When dealing with probabilities in words and letters, one common problem involves determining the chance of certain outcomes. In this case, we aim to find the probability of choosing two 'M' letters from the word 'MAXIMISATION' with replacement. This article will explore how to calculate such a probability and discuss related concepts in greater detail.

Introduction to the Problem

The word 'MAXIMISATION' consists of 11 letters, with the letter 'M' appearing twice. The task is to determine the probability of selecting two 'M' letters when choosing 3 letters with replacement.

Calculation Method 1: Direct Probability Calculation

Let's start with the direct method of calculation. Here’s the step-by-step process:

When choosing the first letter, the probability of selecting an 'M' is 2/11, since there are 2 'M' letters out of a total of 11. Since the draws are with replacement, the probability of selecting an 'M' again in the second draw is also 2/11.

The combined probability of both events happening (selecting two 'M' letters in two draws) is calculated by multiplying the probabilities of each individual event:

$$ frac{2}{11} times frac{2}{11} frac{4}{121} $$

Therefore, the probability of choosing two 'M' letters from 'MAXIMISATION' with replacement is (frac{4}{121}).

Calculation Method 2: Combinatorial Approach

Another way to approach this problem is by using combinatorial mathematics. This method involves calculating the total number of ways to choose 3 letters from the word and then determining how many of those ways involve two 'M' letters.

The total number of ways to choose 3 letters from 12 letters (if we consider each instance of the letter 'M' as a distinct letter) is given by the formula for permutations with repetition and is calculated as follows: $$ frac{12!}{9!3!} 220 $$

Among these, we need to calculate how many ways we can select 2 'M' letters and one other letter. Since there are 10 other letters (A, X, I, S, I, S, A, T), the number of ways to do this is

$$ 2! div 2! times 10 10 $$

The probability is then calculated as the number of favorable outcomes divided by the total number of outcomes:

$$ frac{10}{220} frac{1}{22} approx 0.04545 $$

Alternative Approach: Case-by-Case Analysis

A third approach involves considering each favorable outcome individually: MMX, MXM, and XMM. The probability for each of these cases is as follows:

For MMX, the probability is (1/6 times 1/11 times 10/11) For MXM, the probability is (10/12 times 1/11 times 1/10) For XMM, the probability is (10/12 times 1/11 times 1/10)

Adding these probabilities together, we get:

$$ 3 times frac{1}{6} times frac{1}{11} times frac{10}{11} frac{1}{22} $$

Conclusion

Regardless of the method used, the probability of selecting two 'M' letters from 'MAXIMISATION' with replacement, and the overall probabilities of each favorable outcome, lead to the same result: (frac{1}{22}) or approximately 4.545%. This demonstrates the principle of probability in a straightforward and comprehensible manner.

Keywords

Probability, MAXIMISATION, replacement