HSS.CP.B.7Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.B.7: Applying the Addition Rule for Probability
In plain English: HSS.CP.B.7 is the Common Core statistics and probability standard that asks students to find P(A or B) with the Addition Rule, P(A) + P(B) - P(A and B), in a uniform probability model. The key idea is that outcomes in both events are counted once, not twice, and each answer is read back in the context. It is usually taught in Geometry or Algebra II.
Apply the Addition Rule, P(A or B) = P(A) + P(B) - P(A and B), and interpret the answer in terms of the model.
Common Core State Standards for Mathematics · Domain: Conditional Probability and the Rules of Probability (CP) · Cluster: Use the rules of probability to compute probabilities of compound events in a uniform probability model Also written as HSS-CP.B.7 or S-CP.7 · Official standard
Students learn to find the probability that event A or event B happens, where "or" means at least one of the two. They start from counts in a Venn diagram and a two-way table, see why adding P(A) and P(B) counts the outcomes in both events twice, and arrive at the Addition Rule, P(A or B) = P(A) + P(B) - P(A and B). All of the models in the lesson are uniform: a card drawn from a shuffled deck, two fair number cubes, a random student from a group, a spinner with equal sections.
The second half of the standard is interpretation. After every calculation, students say what the number means in the model: which outcomes it counts, what fraction of the group it describes, and whether the answer is reasonable (an "or" probability can never be smaller than P(A) or P(B), or larger than 1). Students also use the rule backward to find P(A and B) and see that mutually exclusive events are the special case P(A and B) = 0.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why P(A) + P(B) counts the outcomes in both A and B twice, using a Venn diagram or a two-way table
Apply the Addition Rule P(A or B) = P(A) + P(B) - P(A and B) in card, dice, spinner and survey models
Recognize mutually exclusive events and use P(A or B) = P(A) + P(B) when P(A and B) = 0
Solve the Addition Rule for any one of its four quantities when the other three are known
Interpret P(A or B) in terms of the model and judge whether a result is reasonable
Prior Knowledge Required
Students should already be comfortable with:
Finding probabilities of compound events with organized lists, tables and tree diagrams 7.SP.C.8
Describing events as unions, intersections and complements of other events HSS.CP.A.1
Reading counts from two-way frequency tables HSS.CP.A.4
Adding, subtracting and simplifying fractions and decimals
Post the question below and give students two minutes to answer it on their own before they compare with a partner.
Warm-Up Prompt
"A club has 25 members: 12 have a dog, 9 have a cat, and 4 have both a dog and a cat. One member is chosen at random to win a pet-store gift card. A classmate says the probability that the winner has a dog or a cat is 21/25. Do you agree? Draw a picture to decide."
Ask a student who disagreed to draw a Venn diagram on the board: 8 dog only, 4 both, 5 cat only, and 8 with neither. Only 8 + 4 + 5 = 17 members have a dog or a cat, so the probability is 17/25. The 21 counted the 4 members with both pets twice. Tell students that the lesson turns this repair (subtract the overlap once) into a rule that works with probabilities as well as counts.
Direct Instruction20 minutes
Part 1: Where the rule comes from. In a uniform model every outcome has the same probability, so P(A) = (outcomes in A)/(all outcomes). The outcomes in "A or B" are the outcomes in A, plus the outcomes in B, minus the ones that were counted twice. Dividing every count by the total gives the rule. Use Diagram 1 as the picture to return to.
Name the events in words and decide what one outcome is (one card, one pair of rolls, one student).
Find P(A) and P(B) from the counts or from the given probabilities.
Find P(A and B): list or count the outcomes that are in both events. If there are none, the events are mutually exclusive and P(A and B) = 0.
Apply the rule: P(A or B) = P(A) + P(B) - P(A and B).
Interpret: say what fraction of the outcomes the answer describes, and check that it is at least as large as P(A) and P(B) and at most 1.
Cards, overlapping events
One card is drawn from a shuffled standard deck of 52. A = heart, B = face card (jack, queen or king). Three cards are both.
