SOLUTION: A politician is about to give a campaign speech and is holding a stack of eight cue cards, of which the first 3 are the most important. Just before the speech, he drops all of the
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-> SOLUTION: A politician is about to give a campaign speech and is holding a stack of eight cue cards, of which the first 3 are the most important. Just before the speech, he drops all of the
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Question 1104726: A politician is about to give a campaign speech and is holding a stack of eight cue cards, of which the first 3 are the most important. Just before the speech, he drops all of the cards and picks them up in a random order. What is the probability that cards #1, #2, and #3 are still in order on the top of the stack? Answer by ikleyn(52874) (Show Source):
In all, there are 8! permutations of 8 cards.
Of them, the "fortunate" permutations are those that have first three cards as #1, #2 and #3; the rest 5 of the cards may go in an arbitrary order.
Therefore, the probability under the question is = = 0.002976 = 0.2976%.
