An exact sequence fixes positions; a matching count can accept several different arrangements. In the hypothetical model below, AAB and “two As” therefore select different sets of results. Decide whether the question fixes positions before you combine any rows.
Write the eight ordered results
Suppose three independent draws each produce A or B with equal probability. Record the draws from left to right. The full list is AAA, AAB, ABA, ABB, BAA, BAB, BBA and BBB. There are eight entries, each with probability 1/8 under these stated assumptions.
Events group possible outcomes; see the OpenStax terminology reference. Here the eight strings are elementary outcomes. An event is the selection of strings answering one particular question, not a change to the drawing mechanism.
Keep the positional condition intact
If the question is “Was the result exactly AAB?”, the first position must be A, the second must be A and the third must be B. Only AAB passes all three tests. Its probability is consequently 1/8.
ABA fails because its second and third positions differ from the required string. BAA fails at the first and third positions. Both contain two As, but neither can be added to the exact-sequence event. Counting them would answer a broader question than the one asked.
Count all arrangements when only the quantity matters
Now ask “Were there exactly two As?” The accepted strings are AAB, ABA and BAA. Three of the eight equally likely outcomes qualify, giving 3/8. AAA is excluded because it has three As; the remaining four strings have fewer than two.
A useful written answer names the event as well as the fraction: “exact AAB: one string, 1/8” or “any two As: three strings, 3/8”. Neither calculation predicts an actual draw. They differ because the question changed, not because any token became more likely.
For separate background on past observations, see our history-screen reading note. This exercise uses an invented model, not results taken from a game or website.