Equation: P(heart or face card) = 13/52 + 12/52 - 3/52 = 22/52 = 11/26
Venn diagram from survey counts
In an invented group of 200 students, 90 play a school sport, 50 are in the band and 20 do both. One student is chosen at random (Diagram 1).
Two fair number cubes are rolled. A = the sum is 7, B = the sum is 11. No roll has both sums.
Equation: P(sum 7 or sum 11) = 6/36 + 2/36 - 0 = 8/36 = 2/9
Dice, overlapping events
Two fair number cubes are rolled. A = doubles, B = the sum is 8. Only (4, 4) is in both (Diagram 2).
Equation: P(doubles or sum 8) = 6/36 + 5/36 - 1/36 = 10/36 = 5/18
Using the rule backward
For a randomly chosen student in an invented school, P(takes a language) = 0.45, P(takes a computer science course) = 0.30 and P(at least one of the two) = 0.60.
After each example, model the interpretation sentence. For Example 2: "If we pick one of these 200 students at random, the chance of getting someone who plays a sport, is in the band, or both is 3/5. That is 120 of the 200 students." For Example 5: "15% of the students take both a language and a computer science course." Point out that in Example 3 the rule still applies; the subtracted term is simply 0. Stress that "or" in probability always includes the outcomes in both events.
Guided Practice15-20 minutes
Display the invented two-way table below. Pairs answer each question, one at a time, and must write P(A), P(B) and P(A and B) before combining them. Ask a pair to explain each answer in terms of the students in the table.
How 160 juniors and seniors get to school (invented data)
Drive
Bus
Walk
Total
Junior
18
42
20
80
Senior
38
30
12
80
Total
56
72
32
160
Questions for a randomly chosen student: (1) P(senior or drives) = 80/160 + 56/160 - 38/160 = 98/160 = 49/80. (2) P(junior or walks) = 80/160 + 32/160 - 20/160 = 92/160 = 23/40. (3) P(takes the bus or walks) = 72/160 + 32/160 = 104/160 = 13/20, because no student both takes the bus and walks. Then ask: "Why is P(senior or drives) larger than P(senior) but smaller than P(senior) + P(drives)?" Listen for these errors: adding the two totals without subtracting the shared cell, subtracting the overlap twice, and reading "or" as "exactly one".
Independent Practice15 minutes
Students work four problems on their own and write one interpretation sentence for each: (1) one card from a deck, P(king or queen) = 4/52 + 4/52 = 2/13; (2) P(red card or face card) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13; (3) a tile is drawn from tiles numbered 1 to 20, P(even or multiple of 3) = 10/20 + 6/20 - 3/20 = 13/20; (4) P(A) = 0.5, P(B) = 0.35 and P(A and B) = 0.2, so P(A or B) = 0.65. For problem 4, students also find P(neither) = 0.35 and explain why it is 1 - P(A or B).
Closure5 minutes
Exit ticket: A spinner has 8 equal sections numbered 1 to 8. (1) Find P(odd or greater than 5). (Answer: 4/8 + 3/8 - 1/8 = 6/8 = 3/4, since only 7 is in both.) (2) Explain in one sentence why the answer cannot be 7/8. (3) Name two events on this spinner that are mutually exclusive.
Differentiation Strategies
For Struggling Students
Start every problem by drawing a Venn diagram with counts, and fill in the overlap first, then the "only" regions
Keep the answers as fractions with the same denominator (the total number of outcomes) until the last step
Give a sentence frame for interpretation: "If one ___ is chosen at random, the probability that it ___ or ___ is ___, which is ___ out of ___."
