SOLUTION: If a permutation is chosen at random, what is the probability that it begins with at least 2 A's? AAABBBCCC

Question 1168268: If a permutation is chosen at random, what is the probability that it begins with at least 2 A's? AAABBBCCC
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Starting with AT LEAST 2 A's means starting with 3 A's, or starting with 2 A's with some different letter third.

P(AAA...) = (3/9)(2/8)(1/7) = 1/84

P(AAX...) (X NOT the third A) = (3/9)(2/8)(6/7) = 6/84 = 1/14

P(start with AT LEAST 2 A's) = 1/84+6/84 = 7/84 = 1/12

It is faster if your thinking is "first 2 letters A and ANY letter third":

P(AA?) = (3/9)(2/8)(7/7) = 1/12

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