If a permutation is chosen at random from the letters AAABBB

If a permutation is chosen at random from the letters \"AAABBBCC\", what is the probability that it begins with at least 2 A\'s?

Solution

For first letter we have 3 A\'s available out of 8 total letters so

Probability of First letter to be A =3/8

for second letter we have 2 \'A left and total letters left are 7 So

Probability of second letter to be A =2/7

So probability that it starts with at least 2 A\'s = 3/8 *2/7 = 6/56 = 3/28