HSS.CP.A.4Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.A.4: Two-Way Frequency Tables, Independence and Conditional Probability
In plain English: HSS.CP.A.4 is the Common Core statistics and probability standard that asks students to build and read two-way frequency tables for data where each person or object falls into two categories. Students then treat the table as a sample space: they estimate conditional probabilities from its rows and columns and decide whether two events look independent. It is usually taught in Geometry or Algebra II.
Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results.
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.4 or S-CP.4 · Official standard
Students organize data in which every person or object is classified by two categorical variables, such as grade level and favorite subject. They build the two-way frequency table from raw records or from summary statements, and they read what each cell, row total and column total says about the group.
The table then becomes a sample space: if one person is chosen at random from the group, each person is one equally likely outcome. Students estimate conditional probabilities by restricting attention to one row or one column, and they decide whether two events look independent by checking whether knowing one event changes the proportion for the other. Because the data come from samples, students learn to call the results estimates and to treat small differences with care.
Learning Objectives
By the end of this lesson, students will be able to:
Construct a two-way frequency table, with row and column totals, from raw records or from summary counts
Interpret the cells, row totals and column totals of a two-way table in the context of the data
Use the table as a sample space to estimate a conditional probability such as P(science | 10th grade)
Decide whether two events appear independent by comparing a conditional probability with the overall probability
Explain why probabilities estimated from a sample are approximations for the larger population
Prior Knowledge Required
Students should already be comfortable with:
Displaying bivariate categorical data in a two-way table 8.SP.A.4
Probability as a number from 0 to 1 and as a fraction of equally likely outcomes 7.SP.C.5
Writing fractions as decimals and percents, and rounding to a sensible number of places
Describing events as subsets of a sample space HSS.CP.A.1
Ask every student two yes-or-no questions and record the answers on the board as a list of pairs, one pair per student.
Warm-Up Prompt
"Everyone answered two questions: Do you have a pet? Do you have a sibling? How could we organize our answers so that anyone can see, in one glance, how many of us have a pet but no sibling?"
Let pairs sketch an organizer, then share. Steer the class toward a grid with "Pet / No pet" down the side and "Sibling / No sibling" across the top. Fill it in with the class data and add totals. Point out that each student is counted in exactly one cell, so the four cells add up to the class size. Keep this table on the board: you return to it at the end of Direct Instruction.
Direct Instruction20-25 minutes
Part 1: Constructing the table. A two-way frequency table has one row for each category of the first variable and one column for each category of the second. Each cell counts the objects that belong to both categories; the totals in the margins count each category by itself. Work Example 1 from summary statements: students fill the cell they know first, then use subtraction and the totals.
Constructing a table from summary counts
A random sample of 180 high school juniors (invented data): 100 have a driver's license, 70 have a part-time job, and 52 have both.
Equation: License: 52 job, 48 no job (100). No license: 18 job, 62 no job (80). Column totals: 70 job, 110 no job. Grand total 180.
Official example: favorite subject given grade
A school surveys a random sample of 60 students from each grade (invented data, table below). Estimate the probability that a tenth grader favors science, do the same for math and English, and compare.
A sample of 250 airline passengers (invented data): 25 were born in January through June and prefer a window seat, 75 were born in January through June and prefer an aisle seat, 85 born July through December prefer a window seat, and 65 prefer an aisle seat.
Equation: P(window | born Jan-Jun) = 25/100 = 0.25, but P(window) = 110/250 = 0.44, so in this sample the two events are not independent.
Part 2: The table as a sample space. If one person is chosen at random from the 180 juniors, each junior is one equally likely outcome, so a probability is a count divided by 180. A conditional probability such as P(job | license) restricts the sample space to one row: only the 100 license holders count, and 52 of them have a job. Stress the direction: P(license | job) restricts to the job column instead, so its denominator is 70.
Part 3: The official example. Show the table and Diagram 1. Students compute the three conditional probabilities for tenth graders, then for another grade of their choice, and compare with the whole-sample proportions.
Favorite subject by grade, random sample of 60 students per grade (invented data)
Grade
Math
Science
English
Total
9th
22
18
20
60
10th
15
27
18
60
11th
20
20
20
60
12th
19
15
26
60
Total
76
80
84
240
Because the school sampled 60 students per grade, the pooled proportions describe the whole school only if the four grades have about the same size; mention this caveat. The results are estimates: a different random sample would give slightly different numbers.
Part 4: Independence. Two events A and B are independent when knowing that B happened does not change the probability of A, that is, P(A | B) = P(A). With a two-way table, use these steps:
Name the events, for example A = "prefers a window seat" and B = "born January through June".
Find the overall probability P(A) from the column (or row) total for A divided by the grand total.
Find P(A | B) by restricting to the row (or column) for B.
Compare: if the two values are equal, or very close for sample data, the events appear independent; if they differ clearly, they are not independent in this sample.
Apply the steps to Example 4, then to Example 1: P(job | license) = 0.52 while P(job) = 70/180 ≈ 0.39, so having a license and having a job are not independent in that sample. Finish by returning to the warm-up table and asking whether pets and siblings look independent in your class.
Guided Practice15 minutes
Pairs use the table below (invented data from a sample of 160 students) and answer one question at a time before you reveal the answer.
How students get to school and whether they arrived on time (invented data)
Travel mode
On time
Late
Total
Bus
48
12
60
Car
46
4
50
Walk or bike
45
5
50
Total
139
21
160
What does the 4 in the table mean? (Four sampled students came by car and were late.)
Estimate P(late | bus). (12/60 = 0.20.)
