HSS.CP.A.2Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.A.2: Independent Events and the Product Test
In plain English: HSS.CP.A.2 is the Common Core statistics and probability standard that defines independent events: A and B are independent exactly when the probability that both occur equals the product of their probabilities, P(A and B) = P(A) · P(B). Students use this product test on sample spaces, given probabilities and data tables. It is usually taught in Geometry or Algebra II.
Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent.
Common Core State Standards for Mathematics · Domain: Conditional Probability and the Rules of Probability (CP) · Cluster: Understand independence and conditional probability and use them to interpret data Also written as HSS-CP.A.2 or S-CP.2 · Official standard
Students meet the formal definition of independence: two events A and B are independent if P(A and B) = P(A) · P(B). The lesson treats the equation as a test. Students compute the two sides separately, compare them, and state a conclusion, rather than deciding by intuition whether one event "affects" the other.
Students apply the test in three settings: events on a uniform sample space (including two events on a single roll, which surprises many students), events given only by their probabilities, and events defined on data in a two-way table. They also use the definition the other way round, multiplying probabilities when events are known to be independent, and see why events that cannot occur together are never independent when both have positive probability.
Learning Objectives
By the end of this lesson, students will be able to:
State that A and B are independent exactly when P(A and B) = P(A) · P(B)
Decide whether two events in a sample space are independent by computing and comparing P(A and B) and P(A) · P(B)
Apply the same test to events given by probabilities and to events defined on data in a table
Use P(A and B) = P(A) · P(B) to find a probability when events are known to be independent
Explain why independent events and mutually exclusive events are different ideas
Prior Knowledge Required
Students should already be comfortable with:
Describing events with "and", "or" and "not" as intersections, unions and complements HSS.CP.A.1
Finding probabilities in a uniform probability model 7.SP.C.7
Listing sample spaces of compound events 7.SP.C.8
Multiplying fractions and decimals, and converting between them
Pairs answer the three questions, then the class compares.
Warm-Up Prompt
"A coin is flipped twice. (1) List the sample space and find the probability of heads on both flips. (2) Multiply the probability of heads on the first flip by the probability of heads on the second flip. (3) If the first flip is heads, does that change your prediction for the second flip?"
Answers: (1) S = {HH, HT, TH, TT}, so P(HH) = 1/4. (2) 1/2 · 1/2 = 1/4. (3) No. Tell students that today they will turn the match between (1) and (2) into a definition, and that the definition, not the story about coins, is what decides independence. Then ask: "Do you think two events on the same single roll of a number cube could be independent?" Collect predictions and come back to them in Example 1.
Direct Instruction20 minutes
Definition. Two events A and B are independent if P(A and B) = P(A) · P(B). If the two sides are not equal, A and B are not independent (dependent). To use the definition as a test:
Find P(A) and P(B) from the sample space, the given information or the data.
Find P(A and B) from the outcomes that are in both events, not by multiplying.
Multiply P(A) · P(B).
Compare: if P(A and B) equals the product, the events are independent; otherwise they are not.
State the conclusion in the words of the problem.
Stress step 2: the product test only means something if P(A and B) is found on its own. Students who multiply to get P(A and B) will "prove" that every pair of events is independent. Work through the examples, using Diagram 2 for the first two.
Independent events on one roll
A number cube is rolled once. A = "the number is even" = {2, 4, 6} and L = "the number is at most 2" = {1, 2}.
Equation: P(A and L) = P({2}) = 1/6 and P(A) · P(L) = 1/2 · 1/3 = 1/6, so A and L are independent
Not independent on the same roll
Same roll. A = "even" and G = "the number is greater than 3" = {4, 5, 6}.
Equation: P(A and G) = P({4, 6}) = 1/3, but P(A) · P(G) = 1/2 · 1/2 = 1/4, so A and G are not independent
Testing given probabilities
For two events, P(A) = 0.6, P(B) = 0.5 and P(A and B) = 0.35.
Equation: 0.6 · 0.5 = 0.30 and 0.30 ≠ 0.35, so A and B are not independent
Testing events in a data table
In the invented survey of 200 students below, M = "birthday in January-June" and P = "has a pet".
Equation: P(M and P) = 60/200 = 0.30 and P(M) · P(P) = 0.50 · 0.60 = 0.30, so M and P are independent in this data
Using independence to find P(A and B)
A store has two self-checkout scanners that fail independently. On a given day, P(scanner 1 fails) = 0.02 and P(scanner 2 fails) = 0.05.
