Answer:
1. 1120
2, 0.1071
Step-by-step explanation:
1.the number of distinguishable permutations of the given letters
AAABBBCD A is repeated 3, B is also repeated thrice
=
=1120
2.If it begins with at least 2 A's , no. of permutations
=120
Therefore, the probability of random permutations of AA
=120/1120= 3/28= 0.1071