HSS.CP.B.6Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.B.6: Conditional Probability as a Fraction of B's Outcomes
In plain English: HSS.CP.B.6 is the Common Core statistics and probability standard that asks students to find the conditional probability of A given B by counting: out of the outcomes in B, what fraction also belong to A. Students work in uniform models such as dice, coins, cards and randomly chosen people, and they explain what the answer means in the situation. It is usually taught in Geometry or Algebra II.
Find the conditional probability of A given B as the fraction of B's outcomes that also belong to A, 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.6 or S-CP.6 · Official standard
Students find conditional probabilities by counting in uniform probability models, where every outcome is equally likely. The key move is to shrink the sample space: once we know that B happened, only B's outcomes are still possible, and they are still equally likely. So P(A | B) is the number of B's outcomes that also belong to A, divided by the number of outcomes in B.
Students practice the move with two dice, coin tosses, a deck of cards, numbered tickets and randomly chosen people from a group. Every answer ends with an interpretation in terms of the model: "Of the 4 equally likely outcomes with a sum of 9, 2 include a 6", or "over many rolls with a sum of 9, about half will include a 6". Students also compare P(A | B) with P(A) to say whether knowing B makes A more or less likely.
Learning Objectives
By the end of this lesson, students will be able to:
List or count the outcomes of an event B in a uniform probability model
Find P(A | B) as the number of B's outcomes that also belong to A, divided by the number of outcomes in B
Interpret a conditional probability in terms of the model, in a complete sentence that names B's outcomes
Compare P(A | B) with P(A) and explain what knowing B tells us about A
Prior Knowledge Required
Students should already be comfortable with:
Probability of a compound event as the fraction of outcomes in the sample space 7.SP.C.8
Representing sample spaces with organized lists, tables and tree diagrams 7.SP.C.8
Describing events as subsets of a sample space, including intersections HSS.CP.A.1
Simplifying fractions and writing them as decimals
Roll two dice behind a folder where students cannot see them, then give a clue.
Warm-Up Prompt
"I rolled two dice. I will not show you, but I will tell you that the sum is 11. What is the chance that at least one die shows a 6? What if I had told you the sum was 8 instead?"
Let students list the possibilities. With a sum of 11, the only outcomes are (5, 6) and (6, 5), and both contain a 6, so the chance is 1. With a sum of 8, the outcomes are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2), and 2 of these 5 contain a 6, so the chance is 2/5. Ask: why did we stop thinking about all 36 outcomes? Because the clue ruled most of them out. That is the whole idea of today's lesson.
Direct Instruction20-25 minutes
Part 1: Shrinking the sample space. In a uniform model every outcome has the same chance. If we learn that B happened, the outcome must be one of B's outcomes, and those remain equally likely. So:
Describe the uniform model: list or picture all equally likely outcomes (36 for two dice, 8 for three coins, 52 for a card).
Find B: mark every outcome in the condition. Count them; this count is the denominator.
Find the outcomes of B that are also in A: count only inside B. This count is the numerator.
Write P(A | B) = (number of outcomes in both A and B) / (number of outcomes in B), and simplify.
Interpret in the model: "Of the ___ equally likely outcomes in which B happens, ___ also have A", and compare with P(A).
Two dice
Roll two fair dice. B = "the sum is 9" and A = "at least one die shows 6".
Equation: B = {(3, 6), (4, 5), (5, 4), (6, 3)}; 2 of these contain a 6, so P(A | B) = 2/4 = 1/2. Compare P(A) = 11/36.
A standard deck
Draw one card from a well-shuffled 52-card deck. B = "the card is red" and A = "the card is a face card (jack, queen or king)".
Equation: B has 26 cards and 6 of them are red face cards, so P(A | B) = 6/26 = 3/13, the same as P(A) = 12/52.
Three coin tosses
Toss a fair coin three times. B = "at least two heads" and A = "the first toss is heads".
Equation: B = {HHH, HHT, HTH, THH}; 3 of these start with H, so P(A | B) = 3/4, greater than P(A) = 1/2.
A randomly chosen person
A class has 14 juniors (9 take chemistry) and 16 seniors (4 take chemistry). One student is chosen at random.
