eight he has to choose from and install them on the right four wheels?

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Mike took the tires off of two cars. The tires were all identical. He didn't keep them separated, so he got them mixed up. One of the cars is supposed to have the same tires put back on the same wheels he took them from. What is the probability that Mike will pick the right four tires from the eight he has to choose from and install them on the right four wheels?
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Probability he picks the correct four tires:: 1/8C4 = 1/70
Probability he places them in the correct order:: 1/4! = 1/24
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Answer:: (1/70)(1/24) = 0.000595
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Cheers,
Stan H.
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