Describe events as subsets of a sample space using or, and, not

Learning goal: Describe events as subsets of a sample space, the set of all outcomes, either by a characteristic or category of the outcomes, or as unions, intersections and complements of other events, using the words or, and, not. Count the outcomes in each event exactly.

The sample space is a set. A probability model begins with a list of everything that can happen on one trial. That list is the sample space, usually written S, and each entry is an outcome. Two fair six-sided dice, one red and one blue, produce the ordered pair (red value, blue value). The sample space has 6 × 6 = 36 outcomes, from (1, 1) to (6, 6). The pair (2, 5) and the pair (5, 2) are different outcomes, because a different die shows the 2. Writing the sample space out, or at least knowing how to list it, is the first step in every problem in this unit; a probability is meaningless until the set it refers to is fixed.

An event is a subset. An event is any collection of outcomes from the sample space, that is, any subset of S. We usually name an event by a characteristic that its outcomes share. Let A be the event "the sum is 7". As a subset, A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, six outcomes. Let B be the event "the red die shows an even number". B contains every pair whose first entry is 2, 4 or 6: three choices for the red die times six for the blue, so B has 18 outcomes. Let C be the event "doubles", C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, six outcomes. The description in words and the subset are the same event; the words are a rule for deciding which outcomes belong.

Events can also be defined by a category of the outcomes. If a school records, for each student, whether they take a bus and whether they eat school lunch, the sample space for choosing one student at random is the set of students, and "takes a bus" is the subset of students in that category. The subset idea does not depend on dice.

Building new events with or, and, not. Three set operations build new events from old ones, and each has an everyday word.

The union A ∪ B is the event "A or B": every outcome that is in A, in B, or in both. In probability, or is always inclusive. "Sum is 7 or red die is even" includes (2, 5), which satisfies both. To count a union you cannot simply add: A has 6 outcomes and B has 18, but A ∪ B has 21, not 24, because the three outcomes (2, 5), (4, 3), (6, 1) belong to both and must be counted once.

The intersection A ∩ B is the event "A and B": the outcomes that satisfy both conditions at the same time. A ∩ B = {(2, 5), (4, 3), (6, 1)}, three outcomes; each has sum 7 and an even red die. Two events with no outcome in common, such as A (sum 7) and C (doubles), have an empty intersection, A ∩ C = ∅, since a double has an even sum and 7 is odd. Such events are called mutually exclusive or disjoint: they cannot both happen on one trial.

The complement of A, written A′ or "not A", is the event containing every outcome of the sample space that is not in A. Not-A here is "the sum is not 7", with 36 − 6 = 30 outcomes. The complement always depends on the sample space: not-A means everything else in S, so A and A′ together contain every outcome exactly once.

Probabilities from counts. When the outcomes are equally likely, as they are for two fair dice, the probability of an event is the number of outcomes in it divided by the number in S. So P(A) = 6/36 = 1/6, P(B) = 18/36 = 1/2, P(A ∩ B) = 3/36 = 1/12, P(A ∪ B) = 21/36 = 7/12, and P(not A) = 30/36 = 5/6. Notice that P(A) + P(not A) = 1, which is always true, and that P(A ∪ B) = 21/36 is less than P(A) + P(B) = 24/36 by exactly the overlap 3/36. These relationships become the rules of the later lessons; here the point is that they are visible by counting.

A fully worked model. Describe the event "the red die is even and the dice are not doubles" as a subset and find its probability. The event is B ∩ C′. Start from B, the 18 outcomes with red value 2, 4 or 6. Remove the doubles inside B: (2, 2), (4, 4), (6, 6). That leaves 18 − 3 = 15 outcomes, so the probability is 15/36 = 5/12. Now describe "the red die is odd or the dice are doubles": that is B′ ∪ C. B′ has 18 outcomes (red value 1, 3, 5). C has 6, of which (1, 1), (3, 3), (5, 5) are already in B′. So B′ ∪ C has 18 + 6 − 3 = 21 outcomes and probability 21/36 = 7/12.

Why the misconception fails. Students often read "A or B" as exclusive, meaning one but not the other, and count 6 + 18 − 2 × 3 = 18 outcomes for "sum 7 or red even". That is a different event, sometimes written as A Δ B, and it is not what or means in probability. Test any doubtful reading against a specific outcome: does (2, 5) belong to "sum is 7 or red die is even"? It satisfies both conditions, so it certainly satisfies at least one, and it belongs. Another frequent error is to take the complement of the wrong set. "Not doubles" is 30 outcomes of the 36-outcome sample space, not 30 outcomes of some smaller list, because the complement is taken inside S.

Describing in words from a subset. The skill runs both ways. Given the subset {(1, 1), (1, 2), (2, 1)}, a description is "the sum is at most 3". Given {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)}, a description is "at least one die shows 6"; it is the union of "red shows 6" and "blue shows 6", and it has 6 + 6 − 1 = 11 outcomes because (6, 6) is in both. A good description makes the membership test easy to apply to every outcome.

Worked union count: Event D is "the blue die shows 5" (6 outcomes: red 1 to 6 with blue 5) and event A is "the sum is 7". D ∩ A is the single outcome (2, 5). So D ∪ A has 6 + 6 − 1 = 11 outcomes and P(D or A) = 11/36. Adding 6 + 6 = 12 would count (2, 5) twice.

Worked complement: Event E is "at least one die shows 6", which has 11 outcomes (row six, column six, with (6, 6) counted once). Not-E is "neither die shows 6", every pair with both values from 1 to 5: 5 × 5 = 25 outcomes. Check: 11 + 25 = 36, the whole sample space. So P(not E) = 25/36 and P(E) = 1 − 25/36 = 11/36, agreeing with the direct count.

Worked description from a category: A hypothetical class of 30 students is the sample space for drawing one name at random. 18 play an instrument, 12 are on a sports team and 8 do both. "Instrument and team" is the intersection with 8 students. "Instrument or team" is the union with 18 + 12 − 8 = 22 students. "Neither" is the complement of the union, 30 − 22 = 8 students, so P(neither) = 8/30 = 4/15. The three numbers 8, 22 and 8 come from set reasoning, not from separate surveys.

Represent and record: Draw the 6-by-6 grid of outcomes for the red and blue dice. Shade "sum is 7" one way, "red die is even" another, and mark the three outcomes in both. Write each of the events A, B, A ∪ B, A ∩ B, A′ and B′ as a list or a count, and write its probability as a fraction in lowest terms. Then write two more events of your own by a characteristic (for example "product is a multiple of 10"), list them as subsets, and form their union, intersection and complements.

Independent performance: Choose a different sample space of at least 20 equally likely outcomes, for example a spinner numbered 1 to 20, or drawing one card from a small labeled set. Define three events by characteristics of the outcomes. For every pair of your events, list the union and the intersection as subsets and state whether the pair is mutually exclusive. Write each complement as a subset. Success criteria: every event is written both in words and as a subset; every union count equals the two counts added minus the overlap; every complement count plus the event count equals the size of the sample space; every probability is a fraction whose denominator is the sample-space size and which is then reduced.