HSS.CP.B.9Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.B.9: Probability with Permutations and Combinations
In plain English: HSS.CP.B.9 is an advanced (+) Common Core statistics and probability standard that asks students to use permutations and combinations to count outcomes, compute probabilities of compound events and solve problems. The key step is deciding whether order matters, then counting favorable and total outcomes the same way. It is usually taught in Algebra II, Precalculus or a statistics course.
(+) Use permutations and combinations to compute probabilities of compound events and solve problems.
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.9 or S-CP.9 · Official standard
Students use permutations (arrangements, where order matters) and combinations (groups, where order does not matter) to count the outcomes of a chance process, and then use those counts to find probabilities of compound events. In a uniform model, P(event) = (number of favorable outcomes)/(number of possible outcomes), and permutations and combinations make both counts possible when listing every outcome is not.
The lesson builds the formulas P(n, r) = n!/(n - r)! and C(n, r) = n!/(r!(n - r)!) from slot diagrams, then applies them to committees, line-ups, codes and card hands. Throughout, students must decide whether order matters and count the favorable and total outcomes in the same way. They check some answers with the general Multiplication Rule and interpret each result in the situation.
Learning Objectives
By the end of this lesson, students will be able to:
Decide whether a counting situation calls for a permutation or a combination and justify the choice
Compute P(n, r) and C(n, r) with the formulas, by hand and with a calculator, and explain why C(n, r) = P(n, r)/r!
Use permutations to find probabilities of compound events involving order, such as line-ups and ranked results
Use combinations to find probabilities of compound events involving groups, such as committees and card hands
Solve multi-step problems with these counts and interpret the probability in context
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 subsets of a sample space HSS.CP.A.1
The general Multiplication Rule for P(A and B) HSS.CP.B.8, for checking answers
Post the questions and have students list every outcome on paper.
Warm-Up Prompt
"Ava, Ben and Cal sit in a row of 3 seats. (1) In how many different ways can they sit? (2) Two of the three will be sent to the store. How many different pairs are possible? (3) If they sit in a random order, what is the probability that Ava sits in the middle?"
Collect the lists. Part (1) has 6 seatings: ABC, ACB, BAC, BCA, CAB, CBA. Part (2) has only 3 pairs, because "Ava and Ben" is the same pair as "Ben and Ava". For part (3), Ava is in the middle in 2 of the 6 seatings, so the probability is 2/6 = 1/3. Ask: what made part (1) larger than part (2)? Name the ideas: when order matters we count permutations, and when it does not we count combinations. Explain that the lesson is about counting like this when the lists would be far too long to write.
Direct Instruction20 minutes
Part 1: The counting formulas. Use Diagram 2. For r ordered choices from n different objects, the slot diagram gives n(n - 1)(n - 2)... with r factors, which is P(n, r) = n!/(n - r)!. Every group of r objects can be ordered in r! ways (Diagram 1), so the number of groups is C(n, r) = P(n, r)/r! = n!/(r!(n - r)!). Remind students that 0! = 1, so P(n, n) = n! and C(n, n) = 1.
Part 2: From counts to probability. In a uniform model, follow the steps below.
Describe one outcome: is it an ordered arrangement (a line-up, a ranking, a code) or an unordered group (a committee, a hand of cards)?
Count all possible outcomes with P(n, r) or C(n, r).
Count the favorable outcomes the same way. For "exactly k of one kind", multiply the ways to choose from each kind, for example C(6, 2) · C(4, 1).
Divide: P(event) = favorable/total, and simplify.
Interpret and check: say what the probability means, and, where it is quick, check with the Multiplication Rule.
Permutation count
Eight runners are in a race. In how many ways can gold, silver and bronze be awarded? (Diagram 2)
Equation: P(8, 3) = 8 · 7 · 6 = 336
Combination count
A class chooses 4 of its 10 volunteers for a planning committee with no roles.
Equation: C(10, 4) = 10!/(4! 6!) = 210
Probability with combinations
A committee of 3 is chosen at random from 6 juniors and 5 seniors. Find P(all three are juniors).
