Two letters are chosen at random without replacement from the word TIME. What is the probability that the selected letters are exactly T and M (in any order)?

Introduction / Context:
This question introduces basic probability with simple combinations. We are given the four-letter word TIME and we randomly select two letters from it without replacement. The task is to find the probability that the chosen letters are exactly T and M, regardless of the order in which they are drawn. Because all letters in TIME are distinct, this turns into a straightforward counting problem using combinations.

Given Data / Assumptions:

The word TIME has 4 distinct letters: T, I, M and E.

Two letters are selected at random without replacement.

All 2-letter selections are equally likely.

We want the selection that consists of the letters T and M, in any order.

Concept / Approach:
When order does not matter, we use combinations. The total number of ways to choose 2 letters from 4 distinct letters is given by 4C2. For the favourable outcomes, there is exactly one pair containing T and M together, because there is only one way to select these two specific letters when order is ignored. The probability is therefore the ratio of favourable combinations to total combinations.

Step-by-Step Solution:
Step 1: Count total ways to choose 2 letters from 4 distinct letters = 4C2.Step 2: Compute 4C2 = (4 * 3) / (2 * 1) = 6.Step 3: Count favourable selections. The pair we desire is {T, M}. There is exactly one combination containing T and M.Step 4: Number of favourable combinations = 1.Step 5: Probability = favourable / total = 1 / 6.

Verification / Alternative check:
We could list all 2-letter combinations: {T, I}, {T, M}, {T, E}, {I, M}, {I, E}, {M, E}. There are clearly 6 pairs and only one of them is {T, M}, which confirms that the probability is 1 out of 6. This listing approach is convenient when there are only a few possibilities and helps to cross-check the combination formula.

Why Other Options Are Wrong:
The fraction 1/4 might arise if someone incorrectly divides by the number of letters in the word instead of counting pairs. The value 1/8 and 1/12 are too small and come from confusing ordered pairs with unordered pairs or incorrectly squaring probabilities. The fraction 1/3 is too large and suggests that two different pairs are being treated as favourable by mistake. Only 1/6 correctly represents the ratio of the single favourable pair to the six possible pairs.

Common Pitfalls:
One common error is to treat the draws as ordered and to count TM and MT separately while also comparing to a total of 4 * 3 ordered pairs. Although this can be done correctly, many learners mix ordered and unordered thinking and arrive at inconsistent denominators. Another pitfall is to apply 2/4 * 1/3 reasoning, which does not directly represent the event that the unordered pair of letters is exactly T and M. The cleanest approach here is to use combinations and count unordered selections.

Final Answer:
The probability that the two chosen letters are T and M is 1/6.