For Advanced Students
Ask students to write P(A or B) for three events and test their formula on the two-dice grid with A = doubles, B = sum 8 and C = first cube shows 4 (a challenge beyond the standard)
Ask: if P(A) = 0.55 and P(B) = 0.75, what are the largest and smallest possible values of P(A and B)? Explain with a Venn diagram
Ask students to prove that P(A or B) is never more than P(A) + P(B) and never less than the larger of P(A) and P(B)
Assessment Guidance
What to Look For
Check that students identify the outcomes in both events before they compute, rather than subtracting a number they guessed. A correct answer should come with a sentence that names the model (which card, which students, which roll) and says that "or" includes both. Watch for answers greater than 1 or smaller than one of the single-event probabilities: students who notice these on their own understand what the rule is doing. When events are mutually exclusive, ask students to justify P(A and B) = 0 from the outcomes, not from the fact that the answer "looks right".
02
Classroom Activities
3 Activities
1
Our Class in a Venn Diagram
20 minWhole class, then pairs
Students collect real yes/no data from the class, build a Venn diagram, and find P(A or B) for a randomly chosen classmate in two ways: by counting the union directly and with the Addition Rule. Using their own data makes the interpretation sentence concrete.
Procedure
Choose two yes/no questions, for example "Do you have a younger sibling?" and "Did you eat breakfast today?" Each student answers both on a sticky note
Students place their notes in a large two-circle Venn diagram on the board, with a space outside the circles for "neither"
Pairs record the four counts, then find P(A), P(B), P(A and B) and P(A or B) for a classmate picked at random
Each pair checks that P(A or B) from the rule equals the count of notes inside the circles divided by the class size
Discussion Questions
Which notes would be counted twice if we added the two circle totals?
Is there a pair of questions for which the circles could not overlap? What would the rule look like then?
Write a sentence that explains your P(A or B) to someone who was absent today
Modification for Distance Learning
Collect the answers with a two-question online poll and share the four counts on a slide. Pairs build the Venn diagram in a shared drawing tool.
2
Overlap or No Overlap? Card Sort
20 minGroups of 3-4
Groups receive 8 index cards, each describing a pair of events in a uniform model. They sort the cards into "mutually exclusive" and "overlapping", then compute P(A or B) for every card. The goal is to decide from the outcomes, not from the wording, whether P(A and B) is 0.
The 8 Cards
Card 1, one number cube: roll a 2 or a 5 (mutually exclusive, 2/6 = 1/3)
Card 2, one number cube: roll an odd number or a number less than 3 (overlap {1}, 3/6 + 2/6 - 1/6 = 2/3)
Card 3, one card from a deck: a club or a diamond (mutually exclusive, 26/52 = 1/2)
Card 4, one card from a deck: an ace or a black card (overlap 2 cards, 4/52 + 26/52 - 2/52 = 7/13)
Card 5, two number cubes: sum of 2 or sum of 3 (mutually exclusive, 1/36 + 2/36 = 1/12)
Card 6, two number cubes: both cubes even or sum of 6 (overlap (2, 4) and (4, 2), 9/36 + 5/36 - 2/36 = 1/3)
Card 7, tiles numbered 1 to 10: a multiple of 4 or an odd number (mutually exclusive, 2/10 + 5/10 = 7/10)
Card 8, tiles numbered 1 to 10: greater than 6 or even (overlap {8, 10}, 4/10 + 5/10 - 2/10 = 7/10)
Procedure
Each student takes two cards, lists the outcomes in A, in B and in both, and reports to the group
The group sorts the cards into two piles and computes P(A or B) for each card
Groups compare Cards 7 and 8, which have the same answer, and explain why one uses the subtraction and the other does not
Challenge Variation
Groups write two new cards for the same model: one pair of mutually exclusive events and one overlapping pair with the same P(A or B). Groups trade and check each other's cards.
3
Design a Spinner Game
25 minPairs
Pairs design a paper spinner with 12 equal sections, each labeled with a color and a number, and write an "or" event that wins the game. They compute the probability of winning with the Addition Rule and interpret it as the expected share of wins in 120 spins, then test it.