Equation: C(6, 3)/C(11, 3) = 20/165 = 4/33
Probability with permutations
Five students, including Ana and Ben, line up in a random order. Find P(Ana is first and Ben is second).
Equation: 3!/5! = 6/120 = 1/20
Card hand
A 5-card hand is dealt from a shuffled standard deck. Find P(all five cards are hearts).
After Example 3, check it with the Multiplication Rule: (6/11)(5/10)(4/9) = 120/990 = 4/33, the same answer. Interpret: "About 12% of randomly chosen committees would be all juniors." After Example 4, point out that the favorable line-ups fix two places and arrange the other 3 students in 3! ways. After Example 5, discuss why combinations fit: a hand is the same hand no matter the order in which the cards were dealt.
Guided Practice15-20 minutes
Pairs work four problems. Before computing, each pair writes "order matters" or "order does not matter" and a one-line reason.
(a) A museum hangs 5 of its 8 paintings in a row on one wall: P(8, 5) = 6720 arrangements.
(b) A reader chooses 2 of 7 books to take on a trip: C(7, 2) = 21 choices.
(c) A playlist of 12 songs, including 3 favorites, is shuffled: P(the first two songs are both favorites) = P(3, 2)/P(12, 2) = 6/132 = 1/22.
(d) Three people are chosen at random from 4 boys and 5 girls: P(exactly 2 girls) = C(5, 2) · C(4, 1)/C(9, 3) = 40/84 = 10/21.
Listen for these errors: using a permutation for a group with no roles, counting the favorable outcomes as a combination but the total as a permutation, and adding C(5, 2) + C(4, 1) instead of multiplying.
Independent Practice15 minutes
Students work on their own: (1) P(7, 2) = 42 and C(8, 3) = 56, with one sentence each describing a situation that the count fits; (2) a 3-person committee is chosen at random from 8 people, including Jordan: P(Jordan is on it) = C(7, 2)/C(8, 3) = 21/56 = 3/8; (3) a 5-card hand: P(it contains all four aces) = C(4, 4) · C(48, 1)/C(52, 5) = 48/2598960 = 1/54145; (4) the letters M, A, T, H are arranged in a random order: P(they spell MATH) = 1/4! = 1/24. Students interpret problems 2 and 3 in words.
Closure5-10 minutes
Exit ticket: 20 raffle tickets are sold, each to a different person. (1) Three different prizes (first, second, third) are drawn: how many results are possible? (Answer: P(20, 3) = 6840.) (2) Instead, three identical gift cards are drawn: how many results are possible? (Answer: C(20, 3) = 1140.) (3) In the gift-card version, what is the probability that one given ticket holder wins a card? (Answer: C(19, 2)/C(20, 3) = 171/1140 = 3/20.) Students explain in one sentence why answers (1) and (2) differ by a factor of 6.
Differentiation Strategies
For Struggling Students
Start each problem by listing a few outcomes and asking, "Is AB the same outcome as BA here?"
Use slot diagrams for every permutation before writing the formula, and divide by r! only after naming the r objects that could be reordered
Provide a two-column organizer: "all outcomes" and "favorable outcomes", both counted with the same method
For Advanced Students
Ask students to find the probability that a 5-card hand contains exactly one pair and compare it with a simulation of 100 dealt hands
Ask students to explain why C(n, r) = C(n, n - r) using committees ("choosing who is in is the same as choosing who is out")
Ask students to count arrangements of a word with repeated letters, such as LEVEL, and explain the division by 2! twice (a challenge beyond the standard)
Assessment Guidance
What to Look For
Look first at the decision: can students say whether order matters, and why, before they compute? A correct final answer with a mismatched method (a permutation on top and a combination on the bottom) is a sign of luck, not understanding. For "exactly k" problems, check that students multiply the counts for each kind and that the chosen numbers add up to the group size. Finally, ask students to interpret a probability in words and to check a simple case with the Multiplication Rule or with a short list.
02
Classroom Activities
3 Activities
1
Order or No Order? Card Sort
20 minGroups of 3-4
Groups sort 8 scenario cards into "order matters" and "order does not matter", then compute each count. The sort makes students justify the choice between a permutation and a combination before they compute anything.