Equation: P(both fail) = 0.02 · 0.05 = 0.001
Invented survey of 200 students: birthday half-year and pets
Has a pet
No pet
Total
Birthday January-June
60
40
100
Birthday July-December
60
40
100
Total
120
80
200
Use Diagram 1 to show what the test means geometrically: when the events are independent, B takes up the same share of A as it does of the whole sample space. Finish with a warning: independent does not mean "cannot happen together". Events that cannot happen together (mutually exclusive events) have P(A and B) = 0, so if both have positive probability, the product test fails and they are not independent.
Guided Practice15 minutes
Pairs test each table, then two pairs compare conclusions. First, an invented record of 160 coffee orders:
Invented record of 160 coffee orders
Large
Small
Total
Iced
40
24
64
Hot
60
36
96
Total
100
60
160
Test I = "iced" and Z = "large": P(I) = 64/160 = 0.4, P(Z) = 100/160 = 0.625, P(I and Z) = 40/160 = 0.25, and 0.4 · 0.625 = 0.25, so I and Z are independent. Second, an invented survey of 150 students: 60 play a school sport (S), 50 have a part-time job (J), and 18 do both. Here P(S and J) = 18/150 = 0.12, but P(S) · P(J) = 0.4 · (1/3) ≈ 0.133, so S and J are not independent. Ask pairs why the second conclusion depends on having the exact counts. Note for the class: with real survey data the two sides are rarely exactly equal, and later work (HSS.CP.A.4) uses tables to judge approximate independence. Today the table is treated as the whole sample space, so the test is exact.
Independent Practice10-15 minutes
A spinner has 8 equal sections numbered 1-8. Let A = {1, 2, 3, 4}, B = "even", C = "prime" = {2, 3, 5, 7} and D = {7, 8}. Students test five pairs and show both sides of the equation each time. Answers: A and B are independent (1/4 = 1/2 · 1/2); A and C are independent (P({2, 3}) = 1/4 = 1/2 · 1/2); A and D are not independent (they share no outcomes, so 0 ≠ 1/2 · 1/4); B and C are not independent (P({2}) = 1/8 ≠ 1/4); B and D are independent (P({8}) = 1/8 = 1/2 · 1/4).
Closure5-10 minutes
Exit ticket: (1) P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2. Are A and B independent? Show the test. (Yes: 0.5 · 0.4 = 0.2.) (2) Two events have P(E) = 0.3 and P(F) = 0.2 and cannot happen at the same time. Are they independent? (No: P(E and F) = 0, but 0.3 · 0.2 = 0.06.)
Differentiation Strategies
For Struggling Students
Give a three-column organizer for every test: "P(A and B) from the outcomes", "P(A) · P(B)", "Equal?"
Keep probabilities as fractions with the same denominator as the sample space (for example sixths for a number cube) until the comparison is made
Shade events on a strip of outcome cells, as in Diagram 2, before computing anything
For Advanced Students
Ask students to prove that if A and B are independent, then A and Bᶜ are also independent, starting from P(A) = P(A and B) + P(A and Bᶜ)
On a 10-section spinner numbered 1-10, ask for every event B with exactly 4 outcomes that is independent of A = "odd", and to explain the pattern
Ask whether an event can be independent of itself, and for which probabilities that happens
Assessment Guidance
What to Look For
Check that students find P(A and B) from the outcomes or the data before they multiply, and that every conclusion is backed by both sides of the equation written out. Listen for reasoning such as "they are independent because they are different things" or "they are dependent because they overlap"; ask the student to run the test instead. When students use independence to find a probability, check that the problem actually says or implies that the events are independent.
02
Classroom Activities
3 Activities
1
Coin and Cube Experiment
20 minPairs
Pairs run a chance process 60 times and check whether the product rule holds for their relative frequencies, then pool results with the class.
Procedure
Each trial is one coin flip and one roll of a number cube. Let H = "heads" and S = "the cube shows 5 or 6"
Before starting, pairs compute the theoretical values: P(H) = 1/2, P(S) = 1/3 and, since the coin and cube do not influence each other, P(H and S) = 1/6, so about 10 of 60 trials
Pairs run 60 trials and tally H, S and "H and S"
Pairs compute the relative frequencies of H, S and "H and S", and compare the frequency of "H and S" with the product of the other two
The class pools all trials and repeats the comparison
Discussion Questions
Why did no pair get exactly 1/6 for "H and S"? Did the pooled class data come closer?
