SOLUTION: If a permutation is chosen random from the letters “aaabbbccc” what is the probability that it begins with at least 2 a’s

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Question 1184895: If a permutation is chosen random from the letters “aaabbbccc” what is the probability that it begins with at least 2 a’s
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Total number of permutations:

%289%21%29%2F%28%283%21%29%283%21%29%283%21%29%29=1680

(2) Number of permutations starting with all three a's:

%286%21%29%2F%28%283%21%29%283%21%29%29+=+720%2F36=20

(3) Number of permutations starting with exactly two a's:

%287%21%29%2F%281%21%29%283%21%29%283%21%29+=+5040%2F36+=+140

P=%2820%2B140%29%2F1680=2%2F21

ANSWER: 2/21

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Tutor @ikleyn is right -- I double counted some of the permutations.

The 20 permutations in (2) above are counted again in (3); the number of permutations starting with at least 2 a's is 140, as shown in (3).

Then the probability of a permutation starting with at least 2 a's is 140/1680 = 1/12, which agrees with her answer.

Answer by ikleyn(52772) About Me  (Show Source):
You can put this solution on YOUR website!
.
If a permutation is chosen random from the letters “aaabbbccc” what is the probability
that it begins with at least 2 a’s
~~~~~~~~~~~~~~~~~~ 


(1)  We will consider distinguishable arrangements of 9 given letters.

     The total number of all such arrangements is  9%21%2F%283%21%2A3%21%2A3%21%29 = 392880%2F%286%2A6%2A6%29 = 1680.



(2)  The number of all distinguishable arrangements starting with 3 a's is 

         6%21%2F%283%21%2A3%21%29 = 720%2F%286%2A6%29 = 20.



(3)  The number of all distinguishable arrangements starting with  exactly  2 a's is equal (OBVIOUSLY) 

     to the number of all distinguishable arrangements of the 7 (seven letter) "abbbccc", where "a" is not 
     
     in the first (leftmost) position.



     The number of such arrangements is equal to the number of all distinguishable arrangements of these 7 letters

     (which is  7%21%2F%283%21%2A3%21%29 = 140)  highlight%28MINUS%29  the number of all those distinguishable arrangements of these 7 letters,

     where "a" is in the first position. The latter number is  6%21%2F%283%21%2A3%21%29 = 720%2F%286%2A6%29 = 20.



     THUS,  the number of all distinguishable arrangements starting with  exactly  2  a's is equal to 140 - 20 = 120.



(4)  Finally, the number of all distinguishable arrangements starting with at least 2 a's is  20 + 120 = 140.



(5)  THEREFORE, the probability under the problem's question is  P = 140%2F1680 = 14%2F168 = 1%2F12 = 0.08333... = 8.333...% .    ANSWER

Solved.