The 8 Cards
Card 1: choose a president, vice president and treasurer from 10 club members (order matters, P(10, 3) = 720)
Card 2: choose 3 of 10 club members to attend a conference (C(10, 3) = 120)
Card 3: arrange 4 trophies on a shelf (4! = 24)
Card 4: choose 2 pizza toppings from a list of 9 (C(9, 2) = 36)
Card 5: assign first chair and second chair among 6 violinists (P(6, 2) = 30)
Card 6: deal a 5-card hand from a standard deck (C(52, 5) = 2,598,960)
Card 7: set a 3-digit lock code with no repeated digit (P(10, 3) = 720)
Card 8: choose 5 of the 12 players on a team to start a game, with positions ignored (C(12, 5) = 792)
Procedure
Each student reads a card aloud and asks the group, "If we swap two of the chosen items, is it a different outcome?"
The group places the card in a column and computes the count
Groups compare Cards 1 and 2, which use the same people, and explain why one answer is 3! = 6 times the other
Challenge Variation
Groups rewrite two "order does not matter" cards so that order does matter (for example, by adding roles) and compute the new counts.
2
Committee Draw Simulation
25 minPairs, then whole class
Students test a combination-based probability with a physical simulation. A bag holds 12 name slips, 5 of them marked with a star. A "committee" of 4 slips is drawn at random. The class estimates P(at least 3 starred members) and compares it with the exact value.
Procedure
Each pair draws 4 slips at once, records the number of stars, returns the slips and shakes the bag. Repeat 20 times
The class pools its results and compares the relative frequency with 5/33
Discussion Questions
Why do we add the "exactly 3" and "exactly 4" counts? Why are they not overlapping?
Why is drawing 4 slips at once the same as a combination?
How many draws would you need before you trust the class estimate?
Modification for Distance Learning
Use a free online random picker that selects 4 of 12 names without repeats, and pool results in a shared spreadsheet.
3
Card-Hand Investigation
20 minGroups of 3
Groups compute probabilities of three kinds of 5-card hands and rank them from most to least likely before and after computing. Each probability uses C(52, 5) = 2,598,960 as the total.
The Three Hands
All five cards of the same suit: 4 · C(13, 5)/C(52, 5) = 5148/2598960, about 0.002
No face cards: C(40, 5)/C(52, 5) = 658008/2598960, about 0.253
Exactly two hearts: C(13, 2) · C(39, 3)/C(52, 5) = 712842/2598960, about 0.274
Procedure
Before computing, each student ranks the three hands by how likely they seem
Each group member computes one probability and explains the numerator to the others
The group writes one sentence for each result, such as "about 1 hand in 500 is all one suit"
Discussion Questions
Why is the first numerator multiplied by 4?
In "exactly two hearts", where do the other 3 cards come from, and why 39?
Did your ranking before computing match the results?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Arrangements Versus Groups
All 12 ordered pairs of two different letters from A, B, C, D, sorted into columns. Each column holds the 2! = 2 orders of the same group, so there are 12/2 = 6 groups. This is why C(n, r) = P(n, r)/r!.
Diagram 2: From Permutations to Combinations
Example 1 counted with slots: 8 choices for gold, then 7 for silver, then 6 for bronze, so P(8, 3) = 336. Each group of 3 runners appears in 3! = 6 of those orders, so the number of groups is 336/6. The letters X, Y and Z stand for any three runners.
04
Homework Assignment
~30 min
HSS.CP.B.9 Homework: Probability with Permutations and Combinations
Directions: For each count, state whether order matters and why. Write each probability as favorable outcomes over possible outcomes, counted the same way, and simplify. Interpret each final probability in one sentence.
Part 1: Counting (Problems 1-2)
A debate team has 9 members. (a) In how many ways can the team choose a captain and a co-captain? (b) In how many ways can it choose 4 members to travel to a tournament? Explain which part is a permutation and which is a combination.
A bike lock uses a 4-digit code made of 4 different digits from 0 to 9. (a) How many codes are possible? (b) If a code is chosen at random, what is the probability that all four digits are odd?
Part 2: Probabilities of Compound Events (Problems 3-5)
A committee of 5 is chosen at random from 7 teachers and 6 parents. Find the probability that the committee has exactly 3 teachers and 2 parents.