What would it mean if, over many trials, "H and S" happened much more often than the product predicts?
Modification for Distance Learning
Students use a random number generator (1-2 for the coin, 1-6 for the cube) and enter results in a shared spreadsheet that computes the class totals.
2
Independence Detective
20 minGroups of 3-4
Groups test pairs of events that all come from a single roll of a number cube and sort them into "independent" and "not independent" piles.
Event Cards
E = even {2, 4, 6}; O = odd {1, 3, 5}; L = at most 2 {1, 2}
G = at least 5 {5, 6}; P = prime {2, 3, 5}; M = multiple of 3 {3, 6}
Procedure
Groups test these 8 pairs with the product test, writing both sides: E and L, E and G, E and P, E and M, L and G, P and M, O and L, G and P
Key: independent are E and L, E and G, E and M, P and M, O and L, G and P (each shares exactly one outcome, and 1/6 = 1/2 · 1/3); not independent are E and P (1/6 ≠ 1/4) and L and G (0 ≠ 1/9)
Each group writes a rule that explains which pairs passed
Discussion Questions
Every independent pair here is a 1/2 event with a 1/3 event. Why must they share exactly one outcome?
Why did L and G fail even though they "have nothing to do with each other"?
Challenge Variation
Groups make their own event cards for an 8-section spinner numbered 1-8 and find one independent pair and one dependent pair in which both events have probability 1/2.
3
Class Survey: Are These Independent?
20 minWhole class, then groups of 3-4
The class collects its own data on two yes-or-no questions, builds a table, and applies the product test to the class as the sample space.
Procedure
Each student writes two answers on a sticky note: Do you have at least one sibling? Do you prefer mornings to evenings?
Students place their notes on a 2-by-2 grid on the board, and the class writes the counts and totals
Groups compute P(sibling), P(morning), P(sibling and morning) and the product, and decide whether the events are independent for this class
If the class is small, groups also test this invented backup data: 40 students, 30 with a sibling, 16 who prefer mornings, 12 with both. (P(both) = 12/40 = 0.30 and 0.75 · 0.40 = 0.30, so independent.)
Discussion Questions
Our two sides were close but not equal. Are the events independent for this class? What would you want to know before saying anything about all students?
Suggest two survey questions whose answers you would expect not to be independent, and explain why
Modification for Distance Learning
Collect the two answers with a short online form and share the summary table on screen.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Independence as an Area Model
The unit square is the sample space, drawn to scale. The A column has width 0.4, and the shaded region B has total area 0.25 in both squares. On the left, B takes up the same share of the A column as of the rest, so the dark rectangle A and B has area 0.4 · 0.25 = 0.10 and the events are independent. On the right, B is concentrated inside A, the dark rectangle has area 0.4 · 0.4 = 0.16, and the product test fails.
Diagram 2: Testing Two Pairs of Events on One Roll
Shaded cells are the outcomes in each event. Knowing that the roll was at most 2 leaves one even outcome out of two, the same one-half chance as before, and the product test holds. Knowing that the roll was greater than 3 makes "even" more likely (two out of three), and the product test fails. These are Worked Examples 1 and 2.
04
Homework Assignment
~30 min
HSS.CP.A.2 Homework: Testing for Independence
Directions: For every test, write P(A and B) and P(A) · P(B) as separate values, compare them, and state your conclusion in a sentence. Give probabilities as fractions or decimals. Use independence to multiply only when the problem says the events are independent.
Part 1: Using the Definition (Problems 1-3)
Decide whether each pair of events is independent. (a) P(A) = 0.7, P(B) = 0.2 and P(A and B) = 0.14. (b) P(C) = 0.5, P(D) = 0.5 and P(C and D) = 0.3.
A card is drawn from 20 cards numbered 1-20. Let A = "a multiple of 4", B = "at most 8", C = "a multiple of 5" and D = "greater than 15". Test A and B, A and C, and A and D for independence.
Two number cubes are rolled. Let A = "the first cube shows 6", B = "the sum is 7" and C = "the sum is 8". (a) Are A and B independent? (b) Are A and C independent? Show the test for each.
