Probability of Picking 3 Similar Letters from the Word 'Enlargement'

Have you ever wondered about the probability of picking 3 letters without replacement such that exactly two are the same from the word 'enlargement'? This intriguing question delves into the realms of combinatorics and probability. Let's explore the steps and calculations involved in solving this problem.

Analysis of the Word 'Enlargement'

The word 'enlargement' consists of 11 unique letters, with several repeated letters that play a crucial role in our analysis. Specifically, we have:

Total Ways to Choose 3 Letters from 11

We begin by calculating the total number of ways to choose any 3 letters from the 11 available letters. This is given by the combination formula:

(text{Total ways} binom{11}{3} frac{11 times 10 times 9}{3 times 2 times 1} 165)

Counting Favorable Outcomes

The next step is to determine the number of favorable outcomes where exactly two letters are the same and one is different. We need to consider the scenarios where the repeated letters can be 'e' or 'n', as these are the only letters that appear more than once.

Choosing the Repeated Letter

If we choose 'e' as the repeated letter, the different letter can be any of: {n, l, a, r, g, t, m} - a total of 7 options.

Similarly, if we choose 'n' as the repeated letter, the different letter can be any of: {e, l, a, r, g, t, m} - also a total of 7 options.

Thus, the total number of ways to choose exactly two letters that are the same and one that is different is:

(text{Favorable outcomes} 7 7 14)

Calculating the Probability

To find the probability, we use the ratio of favorable outcomes to the total number of ways:

(P frac{text{Favorable outcomes}}{text{Total ways}} frac{14}{165})

Conclusion

Therefore, the probability of picking 3 letters from the word 'enlargement' such that exactly two are the same is:

(boxed{frac{14}{165}})

Step-by-Step Calculation

We can also break down the calculation into smaller steps:

Total Ways to Choose 3 Letters from 11

The total number of ways to choose 3 letters from 11 is:

(binom{11}{3} frac{11!}{8!3!} 165)

Ways to Select 2 'e's from 2

The number of ways to select 2 'e's from 2 is:

(binom{2}{2} frac{2!}{0!2!} 1)

The number of ways to select 1 letter from the remaining 9 letters (excluding 'e's) is:

(binom{9}{1} frac{9!}{8!1!} 9)

Thus, the number of ways to choose 2 'e's plus 1 of the 9 other letters is:

(1 times 9 9)

Ways to Select 2 'n's from 2

The number of ways to select 2 'n's from 2 is:

(binom{2}{2} frac{2!}{0!2!} 1)

The number of ways to select 1 letter from the remaining 9 letters (excluding 'n's) is:

(binom{9}{1} frac{9!}{8!1!} 9)

Thus, the number of ways to choose 2 'n's plus 1 of the 9 other letters is:

(1 times 9 9)

Final Probability Calculation

The total number of favorable outcomes is the sum of the ways to choose 2 'e's and 1 other letter, plus the ways to choose 2 'n's and 1 other letter:

(9 9 18)

The probability of picking 3 letters such that exactly two are the same is:

(frac{18}{165} frac{6}{55})

Alternatively, expressing this as a decimal:

(frac{6}{55} approx 0.1091)

Conclusion Recap

By following these steps, we have determined the probability of picking 3 letters from the word 'enlargement' such that exactly two are the same is:

(boxed{frac{14}{165}})