The letters of the word GARDEN are arranged in a random order. Find the probability that the arrangement starts and ends with a vowel.
A 5-card hand is dealt from a shuffled standard deck. Find the probability that the hand contains exactly two aces. Give the answer as a fraction and as a decimal to three places.
Part 3: Solving a Problem (Problem 6)
A class raffle sells 30 tickets, and you buy 3 of them. Two different prizes are drawn, and the same ticket cannot win twice. Find the probability that you win both prizes and the probability that you win at least one prize. Solve it once with permutations or combinations and once with the Multiplication Rule, and explain whether buying 3 tickets makes winning something likely.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Order Decision
Permutation or combination chosen and justified every time
Correct choice without justification
Wrong method in most problems
Consistent Counting
Favorable and total outcomes counted the same way
One mismatch
Counts not comparable
Accuracy
All counts and probabilities correct and simplified
Most correct
Most incorrect
Interpretation and Check
Each probability interpreted; Problem 6 checked two ways
Interpretation vague or check missing
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
In which situation does order matter?
Answer: C
Swapping the first-prize and second-prize winners gives a different result, so the outcomes are ordered and the count is a permutation. In choices A and B, the same students or books in a different order are the same group. In choice D, a lottery ticket with the same 6 numbers is the same ticket in any order.
Question 2 of 20 · Multiple Choice
A club of 11 members elects a president, a vice president and a treasurer, and no one holds two offices. How many results are possible?
Answer: D
Order matters because the offices are different: P(11, 3) = 11 · 10 · 9 = 990. Choice A is C(11, 3), which ignores the offices. Choice B is 11³, which lets one person hold several offices.
Question 3 of 20 · Multiple Choice
A restaurant lets you choose 3 different side dishes from a menu of 12. How many different choices of 3 sides are there?
Answer: B
The order in which you name the sides does not matter, so C(12, 3) = 12 · 11 · 10/3! = 1320/6 = 220. Choice A is P(12, 3), which counts every order of the same three sides as different. Choice D is 12³ and allows repeats.
Question 4 of 20 · Multiple Choice
A committee of 4 is chosen at random from 7 women and 5 men. What is the probability that all 4 members are women?
Answer: A
P = C(7, 4)/C(12, 4) = 35/495 = 7/99. Choice B is the probability for one person only. Choice C counts only one committee as favorable, but C(7, 4) = 35 committees are all women. Choice D is (7/12)⁴, which treats the choices as if the same person could be picked again.
Question 5 of 20 · Multiple Choice
Seven books, including 3 math books, are placed on a shelf in a random order. What is the probability that the 3 math books take the 3 leftmost spots, in any order?
Answer: B
Favorable arrangements: 3! ways to order the math books on the left times 4! ways to order the other books, so P = 3! · 4!/7! = 144/5040 = 1/35. Choice C, 1/P(7, 3), counts only one order of the math books. Choice A is the chance that one given spot holds a math book.
Question 6 of 20 · Multiple Choice
A lottery draws 4 different numbers from 1 to 25, and order does not matter. What is the probability that a single ticket matches all 4 numbers?
Answer: D
There are C(25, 4) = 12650 possible sets of 4 numbers, and one ticket matches exactly one of them. Choice A uses P(25, 4) = 303600, which counts each set 4! = 24 times. Choice C, 1/25⁴, allows repeated numbers.
Question 7 of 20 · Multiple Choice
A president and a vice president are chosen at random from 10 club members. What is the probability that Kim is president and Lee is vice president?
Answer: B
There are P(10, 2) = 90 ordered results, and exactly one has Kim as president and Lee as vice president, so P = 1/90. Choice A, 1/C(10, 2), ignores which of the two gets which office; it is the probability that Kim and Lee are the two officers in either order.
Question 8 of 20 · Multiple Choice
A bag holds 6 red and 4 blue marbles. Three marbles are drawn at random at the same time. What is the probability of exactly 2 red marbles?