Part 2: Independence in Data (Problems 4-5)
In an invented survey of 300 adults, 120 exercise at least three times a week (E), 180 sleep at least 7 hours a night (S), and 72 do both. Using these 300 adults as the sample space, are E and S independent? Show the test.
In an invented survey of 80 students, 32 live more than 5 miles from school (F), 40 were late at least once this month (T), and 20 are in both groups. (a) Make a two-way table of the counts. (b) Are F and T independent for these students? (c) Explain what your answer means in context.
Part 3: Using Independence (Problem 6)
A home has two smoke detectors that fail independently of each other. Each one works with probability 0.95. (a) Find the probability that both work. (b) Find the probability that both fail. (c) Explain why you could not answer (a) this way if the two detectors ran on the same battery.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
P(A and B) Found Directly
Found from outcomes or data in every test
Found directly in most tests
Found by multiplying
Product Test
Both sides shown and compared correctly
Both sides shown, one comparison wrong
Test not shown
Conclusion
Correct and stated in context
Correct but not in context
Missing or incorrect
Using Independence
Multiplies only when independence is given, with a reason
Correct values, reason missing
Incorrect or unjustified
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Try every question on paper first, then pick an option to check it. The counter tracks multiple-choice answers, and Reset quiz clears your choices.
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
Events A and B are independent exactly when which equation is true?
Answer: D
By definition, A and B are independent when the probability that both occur equals the product of their probabilities. Choice A adds the probabilities, which does not describe independence. Choice B describes events that cannot happen together, which are not independent when both have positive probability.
Question 2 of 20 · Multiple Choice
A and B are independent events with P(A) = 0.3 and P(B) = 0.5. What is P(A and B)?
Answer: B
For independent events, P(A and B) = P(A) · P(B) = 0.3 · 0.5 = 0.15. Choice A adds the probabilities. Choice C subtracts them, and choice D divides 0.3 by 0.5.
Question 3 of 20 · Multiple Choice
For two events, P(A) = 0.25, P(B) = 0.6 and P(A and B) = 0.15. Which conclusion is correct?
Answer: C
The product 0.25 · 0.6 = 0.15 equals P(A and B), so the events are independent. Choice A tests a sum, which is not the definition. Choice B is irrelevant: independent events can have different probabilities. Choice D is true of every intersection, so it proves nothing.
Question 4 of 20 · Multiple Choice
For two events, P(A) = 0.55, P(B) = 0.3 and P(A and B) = 0.1. Which conclusion is correct?
Answer: B
P(A) · P(B) = 0.55 · 0.3 = 0.165, but P(A and B) = 0.1, so the test fails and the events are not independent. Choice A uses a fact that is true of every intersection. Choice C computes a quantity that has nothing to do with the definition. Choice D is wrong because the three probabilities are all the test needs.
Question 5 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. Are "the card is an ace" and "the card is a heart" independent?
Answer: A
Only the ace of hearts is in both events, so P(ace and heart) = 1/52, and 4/52 · 13/52 = 1/13 · 1/4 = 1/52. The two sides match, so the events are independent. Choice B confuses independence with having no overlap. Choice C multiplies incorrectly, and choice D is false because the ace of hearts is both.
Question 6 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. Are "the card is red" and "the card is a heart" independent?
Answer: C
Every heart is red, so "red and heart" is just "heart", with probability 13/52 = 1/4. The product is 1/2 · 1/4 = 1/8, which is different, so the events are not independent. Choice A computes the product but never compares it with P(red and heart). Choice D is false: a heart is both red and a heart.
Question 7 of 20 · Multiple Choice
A fair 12-sided die (numbers 1-12) is rolled. Are A = "a multiple of 3" and B = "even" independent?
Answer: B
A = {3, 6, 9, 12} and B = {2, 4, 6, 8, 10, 12}, so A and B = {6, 12} with probability 1/6, which equals 1/3 · 1/2. Choice A treats overlap as evidence of dependence. Choice C uses 1/2 for P(A) instead of 4/12 = 1/3, and choice D is false because 6 and 12 are shared.
Question 8 of 20 · Multiple Choice
A fair 12-sided die is rolled. Let A = "a multiple of 4" and C = "greater than 6". What is P(A and C), and are A and C independent?