Part 2: Interpreting the answer. Model two kinds of sentences for Example 1. A counting sentence: "Of the 4 equally likely ways to roll a sum of 9, 2 include a 6." A long-run sentence: "If we rolled two dice many times and kept only the rolls with a sum of 9, about half of those would include a 6." Then compare: P(A) = 11/36 ≈ 0.31, so learning that the sum is 9 makes a 6 more likely. In Example 2 the conditional and unconditional values match: knowing the card is red tells us nothing about whether it is a face card.
Point out that dividing each count by 36 first, as in (2/36)/(4/36), gives the same 1/2. That is the formula P(A and B)/P(B), which students see in HSS.CP.A.3; counting inside B is the same idea without the extra step. In Example 4 the model is "each of the 30 students is equally likely to be chosen", so a class list works like a set of equally likely outcomes.
Guided Practice15 minutes
Work these with the class, one step at a time. Ask a different student for each step (model, B, count in B, fraction, sentence).
Two dice: B = "the first die is less than 3" and A = "the sum is at least 7". (B has 12 outcomes; inside B, only (1, 6), (2, 5) and (2, 6) have a sum of 7 or more, so P(A | B) = 3/12 = 1/4. Compare P(A) = 21/36: knowing the first die is small makes a large sum less likely.)
One card: B = "the card is not a heart" and A = "the card is a king". (39 cards, 3 kings, so 3/39 = 1/13, equal to P(king) = 4/52.)
Example 4 class: find P(senior | chemistry). (4/13 ≈ 0.31: of the 13 chemistry students, 4 are seniors.)
Listen for students who put 36 or 52 in the denominator: ask them which outcomes are still possible after the clue.
Independent Practice15 minutes
Students solve each problem and write one interpretation sentence for it.
Two dice: P(doubles | the sum is 8). (Only (4, 4) of the 5 outcomes: 1/5.)
An integer is chosen at random from 1 to 20: P(even | the number is greater than 12). (13 to 20 gives 8 numbers, 4 of them even: 1/2.)
Two coin tosses: P(two heads | at least one head). (HH, HT, TH remain; 1/3.)
One card: P(heart | face card). (3 of the 12 face cards: 1/4.)
Closure5 minutes
Exit ticket: Two dice are rolled. (1) Find P(the sum is at least 11 | at least one die shows 6). (B has 11 outcomes; (5, 6), (6, 5) and (6, 6) are in A, so 3/11.) (2) Write one sentence that interprets your answer in terms of the 36 equally likely outcomes.
Differentiation Strategies
For Struggling Students
Give printed outcome grids and lists so students can shade B in one color and then circle A inside the shaded part
Use the sentence frame "Out of the ___ outcomes in B, ___ are also in A" before writing any fraction
Start with one die and simple conditions such as "the number is odd" before moving to two dice
For Advanced Students
Ask students to find two events A and B for two dice with P(A | B) = P(A), and two with P(A | B) = 0 even though P(A) is not 0
Have students find P(sum is 7 | the dice differ by 1) and explain why the answer does not depend on which die is which
Ask students to extend a coin model to four tosses and find a condition B that makes P(first toss is heads | B) exactly 3/4
Assessment Guidance
What to Look For
The denominator is the clearest sign of understanding: it must be the number of outcomes in B, not the size of the whole sample space. Ask students to point to B's outcomes on a grid or list before they divide. For interpretation, look for sentences that name B's outcomes and the model ("of the 4 equally likely rolls with a sum of 9"), not just "the probability is 1/2". Strong students also say whether knowing B raised, lowered or did not change the chance of A.
02
Classroom Activities
3 Activities
1
Shade the Grid
20 minPairs
Pairs use printed 6-by-6 grids of the 36 outcomes for two dice. For each event card they shade B, circle the outcomes of B that are also in A, write P(A | B) as a fraction and interpret it in one sentence.