Answer: A
Favorable: C(6, 2) · C(4, 1) = 15 · 4 = 60. Total: C(10, 3) = 120. P = 60/120 = 1/2. Choice C is P(all three red) = C(6, 3)/C(10, 3) = 20/120. Choice D is the probability that a single marble is red.
Question 9 of 20 · Multiple Choice
Which expression gives the probability that a 5-card hand from a standard deck contains exactly 3 kings?
Answer: C
Choose 3 of the 4 kings and 2 of the 48 other cards, and divide by all C(52, 5) hands. Choice A forgets the other 2 cards in the hand. Choice B divides by the number of 3-card hands. Choice D counts ordered deals but only with the kings dealt first, so it misses the other positions for the kings.
Question 10 of 20 · Multiple Choice
Which statement about P(7, 3) and C(7, 3) is true?
Answer: D
Each group of 3 can be arranged in 3! = 6 orders, so the 35 groups give 35 · 6 = 210 arrangements: P(7, 3) = C(7, 3) · 3!. Choice A reverses the relationship. Choice C is false: C(7, 4) = 35 = C(7, 3), not 210.
Question 11 of 20 · Multiple Choice
A password is made of 3 different letters from the 26-letter alphabet. If a password is chosen at random, what is the probability that it contains no vowels (A, E, I, O, U)?
Answer: A
Favorable: P(21, 3) = 21 · 20 · 19 = 7980. Total: P(26, 3) = 26 · 25 · 24 = 15600. P = 7980/15600 = 133/260. Choice B is (21/26)³, which allows repeated letters. Choice C is the probability for one letter only.
Question 12 of 20 · Multiple Choice
A student counts the ways to choose a 4-person committee from 11 people as P(11, 4) = 7920. What is the correct count, and why?
Answer: C
A committee with no roles is a group, so each group of 4 was counted 4! = 24 times: C(11, 4) = 7920/24 = 330. Choice A keeps the permutation, which would be correct only if the 4 seats had different roles. Choice D divides by the wrong factorial.
Question 13 of 20 · Multiple Choice
Eight students line up in a random order. What is the probability that Maya is first in line?
Answer: B
Favorable line-ups put Maya first and arrange the other 7 in 7! ways, so P = 7!/8! = 1/8. Choice A is 1/8!, the probability of one specific complete line-up. Choice C is the probability that Maya is not first.
Question 14 of 20 · Multiple Choice
Which expression has the same value as C(15, 11)?
Answer: D
Choosing 11 of 15 people to include is the same as choosing the 4 to leave out, so C(15, 11) = C(15, 4) = 1365. Choices B and C are equal to each other: both are P(15, 4) = 32760, which counts ordered choices.
Question 15 of 20 · Short Answer
Fourteen runners are in a race. How many different results are possible for first, second and third place? If the top three places are equally likely to be any ordered result, what is the probability that your three favorite runners take the top three places in any order?
Results: P(14, 3) = 14 · 13 · 12 = 2184. Favorable: the 3 favorites can fill the top three places in 3! = 6 orders. P = 6/2184 = 1/364, which also equals 1/C(14, 3).
Question 16 of 20 · Short Answer
A jar holds 8 chocolate and 6 caramel candies of the same size. You take 4 at random. Find the probability that you get exactly 2 of each kind.
A 5-card hand is dealt from a shuffled standard deck. Find the probability that all five cards are red, and interpret the result.
P = C(26, 5)/C(52, 5) = 65780/2598960 = 253/9996, about 0.025. Interpretation: only about 1 hand in 40 is all red, far less than the 1/2 chance for a single card.
Question 18 of 20 · Short Answer
Six friends, including Jo and Sam, sit in a row of 6 seats in a random order. Find the probability that Jo and Sam sit next to each other.
Treat Jo and Sam as one block: 5 units can be arranged in 5! = 120 ways, and the block can be Jo-Sam or Sam-Jo, so 240 favorable seatings. Total: 6! = 720. P = 240/720 = 1/3.
Question 19 of 20 · Short Answer
A shipment of 15 phones includes 3 defective phones. An inspector tests 4 phones chosen at random. Find the probability that at least one tested phone is defective, and interpret it.
