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A carnival swing ride swings to the left with probability 0.4 and to the right with probability. If the ride stops after 10 swings, what is the probability that it is exactly at the place it started?

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If probability of moving right is also given , say p(r), then the swing will be at the same place, if the swing moves left and right equal no. of times=10/2 =5 . The answer should be $10C_5\times{p(r)}^5\times p(l)^5$

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