Answer: D
A = {4, 8, 12} and C = {7, 8, 9, 10, 11, 12}, so A and C = {8, 12} and P(A and C) = 2/12 = 1/6. The product is 1/4 · 1/2 = 1/8. Since 1/6 ≠ 1/8, the events are not independent. Choices A and C give the product instead of the probability of the intersection.
Question 9 of 20 · Multiple Choice
Two events have P(A) = 0.3 and P(B) = 0.4 and cannot happen at the same time. Are they independent?
Answer: B
Because the events cannot happen together, P(A and B) = 0, while P(A) · P(B) = 0.3 · 0.4 = 0.12. The test fails. Choice A is a frequent confusion of "mutually exclusive" with "independent". Choices C and D use facts that have nothing to do with the definition.
Question 10 of 20 · Multiple Choice
In an invented group of 100 people, 40 own a bike, 50 live in the city, and 20 do both. For one person chosen at random, are "owns a bike" and "lives in the city" independent?
Answer: A
P(both) = 20/100 = 0.20 and the product of the separate probabilities is 0.40 · 0.50 = 0.20, so the events are independent for this group. Choice C tests a sum instead of a product. Choice D is wrong because counts give all three probabilities.
Question 11 of 20 · Multiple Choice
A and B are independent, P(A) = 0.8 and P(A and B) = 0.2. What is P(B)?
Answer: C
Independence gives 0.2 = 0.8 · P(B), so P(B) = 0.2/0.8 = 0.25. Choice A multiplies 0.8 · 0.2. Choice B subtracts, and choice D divides the wrong way and is not a probability.
Question 12 of 20 · Multiple Choice
A laptop has a main battery and a backup battery that work independently. On a given day the main battery works with probability 0.9 and the backup with probability 0.95. What is the probability that both work?
Answer: A
Because the batteries work independently, P(both work) = 0.9 · 0.95 = 0.855. Choice B adds and is greater than 1, so it cannot be a probability. Choice C averages the two values, and choice D subtracts them.
Question 13 of 20 · Multiple Choice
A coin is flipped three times. Let A = "the first flip is heads" and B = "exactly two of the three flips are heads". Which is correct?
Answer: C
Of the 8 outcomes, A and B = {HHT, HTH}, so P(A and B) = 2/8 = 1/4. P(A) = 1/2 and P(B) = 3/8 ({HHT, HTH, THH}), so the product is 3/16. The two sides differ. Choice A assumes that independent flips make every pair of events independent; B depends on the first flip. Choice D gives P(B), not P(A and B).
Question 14 of 20 · Multiple Choice
In a group of 60 students, 24 play an instrument and 15 are in grade 9. How many grade-9 students must play an instrument for the two events to be independent?
Answer: D
Independence needs P(both) = (24/60)(15/60) = 0.4 · 0.25 = 0.1, and 0.1 · 60 = 6 students. Choice A is 24 - 15, choice B assumes every grade-9 student plays, and choice C adds the two counts.
Question 15 of 20 · Short Answer
In an invented study of 250 shoppers, 100 used a coupon (C), 150 bought online (O), and 70 did both. Using these shoppers as the sample space, are C and O independent? Show the test.
P(C and O) = 70/250 = 0.28. P(C) · P(O) = 0.4 · 0.6 = 0.24. Since 0.28 ≠ 0.24, C and O are not independent: in this data, using a coupon and buying online happen together more often than independence would predict.
Question 16 of 20 · Short Answer
A card is drawn from 10 cards numbered 1-10. Let A = "even" and B = "at most 4". Are A and B independent? Show the test.
A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4}, so A and B = {2, 4}. P(A and B) = 2/10 = 1/5 and P(A) · P(B) = 1/2 · 4/10 = 1/5. They are independent.
Question 17 of 20 · Short Answer
A and B are independent, with P(A) = 0.35 and P(B) = 0.6. Find P(A and B) and P(A and not B).
P(A and B) = 0.35 · 0.6 = 0.21. The outcomes of A are either in B or not in B, so P(A and not B) = P(A) - P(A and B) = 0.35 - 0.21 = 0.14.
Question 18 of 20 · Short Answer
A fair 12-sided die is rolled, and A = "the number is at most 6". (a) Find an event B with exactly 4 outcomes that is independent of A, and show the test. (b) Explain why no event with exactly 3 outcomes can be independent of A.
