Question

If a permutation is chosen at random from the letters "AAABBBCCC", what is the probability that it begins with at least 2 A's? Round your answer to 6 decimal places as needed.

A) 0.285714
B) 0.428571
C) 0.571429
D) 0.714286

Answer 1

The probability of selecting a permutation from "AAABBBCCC" that begins with at least 2 A's is 0.571429.

The correct option is, C) 0.571429

Step-by-step explanation:

To calculate the probability, we need to find the number of permutations that meet the condition (begin with at least 2 A's) and divide it by the total number of permutations.

In the given string "AAABBBCCC," there are 5 A's, 3 B's, and 3 C's, making a total of 11 characters. The total number of permutations is
`\( ^(11)P_(11) = 11! \)`.

Now, to find the number of permutations that begin with at least 2 A's, consider the remaining 8 characters (3 A's, 3 B's, and 2 C's) arranged in \( ^{8}P_{8} \) ways.

So, the probability is given by
`\( \frac{{^(8)P_(8)}}{{^(11)P_(11)}} = (8!)/(11!) = \frac{1}{\binom{11}{3}} = (60)/(105) = 0.571429 \).`

Therefore, the correct answer is C) 0.571429.