Procedure
Label the 12 sections with the numbers 1 to 12 and color each section blue or white, in any pattern the pair chooses
Write a winning rule of the form "the spinner lands on blue or on a number that is ___", chosen so that the chance of winning is between 1/2 and 2/3
Compute P(win) with the Addition Rule, showing P(A), P(B) and P(A and B) as fractions with denominator 12
Predict how many wins to expect in 120 spins, then spin 30 times and compare the class's combined results with the prediction
Discussion Questions
How can you change the coloring, without changing the rule, to make the game easier to win?
Why did your 30 spins not match the prediction exactly? What would happen with 3,000 spins?
Could a different pair have the same P(A) and P(B) as yours but a different P(win)? How?
Modification for Distance Learning
Use a free online spinner tool with 12 equal sections. Pairs share a screenshot of their spinner and their computation before spinning.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Why the Overlap Is Subtracted
Venn diagram for Example 2. Adding 90 sport players and 50 band members counts the 20 students in both groups twice, so the Addition Rule subtracts them once: 70 + 20 + 30 = 120 students are in at least one group. The circles are not drawn to scale by area.
Diagram 2: The Two-Dice Sample Space
The 36 equally likely outcomes of rolling two number cubes, drawn as a 6-by-6 grid (Example 4). The six doubles lie on the diagonal and the five sums of 8 lie on an anti-diagonal. They share one square, (4, 4), so it is subtracted once. Counting the shaded squares directly gives the same 10 outcomes.
04
Homework Assignment
~30 min
HSS.CP.B.7 Homework: The Addition Rule
Directions: Show P(A), P(B) and P(A and B) before you combine them. Give each probability as a simplified fraction or a decimal, and write one sentence that interprets each final answer in terms of the situation.
Part 1: Applying the Rule (Problems 1-3)
One card is drawn from a shuffled standard deck of 52 cards. Aces count as 1. Find the probability that the card is a spade or has a value less than 5. List the cards that are in both events.
In an invented group of 180 sophomores, 76 take art, 58 take a coding elective, and 22 take both. One sophomore is chosen at random. Draw a Venn diagram, then find P(art or coding) and P(neither).
Two fair number cubes are rolled. Find the probability that the product of the two numbers is even or the sum is greater than 9. Use a 6-by-6 grid to find P(A and B).
Part 2: Interpreting and Checking (Problems 4-6)
For a randomly chosen student at an invented school, P(plays video games weekly) = 0.62, P(reads for fun weekly) = 0.41, and P(does both weekly) = 0.23. Find the probability that the student does at least one of the two, and the probability that the student does neither. Interpret both answers for a school of 500 students.
A student says the probability of drawing a heart or a red card from a standard deck is 13/52 + 26/52 = 39/52. Explain the error using the outcomes in both events, and give the correct probability.
A bag holds 40 marbles of the same size: 12 red (5 of them striped), 10 blue (3 striped) and 18 green (6 striped). One marble is drawn at random. Find P(red or blue) and P(red or striped). Explain why the rule needs a subtraction for one of these but not for the other.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Identifying the Overlap
P(A and B) found from the outcomes in both events, or shown to be 0
Overlap found with one counting error
Overlap ignored or guessed
Applying the Rule
P(A) + P(B) - P(A and B) used correctly in every problem
Rule used correctly in most problems
Probabilities added or multiplied without the rule
Accuracy
All answers correct and simplified
Most answers correct
Most answers incorrect
Interpretation
Each answer explained in terms of the situation and checked for reasonableness
Interpretations present but vague
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which expression is the Addition Rule for P(A or B)?
Answer: B
The outcomes in both A and B are counted once in P(A) and again in P(B), so P(A and B) is subtracted once. Choice A is correct only for mutually exclusive events. Choice C is the product of the probabilities, which belongs to "and" questions about independent events, not "or" questions.
Question 2 of 20 · Multiple Choice
One card is drawn from a shuffled standard deck of 52 cards. What is the probability that it is a diamond or an ace?
Answer: C
P(diamond) = 13/52, P(ace) = 4/52, and the ace of diamonds is in both, so P = 13/52 + 4/52 - 1/52 = 16/52 = 4/13. Choice A forgets to subtract the ace of diamonds, which is counted twice. Choice B is only P(diamond and ace).