Event Cards (answers for the teacher)
Card 1. B: the sum is 6; A: at least one die shows 2 (2 of 5, so 2/5)
Card 2. B: the second die shows 1; A: the sum is odd (3 of 6, so 1/2)
Card 3. B: both dice are odd; A: the sum is at least 8 (3 of 9, so 1/3)
Card 4. B: the sum is at most 4; A: doubles (2 of 6, so 1/3)
Card 5. B: the dice differ by exactly 2; A: the sum is 8 (2 of 8, so 1/4)
Card 6. B: the first die shows 5; A: the dice differ by at most 1 (3 of 6, so 1/2)
Procedure
Partner A shades B on a fresh grid; Partner B circles A inside the shading and writes the fraction; roles swap on each card
For each card, the pair writes an interpretation: "Of the ___ equally likely rolls in which ___, ___ also ___"
Pairs mark each card with an up arrow, down arrow or equals sign to show whether knowing B raised, lowered or did not change the chance of A
Discussion Questions
On Card 2, is the answer the same as P(sum is odd) for all 36 outcomes? What does that tell you?
Which card has the smallest B? Does a small B make the conditional probability easier or harder to estimate by rolling real dice?
2
Roll, Keep and Compare
20 minGroups of 3
Groups roll two dice many times, keep only the rolls in which B happens, and compare the relative frequency of A among the kept rolls with the model's answer. This connects the counting answer to a long-run interpretation.
Procedure
Event B: the sum is at least 9. Event A: doubles
Before rolling, each group uses a grid to find the model's P(A | B): B has 10 outcomes and 2 of them, (5, 5) and (6, 6), are doubles, so 2/10 = 1/5
One student rolls, one records every roll, and one tallies only the rolls with a sum of 9 or more and whether they were doubles. Make 60 rolls
Groups compute the fraction of kept rolls that were doubles, then the class pools all groups' kept rolls
Discussion Questions
About how many of your 60 rolls did you expect to keep? (The model says 10/36 of them, about 17.)
Why is the pooled class result usually closer to 1/5 than a single group's result?
Finish the sentence: "In the long run, of the rolls with a sum of at least 9, about ___ are doubles."
Modification for Distance Learning
Use a spreadsheet with two random integer columns from 1 to 6 and 1,000 rows. Students filter for sums of at least 9 and count the doubles among the filtered rows.
3
The Club Roster Model
20 minPairs
Pairs treat a club roster as a uniform model: one member is chosen at random to represent the club. They find conditional probabilities by restricting to a row or column of the roster table and interpret each one.
Roster (invented)
60 members: juniors, 4 officers and 26 other members (30 juniors); seniors, 8 officers and 22 other members (30 seniors).
Procedure
Pairs find P(senior | officer), P(officer | senior), P(officer | junior) and P(junior | not an officer). (8/12 = 2/3, 8/30 = 4/15, 4/30 = 2/15 and 26/48 = 13/24.)
For each, pairs write the group they restricted to and one interpretation sentence
Pairs swap sheets with another pair and check each other's denominators
Challenge Variation
Change the roster so that P(officer | senior) = P(officer | junior) while keeping 12 officers and 60 members. How many senior officers must there be if there are still 30 seniors?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Shrinking the Sample Space for Two Dice
The 36 equally likely outcomes of rolling two dice, with each cell showing the sum. B (sum is 9) is shaded; its outcomes that also contain a 6 are dark. Knowing B leaves only 4 possible outcomes, and 2 of them are in A, so P(A | B) = 1/2.
Diagram 2: Three Coin Tosses, Before and After the Clue
Left: the uniform model for three tosses of a fair coin. Right: once we know there were at least two heads, only 4 outcomes remain, still equally likely. Three of them start with H, so the clue raises the chance that the first toss was heads from 1/2 to 3/4.
04
Homework Assignment
~30 min
HSS.CP.B.6 Homework: Conditional Probability by Counting
Directions: For each problem, describe the uniform model, list or count the outcomes in B, then find P(A | B) as a fraction of B's outcomes. Finish every problem with one sentence that interprets your answer in terms of the model and says whether knowing B changes the chance of A.
Part 1: Dice, Coins and Spinners (Problems 1-3)
Two fair dice are rolled. Let B = "the dice show the same number" and A = "the sum is greater than 7". Find P(A | B) and compare it with P(A).
A fair coin is tossed four times. Let B = "exactly two heads" and A = "the first two tosses are both heads". List the outcomes in B, find P(A | B), and compare it with P(A).
