Apply the Addition Rule and the general Multiplication Rule and interpret the results

Learning goal: Apply the Addition Rule, P(A or B) = P(A) + P(B) − P(A and B), and the general Multiplication Rule in a uniform probability model, P(A and B) = P(A) · P(B | A) when P(A) > 0, and = P(B) · P(A | B) when P(B) > 0, and interpret each answer in terms of the model.

Scope note. S.CP.8 (the general Multiplication Rule) is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

The Addition Rule and where it comes from. In the first lesson we counted the union "sum is 7 or red die even" as 6 + 18 − 3 = 21 outcomes, subtracting the overlap so that shared outcomes were counted once. Dividing every count by the 36 outcomes of the sample space turns that count into the Addition Rule:

P(A or B) = P(A) + P(B) − P(A and B).

The rule holds for any two events in any probability model, not only uniform ones, because it is just the statement that outcomes in both events must not be counted twice. When A and B are mutually exclusive, P(A and B) = 0 and the rule shortens to P(A or B) = P(A) + P(B). That shortened form is the one students remember, and it is wrong whenever the events overlap; the full rule is the safe one.

For the dice, P(sum 7 or red even) = 6/36 + 18/36 − 3/36 = 21/36 = 7/12. Interpretation in the model: of the 36 equally likely rolls, 21 have a sum of 7 or an even red die or both, so a fair roll satisfies at least one of the conditions with probability 7/12. For the mutually exclusive pair "sum 7" and "doubles", P(sum 7 or doubles) = 6/36 + 6/36 − 0 = 12/36 = 1/3.

Using the rule backwards. The rule has four quantities, and knowing any three gives the fourth. If P(A) = 0.5, P(B) = 0.4 and P(A or B) = 0.7, then P(A and B) = 0.5 + 0.4 − 0.7 = 0.2. In a class where 60% play an instrument, 40% are on a team and 73.3% do at least one (the 22 of 30 from earlier lessons, 22/30 ≈ 0.733), the share doing both is 0.6 + 0.4 − 22/30 = 18/30 + 12/30 − 22/30 = 8/30, the 8 students who do both. The rule recovers the overlap from the union.

The general Multiplication Rule. The conditional probability definition P(B | A) = P(A and B) / P(A) can be rearranged, when P(A) > 0, into

P(A and B) = P(A) · P(B | A),

and by symmetry, when P(B) > 0, P(A and B) = P(B) · P(A | B). This is the general Multiplication Rule. It says: the probability that both happen equals the probability of one of them times the probability of the other given the first. Unlike the independence product P(A)P(B), which is only valid when the events are independent, the general rule is always valid; it simply uses the conditional probability in place of the unconditional one. When A and B are independent, P(B | A) = P(B) and the general rule reduces to the special one.

Drawing without replacement. The rule is the natural tool for a two-stage process where the second stage depends on the first. A bag holds 5 red and 3 blue counters, and two are drawn at random without replacement. In the uniform model every ordered pair of distinct counters is equally likely: 8 × 7 = 56 ordered pairs. P(first red) = 5/8. Given that the first was red, 4 red and 3 blue remain, so P(second blue | first red) = 3/7. Then P(first red and second blue) = (5/8)(3/7) = 15/56. Interpretation: of the 56 equally likely ordered draws, 15 (5 choices of red times 3 choices of blue) begin red and end blue. Conditioning gives the same number that a direct count gives.

Both orders of the rule give the same number. Return to the dice with B "red die even" and E "sum is 8". Using P(B) · P(E | B): P(B) = 1/2 and, given an even red die (18 outcomes), the sums of 8 are (2, 6), (4, 4), (6, 2), so P(E | B) = 3/18 = 1/6; product 1/12. Using P(E) · P(B | E): P(E) = 5/36 and, among the five sum-8 outcomes, three have an even red die, so P(B | E) = 3/5; product (5/36)(3/5) = 15/180 = 1/12. Both routes reach 3/36 = 1/12, the direct count of the intersection. Choose whichever conditional is easier to see.

