Exercise


Apr 30'23

Answer

Solution: D

Consider the two mutually exclusive events “first envelope correct” and “first envelope incorrect.” The probability of the first event is 1/4 and meets the requirement of at least one correct. For the 3/4 of the time the first envelope is incorrect, there are now 3 more envelopes to fill. Of the six permutations, three will place one letter correctly. The total probability is 1/4 + 3/4(3/6) = 5/8.

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