A spinner has 5 equal sections numbered 1 to 5 and is spun twice. Let B = "the product of the two numbers is even" and A = "the sum of the two numbers is 6". Find P(A | B).
Part 2: Cards, Tickets and Errors (Problems 4-6)
One card is drawn from a well-shuffled standard deck. Find P(the card is a 2, 3, 4 or 5 | the card is a heart) and P(the card is a heart | the card is a 2, 3, 4 or 5). Explain why the two answers are different.
A school sold 200 raffle tickets numbered 1 to 200. The soccer team sold tickets 1-80 and the band sold tickets 81-200. One ticket is drawn at random. Let B = "the ticket number is a multiple of 10" and A = "the soccer team sold the ticket". Find P(A | B) and P(B | A), and interpret both.
A student rolls two dice and finds P(the sum is 7 | the first die shows 4) by writing "only (4, 3) works, so the answer is 1/36". Explain the error, give the correct probability, and interpret it in terms of the model.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Model and Condition
Uniform model described and all outcomes of B listed or counted correctly
B counted with one error
B not identified
Fraction of B
Numerator counted inside B and denominator is B's size, in every problem
Correct method with one counting error
Divides by the whole sample space
Interpretation
Every answer interpreted in terms of the model, with a comparison to P(A)
Interpretations present but vague
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Every question uses a uniform model: all outcomes equally likely. Pick or write an answer, then open the explanation. Your score updates as you go, and Reset quiz starts the quiz over.
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
Two fair dice are rolled. Find P(doubles | the sum is 10).
Answer: C
B = {(4, 6), (5, 5), (6, 4)} has 3 outcomes, and only (5, 5) is a double: 1/3. Choice A divides by all 36 outcomes. Choice B is P(doubles) without the condition, and choice D is P(sum is 10) = 3/36.
Question 2 of 20 · Multiple Choice
Two fair dice are rolled. Find P(the sum is 5 | at least one die shows 1).
Answer: A
B has 11 outcomes (6 with a 1 on the first die, 6 on the second, minus the double-counted (1, 1)). Of these, (1, 4) and (4, 1) have a sum of 5, so the answer is 2/11. Choice B divides those 2 outcomes by 36. Choice C is P(sum is 5) = 4/36, which ignores the condition.
Question 3 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. Find P(ace | the card is not a face card).
Answer: B
The 40 cards that are not jacks, queens or kings include all 4 aces: 4/40 = 1/10. Choice A is P(ace) out of all 52 cards. Choice C is P(not a face card), and choice D wrongly removes the aces from the denominator (4/36).
Question 4 of 20 · Multiple Choice
A fair coin is tossed three times. Find P(the last toss is tails | exactly one head).
Answer: D
B = {HTT, THT, TTH}. The last toss is tails in HTT and THT, so the answer is 2/3. Choice A is P(last toss is tails) without the clue, and choice C is P(exactly one head) = 3/8.
Question 5 of 20 · Multiple Choice
A student is chosen at random from a class. For this class, P(plays soccer | sophomore) = 5/12. Which sentence interprets this correctly?
Answer: B
The condition, sophomore, is the group we restrict to, so the fraction describes the sophomores. Choice C reverses the condition, and choice D describes a joint probability out of the whole class.
Question 6 of 20 · Multiple Choice
A bakery filled 80 orders one day: 30 were for delivery, 20 were cakes, and 12 were cakes for delivery. An order is picked at random. Find P(cake | delivery).
Answer: C
Restrict to the 30 delivery orders: 12/30 = 0.40. Choice A divides by all 80 orders, choice B is P(delivery) = 30/80, and choice D reverses the condition: 12/20 is the share of cakes that were delivered.
Question 7 of 20 · Multiple Choice
In a uniform model, you find P(A | B) by counting. Which number belongs in the denominator?
Answer: A
Knowing B happened rules out every outcome outside B, so B's outcomes form the new sample space. Choice C gives P(A and B) instead, and choice D is the numerator, not the denominator.
Question 8 of 20 · Multiple Choice
Two fair dice are rolled. Find P(the sum is greater than 9 | the first die shows 6).