Question 3 of 20 · Multiple Choice
A fair 12-sided die has faces numbered 1 to 12. What is the probability of rolling a multiple of 3 or a number greater than 8?
Answer: A
Multiples of 3: {3, 6, 9, 12}, 4 outcomes. Greater than 8: {9, 10, 11, 12}, 4 outcomes. Both: {9, 12}. P = 4/12 + 4/12 - 2/12 = 6/12 = 1/2. Choice B adds 4 + 4 = 8 and counts 9 and 12 twice. Choice C is only the overlap, and choice D is only the multiples of 3.
Question 4 of 20 · Multiple Choice
An invented gym has 400 members: 220 take a yoga class, 150 take a spin class, and 100 take both. What is the probability that a randomly chosen member takes yoga or spin?
Answer: D
P = 220/400 + 150/400 - 100/400 = 270/400 = 27/40. Choice A adds the two class totals (220 + 150 = 370) without subtracting the 100 members counted twice. Choice B counts only the members who take spin but not yoga (50). Choice C is only P(yoga and spin).
Question 5 of 20 · Multiple Choice
One fair number cube is rolled. Which pair of events is mutually exclusive?
Answer: B
Numbers less than 3 are {1, 2} and numbers greater than 4 are {5, 6}: no outcome is in both, so P(A and B) = 0. Choice C is not mutually exclusive, because 3 and 5 are both odd and prime. In choices A and D, the single outcome 4 or 6 belongs to both events.
Question 6 of 20 · Multiple Choice
For two events in a probability model, P(A) = 0.6, P(B) = 0.5 and P(A or B) = 0.8. What is P(A and B)?
Answer: C
Solve the rule for the overlap: 0.8 = 0.6 + 0.5 - P(A and B), so P(A and B) = 1.1 - 0.8 = 0.3. Choice B adds all three probabilities. Choice D is 1 - 0.8, which is P(neither A nor B), not P(A and B). Choice A is P(A) - P(B), which does not use P(A or B) at all.
Question 7 of 20 · Multiple Choice
At an invented community center, the probability that a randomly chosen member uses the pool or the weight room is 0.72. Which statement interprets this correctly?
Answer: B
"Or" in probability is inclusive: the event contains members who use the pool only, the weight room only, and both. Choice A describes "and". Choice C leaves out the members who use both, and choice D is the complement, which is 1 - 0.72 = 0.28.
Question 8 of 20 · Multiple Choice
A student finds P(A) = 0.7 and P(B) = 0.6 in a probability model and reports P(A or B) = 1.3. What is the best response?
Answer: D
Because P(A or B) is at most 1, 0.7 + 0.6 - P(A and B) ≤ 1, so P(A and B) is at least 0.3 and the events must overlap. Choice A is wrong because two events with these probabilities cannot be mutually exclusive: their probabilities would add to more than 1. Choice B confuses "or" with the product rule for independent events.
Question 9 of 20 · Multiple Choice
Two fair number cubes are rolled. What is the probability that the sum is 5 or at least one cube shows a 1?
Answer: C
Sum of 5: (1, 4), (2, 3), (3, 2), (4, 1), 4 outcomes. At least one 1: 11 outcomes. Both: (1, 4) and (4, 1). P = 4/36 + 11/36 - 2/36 = 13/36. Choice A adds 4 + 11 without removing the two shared outcomes. Choice B counts only the outcomes with a 1, and choice D is only the overlap.
Question 10 of 20 · Multiple Choice
A tile is drawn at random from 30 tiles numbered 1 to 30. What is the probability that the number is a multiple of 3 or a multiple of 5?
Answer: A
There are 10 multiples of 3 and 6 multiples of 5; 15 and 30 are multiples of both. P = 10/30 + 6/30 - 2/30 = 14/30 = 7/15. Choice B uses 16/30 and counts 15 and 30 twice. Choice C is only the overlap, and choice D is only the multiples of 3.