(a) Independence needs P(A and B) = 1/2 · 4/12 = 1/6 = 2/12, so exactly 2 of the 4 outcomes of B must be at most 6. One answer: B = {1, 2, 7, 8}, with P(A and B) = P({1, 2}) = 2/12 = 1/6 = P(A) · P(B). (b) For a 3-outcome event, P(A) · P(B) = 1/2 · 3/12 = 1/8 = 1.5/12, but P(A and B) is always a whole number of twelfths, so the two sides can never be equal.
Question 19 of 20 · Short Answer
A basketball player makes 80% of her free throws. Assume the result of each shot is independent of the other. Find the probability that she (a) makes both of two free throws and (b) makes the first and misses the second.
(a) 0.8 · 0.8 = 0.64. (b) P(miss) = 1 - 0.8 = 0.2, so 0.8 · 0.2 = 0.16. Both answers use the product rule, which is allowed only because the shots are assumed independent.
Question 20 of 20 · Short Answer
A survey company reports that P(A) = 0.45, P(B) = 0.8 and P(A and B) = 0.36 for two events in its data. Are A and B independent? Explain what the result means.
P(A) · P(B) = 0.45 · 0.8 = 0.36, which equals P(A and B), so A and B are independent in this data. The share of A among the whole group is the same as it would be if the two events had no connection.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.A.2 mean?
HSS.CP.A.2 means students know the definition of independent events and can use it. Two events A and B are independent when P(A and B) = P(A) · P(B), and students check this equation to decide whether a given pair of events is independent.
What is the formula for independent events?
The formula is P(A and B) = P(A) · P(B). It works in two directions. As a test, you compute both sides separately and compare them. As a rule, when a problem says two events are independent, you multiply their probabilities to get the probability that both happen.
Is independent the same as mutually exclusive?
No. Mutually exclusive events cannot happen together, so P(A and B) = 0. Independent events can happen together, and they do so at exactly the rate P(A) · P(B). If both events have positive probability, mutually exclusive events are never independent: knowing that one happened tells you the other did not.
How do you check whether two events are independent using a two-way table?
Use the table as the sample space and find three probabilities from the counts:
P(A) = row or column total for A divided by the grand total
P(B) = total for B divided by the grand total
P(A and B) = the cell count for both divided by the grand total
Then compare P(A and B) with P(A) · P(B).
Can two events on the same roll of a number cube be independent?
Yes. On one roll, "even" = {2, 4, 6} and "at most 2" = {1, 2} are independent, because P({2}) = 1/6 and 1/2 · 1/3 = 1/6. Independence is a property of the probabilities, not of whether the events come from separate trials.
What if P(A and B) is close to P(A) · P(B) but not equal?
Under the definition, the events are not independent unless the two sides are exactly equal. With real data, the sides are rarely exactly equal even when there is no real connection, so statisticians speak of events being approximately independent in a sample. Judging that from a two-way table is part of HSS.CP.A.4.
Is HSS.CP.A.2 on the SAT?
Not by name. The digital SAT does not list Common Core codes, but its Problem-Solving and Data Analysis domain includes probability and conditional probability questions, many of them based on two-way tables. Reading a table and computing P(A and B) are skills this standard practices.
Why does the standard define independence with a product instead of "one event does not affect the other"?
The product gives a test that can be checked exactly. "Does not affect" is a good intuition, but it is vague, and it leads students to judge by the story instead of the numbers. The next standard, HSS.CP.A.3, connects the two ideas: independence also means that knowing B happened does not change the probability of A.
What mistakes do students make with independent events?
Finding P(A and B) by multiplying and then "confirming" independence with the same product
Adding P(A) and P(B) instead of multiplying
Calling overlapping events dependent, or non-overlapping events independent
Assuming independence in a problem that does not state it
How does HSS.CP.A.2 connect to later math?
It is the basis for the multiplication rules of HSS.CP.B.8, and for later probability models in which repeated independent trials are multiplied, such as the chance of a particular sequence of coin flips. It also prepares students for statistics courses, where tests of independence in two-way tables compare observed counts with the counts that the product rule predicts.
07
Related Standards
6 standards
These standards connect to HSS.CP.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSS.CP.A.1Prerequisite
Describe events as subsets of a sample space and as unions, intersections, complements