Answer: D
B = {(6, 1), ..., (6, 6)} has 6 outcomes, and (6, 4), (6, 5) and (6, 6) have sums of 10 or more: 3/6 = 1/2. Choice A divides 3 by 36, and choice B is P(sum greater than 9) = 6/36 without the condition.
Question 9 of 20 · Multiple Choice
A spinner has 8 equal sections numbered 1 to 8. Find P(even | the number is greater than 3).
Answer: C
B = {4, 5, 6, 7, 8} has 5 outcomes, and 4, 6 and 8 are even: 3/5. Choice A divides by all 8 sections, choice B is P(even) without the condition, and choice D is P(greater than 3).
Question 10 of 20 · Multiple Choice
In a uniform model, P(A | B) = 0 while B has several outcomes. What does this mean?
Answer: A
A zero numerator means that no outcome of B is in A. A can still happen outside B, so choice B does not follow. Choice D is wrong unless P(A) is also 0.
Question 11 of 20 · Multiple Choice
In a uniform model, P(A | B) = 1. What does this mean?
Answer: D
The numerator equals the denominator, so all of B's outcomes are in A: B sits inside A. Choice C reverses this. A can have extra outcomes outside B, so choices A and B need not be true.
Question 12 of 20 · Multiple Choice
A student finds P(the sum is 4 | doubles) for two dice by counting (2, 2) and writing 1/36. What is the correct probability?
Answer: B
Given doubles, only the 6 double outcomes remain, and (2, 2) is one of them: 1/6. The student divided by all 36 outcomes, which gives P(sum is 4 and doubles). Choice A is P(sum is 4) = 3/36.
Question 13 of 20 · Multiple Choice
An integer is chosen at random from 1 to 30. Find P(multiple of 5 | multiple of 3).
Answer: A
The multiples of 3 are 3, 6, ..., 30, which is 10 numbers, and 15 and 30 are also multiples of 5: 2/10 = 1/5. Choice B divides by all 30 numbers, and choice C is P(multiple of 3).
Question 14 of 20 · Multiple Choice
A fair coin is tossed three times. Compared with P(the first toss is heads) = 1/2, the value of P(the first toss is heads | at least one tail) is:
Answer: C
B contains every outcome except HHH, so it has 7 outcomes; HHT, HTH and HTT start with H, giving 3/7, which is less than 1/2. Learning that a tail appeared removes the all-heads outcome and makes a first-toss head less likely.
Question 15 of 20 · Short Answer
Two fair dice are rolled. Find P(at least one die shows 2 | the sum is odd). Then interpret your answer in terms of the model.
An odd sum happens in 18 of the 36 outcomes. Of these, (2, 1), (2, 3), (2, 5), (1, 2), (3, 2) and (5, 2) contain a 2, so P = 6/18 = 1/3. Interpretation: of the 18 equally likely rolls with an odd sum, 6 include a 2; in the long run, about one in three rolls with an odd sum will show a 2.
Question 16 of 20 · Short Answer
A robotics program has 48 students: 20 girls and 28 boys. On the build team there are 9 girls and 14 boys. One student is chosen at random. Find P(girl | build team) and P(build team | girl), and interpret each.
The build team has 9 + 14 = 23 students, so P(girl | build team) = 9/23 ≈ 0.39: of the 23 build team members, 9 are girls. P(build team | girl) = 9/20 = 0.45: of the 20 girls, 9 are on the build team. The overlap is the same 9 students, but the group we restrict to changes.
Question 17 of 20 · Short Answer
A fair coin is tossed four times. Find P(the first toss is tails | at least three heads) and explain what the answer means in the model.
B = {HHHT, HHTH, HTHH, THHH, HHHH} has 5 equally likely outcomes, and only THHH starts with tails, so P = 1/5. Of the 5 ways to get at least three heads in four tosses, 1 begins with tails, so knowing there were many heads makes a first-toss tail less likely than 1/2.
Question 18 of 20 · Short Answer
Two fair dice are rolled. Let A = "the sum is 6" and B = "at least one die shows 3". Find P(A | B) and P(B | A) and explain why they are different.
B has 11 outcomes and only (3, 3) has a sum of 6, so P(A | B) = 1/11. A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} has 5 outcomes and only (3, 3) contains a 3, so P(B | A) = 1/5. Both use the same single outcome, (3, 3), but divide by different groups: 11 outcomes in B and 5 in A.