A fully worked model with cards. Two cards are dealt from a standard 52-card deck without replacement. What is the probability that both are aces? P(first ace) = 4/52. Given a first ace, 51 cards remain of which 3 are aces, so P(second ace | first ace) = 3/51. The general rule gives P(both aces) = (4/52)(3/51) = 12/2652 = 1/221. Interpretation: in the uniform model of 52 × 51 = 2,652 equally likely ordered two-card deals, 4 × 3 = 12 consist of two aces. If the first card were replaced and the deck shuffled before the second draw, the draws would be independent and the answer would be (4/52)(4/52) = 1/169 instead; the model matters, and the phrase "without replacement" is what makes the conditional probability 3/51 rather than 4/52.

Combining the two rules. From the same bag of 5 red and 3 blue, what is P(one of each color) in two draws? "One of each" is "red then blue or blue then red", and these two orders are mutually exclusive, so by the Addition Rule the probability is P(RB) + P(BR). By the Multiplication Rule P(RB) = (5/8)(3/7) = 15/56 and P(BR) = (3/8)(5/7) = 15/56. Total 30/56 = 15/28. The two rules do different jobs: multiplication handles "and" along one sequence of stages; addition handles "or" across alternative sequences.

Misconceptions. The first is adding probabilities of overlapping events, P(A) + P(B), and getting a number that may exceed 1: P(red even) + P(blue even) = 1/2 + 1/2 = 1 would claim that every roll has an even die, which is false, because the 9 rolls with both even were double counted; the correct union is 1/2 + 1/2 − 1/4 = 3/4. The second is multiplying unconditional probabilities for dependent stages: (4/52)(4/52) for two aces without replacement ignores the removed ace. The remedy for both is to ask, before applying either rule, whether the events overlap and whether the second stage is conditioned on the first.

Worked Addition Rule with cards: One card is drawn. P(king or heart) = P(king) + P(heart) − P(king of hearts) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. Interpretation: 16 of the 52 equally likely cards are kings or hearts: the 13 hearts plus the 3 kings that are not hearts. Adding 4 + 13 = 17 would count the king of hearts twice.

Worked Multiplication Rule with two stages: A drawer has 6 black socks and 4 white socks; two are taken at random without replacement. P(both black) = P(first black) · P(second black | first black) = (6/10)(5/9) = 30/90 = 1/3. Interpretation: of the 10 × 9 = 90 equally likely ordered draws, 6 × 5 = 30 are black-black. With replacement the answer would be (6/10)(6/10) = 9/25, larger, because a black sock would be available again.

Worked rule in reverse: In a hypothetical survey, P(owns a bicycle) = 0.6 and P(owns a bicycle and rides to school) = 0.15. Then P(rides to school | owns a bicycle) = 0.15/0.6 = 0.25: a quarter of bicycle owners ride to school. The Multiplication Rule rearranged is just the definition of conditional probability, so any one of the three quantities follows from the other two.

Represent and record: For the 6-by-6 dice grid, shade "red die even" and "blue die even", count the overlap (9), and write the Addition Rule with all four numbers as fractions over 36. Then draw a two-stage tree for drawing two counters from 5 red and 3 blue without replacement: label every branch with its conditional probability, multiply along each of the four paths, and check that the four path probabilities (20/56, 15/56, 15/56, 6/56) add to 1.

Independent performance: Design one problem whose answer needs the Addition Rule with overlapping events and one whose answer needs the general Multiplication Rule with dependent stages (without replacement). Solve each, then solve it a second way by direct counting in the uniform model. Include one example where a classmate wrongly used P(A) + P(B) or P(A) · P(B), show the wrong number, and explain which outcomes were double counted or which conditioning was ignored. Success criteria: every rule is written with its symbols and then its numbers; every answer is interpreted as a fraction of a named set of equally likely outcomes; the two methods agree for each problem; the wrong method is identified by name and by what it omits.