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

We are tasked with finding the probability that a randomly chosen permutation of the letters "AAABBBCD" begins with at least two A's.

### Step 1: Determine the total number of permutations
The total number of permutations of the word "AAABBBCD" is calculated by taking into account the repeated letters. There are:
- 3 A's,
- 3 B's,
- 1 C,
- 1 D.

The total number of distinct permutations is given by:

\[
\text{Total permutations} = \frac{8!}{3!3!1!1!}
\]

### Step 2: Count the favorable permutations
We want to count the number of permutations where the first two letters are both A's. Once the first two positions are filled with A's, we are left with the following letters to permute:
- 1 A,
- 3 B's,
- 1 C,
- 1 D.

The number of favorable permutations is:

\[
\text{Favorable permutations} = \frac{6!}{3!1!1!1!}
\]

### Step 3: Calculate the probability
The probability is the ratio of favorable permutations to total permutations:

\[
P(\text{at least 2 A's}) = \frac{\frac{6!}{3!1!1!1!}}{\frac{8!}{3!3!1!1!}}
\]

Now, let's calculate this.The probability that a randomly chosen permutation of the letters "AAABBBCD" begins with at least two A's is approximately \( 0.107143 \).

Would you like any further explanation or have any questions? Here are five related questions you might find interesting:

1. What is the probability that a random permutation of "AAABBBCD" starts with exactly 3 A's?
2. How many distinct permutations can be made from the letters "AAABBBCD"?
3. What is the probability that a permutation of "AAABBBCD" does not start with an A?
4. How does the probability change if the letter set was "AAABBBCCD"?
5. What if the condition was that the permutation ends with at least 2 A's?

**Tip**: When calculating permutations with repeated items, always divide by the factorial of the frequency of each repeated letter to account for identical arrangements.

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

Find the probability that a permutation of 'AAABBBCD' starts with at least 2 A's. This solution covers the calculation of total and favorable permutations using combinatorics. Learn how to apply the fundamental principle of counting to solve probability problems.

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