Question 11 of 20 · Multiple Choice
In an invented group of 80 campers, 44 can swim, 30 can paddle a canoe, and 18 can do both. What is the probability that a randomly chosen camper can do neither?
Answer: D
Swim or canoe: 44 + 30 - 18 = 56 campers, so neither is 80 - 56 = 24 campers and P = 24/80 = 3/10. Choice A is P(swim or canoe), the complement of the question asked. Choice B subtracts 44 + 30 = 74 from 80 and forgets that 18 campers were counted twice.
Question 12 of 20 · Multiple Choice
A Venn diagram of an invented class of 50 students shows 16 in A only, 6 in both A and B, 11 in B only, and 17 in neither. What is P(A or B) for a randomly chosen student?
Answer: B
The students in A or B are the three regions inside the circles: 16 + 6 + 11 = 33, so P = 33/50. Using the rule gives the same result: 22/50 + 17/50 - 6/50 = 33/50. Choice A adds the overlap twice (16 + 6 + 11 + 6), and choice C leaves the overlap out. Choice D is P(neither).
Question 13 of 20 · Multiple Choice
At an invented bike shop, every receipt lists exactly one item. For a randomly chosen receipt, P(helmet) = 0.15 and P(lock) = 0.42. What is P(helmet or lock)?
Answer: C
A receipt with one item cannot show both a helmet and a lock, so the events are mutually exclusive and P = 0.15 + 0.42 - 0 = 0.57. Choice A multiplies the probabilities, which answers a different question. Choice D is 1 - 0.57, the probability that the receipt is for some other item.
Question 14 of 20 · Multiple Choice
Which set of values is impossible for two events in a probability model?
Answer: A
The event "A or B" contains every outcome of A, so P(A or B) can never be smaller than P(A) = 0.5. In choice A the rule would give P(A and B) = 0.6, which is larger than P(B) = 0.4 and so impossible. Choice C is possible: it is the mutually exclusive case, with P(A and B) = 0.
Question 15 of 20 · Short Answer
One card is drawn from a shuffled standard deck. Find the probability that it is black or a number card (2 through 10), and interpret the answer.
P(black) = 26/52 and P(number card) = 36/52. The black number cards are 2 through 10 in clubs and spades, 18 cards. P = 26/52 + 36/52 - 18/52 = 44/52 = 11/13. Interpretation: only the 8 red cards that are not number cards (6 red face cards and 2 red aces) are left out, so about 85% of draws give a black card or a number card.
Question 16 of 20 · Short Answer
A spinner has 10 equal sections numbered 1 to 10. Find P(prime or even) and list the outcome that is in both events.
Primes: {2, 3, 5, 7}, so P = 4/10. Even: {2, 4, 6, 8, 10}, so P = 5/10. Both: only 2. P(prime or even) = 4/10 + 5/10 - 1/10 = 8/10 = 4/5. Only the sections 1 and 9 are in neither event.
Question 17 of 20 · Short Answer
At an invented music school of 150 students, 54 play piano, 45 play guitar and 21 play both. One student is chosen at random. Find P(piano or guitar) and P(neither), and interpret P(neither).
P(piano or guitar) = 54/150 + 45/150 - 21/150 = 78/150 = 13/25. P(neither) = 1 - 13/25 = 12/25. Interpretation: 72 of the 150 students play neither piano nor guitar, so a randomly chosen student has a 48% chance of playing some other instrument (or none).
Question 18 of 20 · Short Answer
At an invented coffee shop, for a randomly chosen customer, P(buys a pastry) = 0.38, P(buys an iced drink) = 0.47 and P(buys a pastry or an iced drink) = 0.71. Find the probability that the customer buys both, and interpret it.
0.71 = 0.38 + 0.47 - P(both), so P(both) = 0.85 - 0.71 = 0.14. Interpretation: about 14% of customers buy a pastry and an iced drink. Out of 200 customers, the model predicts about 28 buy both.
Question 19 of 20 · Short Answer
Two fair number cubes, one red and one blue, are rolled. Find the probability that the red cube shows an even number or the sum is 7.
