Question 1184895
<br>
(1) Total number of permutations:<br>
{{{(9!)/((3!)(3!)(3!))=1680}}}<br>
(2) Number of permutations starting with all three a's:<br>
{{{(6!)/((3!)(3!)) = 720/36=20}}}<br>
(3) Number of permutations starting with exactly two a's:<br>
{{{(7!)/(1!)(3!)(3!) = 5040/36 = 140}}}<br>
{{{P=(20+140)/1680=2/21}}}<br>
ANSWER: 2/21<br>
----------------------------------------------------------------<br>
Tutor @ikleyn is right -- I double counted some of the permutations.<br>
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).<br>
Then the probability of a permutation starting with at least 2 a's is 140/1680 = 1/12, which agrees with her answer.<br>