Represent compound sample spaces and count events without duplication

Learning goal: Use organized lists, tables and tree diagrams to represent compound events. Identify favorable outcomes and compute probabilities under explicit equal-likelihood or weighted assumptions.

A compound experiment combines more than one simple outcome, such as tossing a coin and rolling a number cube or rolling a cube twice. An ordered pair records which result belongs to which part of the experiment, such as the first roll and the second roll. For two rolls, (2,5) and (5,2) are different outcomes even though their sum is the same. Decide whether order matters from the experiment rather than from how convenient a list would be.

For a fair coin and an independent fair six-sided cube, list H1 through H6 and T1 through T6. There are twelve equally likely pairs under these assumptions. Heads and an even number includes H2, H4 and H6, giving 3/12 = 1/4. A tree can show two first branches for the coin, each followed by six cube branches. Each complete path represents one elementary compound outcome.

Independence means that one result does not change the model probabilities for the other. Fairness alone for each device does not describe every possible relationship between their results; the model here explicitly treats them as independent. Under those assumptions, each coin-cube pair has probability (1/2)(1/6) = 1/12. Counting pairs works because they have equal probabilities.

For two independent rolls of a fair six-sided cube, use a six-by-six table. Rows represent the first roll and columns the second, giving 36 ordered pairs. The full table is the sample space, including cells outside the highlighted event. Double sixes is only (6,6), so its probability is 1/36, not two out of six merely because the event mentions six twice.

The event at least one six includes every cell in row six and every cell in column six. Each set has six cells, but (6,6) belongs to both. Count it once: 6 + 6 − 1 = 11 favorable pairs, giving probability 11/36. The diagram highlights the full union and distinguishes the overlapping cell. Adding twelve without correcting the overlap would count one outcome twice.

An alternative uses the complement. Neither roll is six gives a five-by-five block of 25 outcomes, so at least one six has probability 1 − 25/36 = 11/36. The counting and complement methods agree because together the event and its complement include all 36 outcomes exactly once.

Sums are not equally likely categories. The possible sums run from two through twelve, but sum two has only (1,1), while sum seven has six ordered pairs: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). Therefore probabilities are 1/36 and 6/36, not one eleventh each. Use the elementary outcomes rather than assigning equal weight to category names.

Not every compound situation permits a simple equal-category count. Suppose a bag contains labeled counters R1, R2 and B, and two are drawn fairly without replacement. The possible ordered pairs of distinct labels are R1R2, R1B, R2R1, R2B, BR1 and BR2, six equally likely outcomes. Two of them contain two reds, giving probability 2/6 = 1/3.

A color-only tree represents the same draw with unequal branches. The first red has probability 2/3. After a red is removed, one red remains among two counters, so the next red has probability 1/2. Their path probability is (2/3)(1/2) = 1/3. Blue then blue is impossible because the only blue was removed. Treating RR, RB, BR and BB as four equally likely outcomes would be wrong.

When elementary outcomes are not equally likely, event probability is found by adding the appropriate outcome probabilities, not by dividing favorable category count by category count. The uniform tables in this lesson permit counting because their assumptions establish equal weights. A well-organized sample space records both what can happen and the probability structure needed to use it.

Worked overlapping event: For the coin-cube model, heads or a six includes all six heads pairs plus T6. H6 is already among the heads pairs, so the event has seven of twelve outcomes. Simply adding six heads pairs and both six pairs would count H6 twice.

Worked tree: Draw first branches R1, R2 and B, then branch to each of the two labels still available. There are six complete paths and no path repeats the same physical counter. Grouping paths by color produces two RR paths, two RB and two BR, confirming the one-third two-red probability.

Worked interpretation: At least one six includes double sixes; exactly one six excludes it. The first probability is 11/36, while exactly one six has ten favorable pairs and probability 10/36. Precise everyday wording determines the event set.

Representation workshop: Make an organized list for the coin-cube experiment, a six-by-six table for two cube rolls and a tree for the without-replacement bag. Check each for omissions, duplicates and impossible outcomes. Name which outcomes are equally likely and why. Highlight one event in each representation.

Independent performance: Create and solve five compound-event questions, including an and event, an overlapping or event, a complement, a sum category and a without-replacement event. Identify favorable elementary outcomes before calculating. Use two representations for one question and explain why they give the same probability. A partner checks that you did not treat unequal categories or impossible paths as equally likely outcomes.