Question 19 of 20 · Short Answer
The 40 seats in a row of a theater are numbered 1 to 40, and a winning seat is chosen at random. Find P(the seat number is 10 or less | the seat number is odd) and interpret it.
There are 20 odd seat numbers, and 1, 3, 5, 7 and 9 are 10 or less, so P = 5/20 = 1/4. Interpretation: if we know the winning seat has an odd number, it is one of 20 equally likely seats, and 5 of those are among the first ten seats. This equals P(seat 10 or less) = 10/40, so knowing the number is odd does not change the chance.
Question 20 of 20 · Short Answer
Describe two events A and B for rolling two fair dice so that P(A | B) = 1/2. Show the count that proves it.
Sample answer: B = "the first die shows 1" (6 outcomes) and A = "the second die is even". Inside B, (1, 2), (1, 4) and (1, 6) are in A, so P(A | B) = 3/6 = 1/2. Any pair of events works if exactly half of B's outcomes are also in A.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.B.6 mean?
HSS.CP.B.6 means students find P(A | B) by looking only at the outcomes where B happens and asking what fraction of them are also in A. They then explain the answer in terms of the situation, such as dice, cards or a randomly chosen person. It belongs to the Common Core cluster on the rules of probability in a uniform model.
What is a uniform probability model?
It is a model in which every outcome has the same chance, such as the 36 outcomes of two fair dice or the 52 cards of a shuffled deck. Choosing one person at random from a list is also a uniform model: each person is one equally likely outcome. Counting works for conditional probability only when the outcomes are equally likely.
How do you find P(A | B) by counting outcomes?
List or mark every outcome in B, count them, then count how many of those are also in A. Divide the second count by the first. For one roll of a die with B = "greater than 2" and A = "even", B = {3, 4, 5, 6}, and 4 and 6 are even, so P(A | B) = 2/4 = 1/2.
What does "interpret the answer in terms of the model" mean?
It means saying what the number tells you about the situation, not just giving the fraction. A good interpretation names the outcomes of B and what share of them have A, for example "of the 6 equally likely outcomes where the second die is 5, three have an even sum". A long-run version is also fine: "over many such rolls, about half will ...". Comparing with P(A) adds meaning: did knowing B make A more or less likely?
Why do we not divide by the total number of outcomes?
Because the condition rules out every outcome that is not in B. Dividing by the full sample space gives P(A and B), a different probability. The two are linked: P(A | B) = P(A and B)/P(B), so dividing by the full total must be followed by dividing by P(B).
How is HSS.CP.B.6 related to the formula P(A and B)/P(B)?
They give the same result in a uniform model. Suppose the model has N equally likely outcomes, B has b of them, and k are in both A and B. Then P(A and B)/P(B) = (k/N)/(b/N) = k/b, which is the counting answer. HSS.CP.A.3 introduces the formula; HSS.CP.B.6 uses the counting form, which is often easier and makes the meaning clear.
Is P(A | B) the same as P(B | A)?
Usually not. Both have the same numerator, the outcomes in both events, but P(A | B) divides by the size of B and P(B | A) divides by the size of A. They are equal only when A and B have the same number of outcomes, or when both are 0.
What are common mistakes on HSS.CP.B.6 problems?
Common errors include dividing by the whole sample space, counting outcomes of A that are outside B, reversing A and B, and miscounting outcomes such as (1, 1) twice when a condition says "at least one die". Shading B on a grid before counting prevents many of these.
Is HSS.CP.B.6 taught in Geometry or Algebra 2?
It depends on the school. The Common Core model course pathways place the conditional probability standards in Geometry (traditional pathway) or Mathematics II (integrated pathway), and many schools teach them in Algebra II or a statistics course. It usually comes right after HSS.CP.A.3 and HSS.CP.A.4.
Can I use a two-way table for HSS.CP.B.6?
Yes. If one person or object is chosen at random from the table, every entry is an equally likely outcome, so the table is a uniform model. P(A | B) is then the count in the A-and-B cell divided by the total for B, exactly the same counting idea as with dice or cards.
07
Related Standards
5 standards
These standards connect to HSS.CP.B.6: 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