Use the complement: P(no defective) = C(12, 4)/C(15, 4) = 495/1365 = 33/91. P(at least one) = 1 - 33/91 = 58/91, about 0.64. Interpretation: this test catches at least one defective phone about 64% of the time, so a shipment with 3 bad phones passes about 36% of the time.
Question 20 of 20 · Short Answer
A club of 13 members wants (a) a 3-person planning team with no roles and (b) a president, vice president and secretary. Count each, explain how the counts are related, and find the probability that a given member, Rosa, is on a randomly chosen planning team.
(a) C(13, 3) = 286. (b) P(13, 3) = 13 · 12 · 11 = 1716. Each team of 3 can fill the three offices in 3! = 6 ways, and 286 · 6 = 1716. Teams that include Rosa: C(12, 2) = 66, so P = 66/286 = 3/13.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.B.9 mean?
HSS.CP.B.9 means students can use permutations and combinations to count outcomes and then compute probabilities of compound events, such as the chance that a random committee has a given makeup or that a hand of cards contains certain cards. It also covers solving counting problems in context. It belongs to the Conditional Probability and the Rules of Probability domain.
Is HSS.CP.B.9 an Algebra 2 or Precalculus standard?
It can be either. HSS.CP.B.9 is marked (+), which Common Core uses for additional mathematics that students should learn to take advanced courses. Many schools teach it in Algebra II, Precalculus or a statistics course, and some include counting principles in Geometry along with the other conditional probability standards.
How do you know whether to use a permutation or a combination?
Ask whether swapping two of the chosen items gives a different outcome. If it does (positions, ranks, roles or codes), use a permutation. If it does not (a committee, a hand of cards, a group of toppings), use a combination. Words such as "first", "president" or "arrange" usually signal order; "group", "team" or "select" often do not, but the situation, not the wording, decides.
What are the formulas for permutations and combinations?
For r objects chosen from n different objects:
Permutations: P(n, r) = n!/(n - r)!, written nPr on many calculators
Combinations: C(n, r) = n!/(r!(n - r)!), written nCr or as a binomial coefficient
By definition 0! = 1, so P(n, n) = n! and C(n, 0) = C(n, n) = 1.
Why do combinations divide by r!?
Because a permutation counts every order of the same group separately. A group of r objects can be ordered in r! ways, so dividing P(n, r) by r! leaves one count per group. With the letters A, B, C, D, there are 12 ordered pairs but only 6 pairs, since AB and BA are the same pair.
How are permutations and combinations used to find probabilities?
In a uniform model, a probability is favorable outcomes divided by possible outcomes. Permutations and combinations count both when a list would be too long. The key rule is to count both the same way: if the total counts unordered committees, the favorable count must also count unordered committees. For "exactly k of one kind", multiply the counts for each kind.
Can the Multiplication Rule give the same answers?
Yes. Many problems can be solved either by counting or with the general Multiplication Rule (HSS.CP.B.8). For a committee of 3 chosen from 6 juniors and 5 seniors, (6/11)(5/10)(4/9) = 4/33 equals C(6, 3)/C(11, 3). Counting is usually easier when the question asks for "exactly k" of a kind in any order, because the rule would need a separate product for each order.
What are common mistakes with permutations and combinations?
Common errors include:
Using a permutation for a group with no roles, which overcounts by a factor of r!
Counting favorable outcomes one way and total outcomes another way
Adding the counts for each kind in an "exactly k" problem instead of multiplying them
Allowing repeats (such as nʳ) when each object can be chosen only once
How do I compute C(n, r) on a calculator?
Most scientific and graphing calculators have nPr and nCr commands, often in a probability or math menu: type n, choose nCr, then type r. For small numbers, simplify by hand first. For example, C(9, 3) = (9 · 8 · 7)/(3 · 2 · 1) = 84, because the 6! in the numerator and denominator cancel.
Is HSS.CP.B.9 on the SAT?
Permutation and combination formulas are not a focus of the digital SAT. Its probability questions appear in the Problem-Solving and Data Analysis domain and usually rely on tables or simple counts. HSS.CP.B.9 is more directly useful in statistics courses, in probability units of Precalculus, and in discrete mathematics.
07
Related Standards
6 standards
These standards connect to HSS.CP.B.9: 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