A raffle has 500 tickets numbered 1 to 500, and one ticket is drawn at random. Find the probability that the winning number ends in 0 or is 451 or greater, and interpret the answer.
Ends in 0: 50 tickets. 451 or greater: 50 tickets. Both: 460, 470, 480, 490, 500, which is 5 tickets. P = 50/500 + 50/500 - 5/500 = 95/500 = 19/100. Interpretation: 95 of the 500 tickets win under this rule, so a buyer of one ticket has a 19% chance that it is one of them.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.B.7 mean?
HSS.CP.B.7 means students can find the probability of "A or B" with the Addition Rule, P(A or B) = P(A) + P(B) - P(A and B), and explain what the answer says about the situation. The models are uniform: every outcome, such as a card, a roll or a student chosen at random, is equally likely. The code stands for High School Statistics, Conditional Probability and the Rules of Probability, cluster B, standard 7.
Why do you subtract P(A and B) in the Addition Rule?
Because the outcomes in both events are counted twice when you add P(A) and P(B). The ace of spades, for example, is counted once among the aces and once among the spades. Subtracting P(A and B) removes the second copy, so every outcome in "A or B" is counted exactly once. A Venn diagram with counts in each region shows this clearly.
Does "or" in probability mean one or the other but not both?
No. In probability, "A or B" is inclusive: it means A happens, B happens, or both happen. In set language it is the union of A and B. If a problem wants "exactly one of the two", it will say so, and that probability is P(A or B) - P(A and B).
When can you just add the two probabilities?
Only when the events are mutually exclusive, meaning no outcome is in both. Then P(A and B) = 0 and the rule becomes P(A or B) = P(A) + P(B). Drawing a heart or a spade on one draw is an example: no card is both. Students should justify P(A and B) = 0 by checking the outcomes, not assume it because the events have different names.
Is HSS.CP.B.7 taught in Geometry or Algebra 2?
It depends on the course sequence. In many traditional sequences the conditional probability standards, including HSS.CP.B.7, are part of Geometry; in others they appear in Algebra II or in an integrated Math II or III course. Unlike HSS.CP.B.8 and HSS.CP.B.9, this standard does not carry the (+) symbol, so it is expected of all students.
What does "interpret the answer in terms of the model" mean?
It means saying what the probability tells you about the actual situation, not only giving a number. For a model of 300 students where P(A or B) = 0.35, a good interpretation is: "A randomly chosen student has a 35% chance of being in at least one of the two groups; that is about 105 of the 300 students." Interpretation also includes a reasonableness check: P(A or B) must be at least as large as each of P(A) and P(B) and no larger than 1.
How do Venn diagrams and two-way tables help with the Addition Rule?
Both show the overlap directly. In a Venn diagram, P(A and B) is the region where the circles overlap. In a two-way table, it is the cell where the row for A meets the column for B. Students can then check the rule by adding the separate regions or cells that make up "A or B" and getting the same answer.
What are common mistakes with the Addition Rule?
Common errors include:
Adding P(A) and P(B) without subtracting the overlap, which can give a probability greater than 1
Subtracting the overlap twice after already counting only the "A only" and "B only" regions
Multiplying the probabilities, which answers an "and" question for independent events
Reading "or" as "exactly one"
Can the Addition Rule be used with three events?
Yes, as an extension beyond the standard. For three events, P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C). The last term is added back because the outcomes in all three events are subtracted too many times. HSS.CP.B.7 itself only asks for two events.
Is HSS.CP.B.7 tested on the SAT?
Probability appears in the Problem-Solving and Data Analysis domain of the digital SAT, often with data in a two-way table. A question may ask for the probability that a randomly chosen person is in one row or one column, which is an Addition Rule question even when the formula is not named. Reading the table carefully and not counting a cell twice is the key skill.
07
Related Standards
6 standards
These standards connect to HSS.CP.B.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.SP.C.8Prerequisite
Find probabilities of compound events using lists, tables, tree diagrams and simulation