Estimate P(bus | late) and explain why it differs from the previous answer. (12/21 ≈ 0.57: now the sample space is the 21 late students, not the 60 bus riders.)
Are "late" and "rides the bus" independent in this sample? (No: P(late) = 21/160 ≈ 0.13, and bus riders were late 0.20 of the time. Car riders: 0.08; walkers and bikers: 0.10.)
Listen for students who divide by the grand total when a conditional probability is asked, and for students who compare counts (12 late bus riders versus 4 late car riders) instead of proportions.
Independent Practice15 minutes
Students work alone on this set, then check with a partner.
A random sample of 300 adults (invented data): 120 drink coffee daily, 90 work night shifts, and 54 of the night-shift workers drink coffee daily. Construct the two-way table with totals. (Night shift: 54 coffee, 36 no coffee, 90 total. Day shift: 66 coffee, 144 no coffee, 210 total. Columns: 120 and 180.)
Estimate P(coffee | night shift) and P(coffee | day shift). (0.60 and 66/210 ≈ 0.31.)
Decide whether daily coffee and shift type appear independent, and justify with numbers. (No: 0.60 is far from P(coffee) = 0.40.)
Closure5 minutes
Exit ticket: A sample of 100 phone owners (invented data) shows Brand X: 12 cracked screens, 48 not cracked; Brand Y: 8 cracked, 32 not cracked. (1) Estimate P(cracked | Brand X). (0.20.) (2) Are brand and a cracked screen independent in this sample? Explain with a comparison. (Yes: P(cracked | Brand Y) = 0.20 and P(cracked) = 20/100 = 0.20 as well.)
Differentiation Strategies
For Struggling Students
Give a blank table template with the row and column labels already written, and have students shade the row or column that the word "given" points to before dividing
Have students say the sample space out loud: "out of the 60 tenth graders" before writing any fraction
Start with 2-by-2 tables with totals of 100 so that proportions and percents match
For Advanced Students
Ask students to fill in a 2-by-2 table with a given grand total and given row and column totals so that the two events are exactly independent, and to explain why only one set of cells works
Have students show that P(A | B) = P(A) and P(B | A) = P(B) are either both true or both false for any 2-by-2 table with nonzero totals
Ask students to take two random samples of 20 from the Activity 1 data cards and see how much P(science | 10th) changes from sample to sample
Assessment Guidance
What to Look For
Check that every table a student builds has cells that add to the row totals, the column totals and the grand total. When a student writes a conditional probability, ask which group is the sample space; the denominator should be that group's total, not the grand total. For independence, look for a comparison of proportions (a conditional probability against the overall probability), not a comparison of raw counts, and for careful language with sample data: "appear independent in this sample" rather than "are independent".
02
Classroom Activities
3 Activities
1
Favorite Subject Survey
25 minGroups of 3-4
Students carry out the official example: they collect or receive data on favorite subject (math, science or English) and grade level, build the two-way table, estimate the probability that a tenth grader favors science, repeat for the other subjects and compare.
Procedure
If your school allows a short survey, each group asks a random sample of students (for example, names drawn from a numbered class list with a random number generator) for their grade and favorite of math, science and English
Otherwise, give each group the 40 printed data cards (invented): 16 tenth graders (4 math, 7 science, 5 English) and 24 eleventh graders (8 math, 6 science, 10 English)
Groups build the two-way table with totals on chart paper
Groups estimate P(science | 10th grade), P(math | 10th grade) and P(English | 10th grade), then the same three for the other grade. With the cards: 7/16 ≈ 0.44 and 6/24 = 0.25 for science
Each group writes two sentences comparing the grades, using the phrase "given that the student is in"
Discussion Questions
Why do we compare 7/16 with 6/24 instead of comparing 7 science fans with 6 science fans?
Does your table suggest that favorite subject and grade are independent? What would the rows look like if they were?
If another class took a new random sample, would they get exactly the same estimates? Why is the word "estimate" in the standard?
Modification for Distance Learning
Share the data cards as a spreadsheet with one row per student. Groups use a filter or a pivot table to build the two-way table, then post their estimates to a shared slide.
2
Independent or Not? Table Sort
20 minPairs
Pairs receive six two-way table cards (invented data) and sort them into "appear independent" and "not independent" by comparing a conditional probability with the overall probability for each table.
Table Cards
Card 1, front half of room by glasses: front 30 glasses, 20 no glasses; back 45 glasses, 30 no glasses (0.60 and 0.60: independent)
Card 2, lives on a farm by owns work boots: farm 40 boots, 10 no boots; not farm 20 boots, 30 no boots (0.80 versus 0.60: not independent)
Card 3, born on an even date by prefers mountains to beach: even 18 mountains, 42 beach; odd 12 mountains, 28 beach (0.30 and 0.30: independent)
Card 4, takes art by takes music: art 25 music, 25 no music; no art 15 music, 35 no music (0.50 versus 0.40: not independent)
Card 5, summer birthday by likes spicy food: summer 9 spicy, 27 not; other 16 spicy, 48 not (0.25 and 0.25: independent)
Card 6, eats breakfast by has first-period PE: breakfast 33 PE, 17 no PE; no breakfast 31 PE, 19 no PE (0.66 versus 0.64: very close)
Procedure
For each card, pairs add the totals, then compute the proportion of the first column within the first row and within the whole table
Pairs place each card on a sorting mat and write the two proportions on a sticky note
Pairs compare mats with another pair and resolve disagreements
Discussion Questions
Card 6 is not exactly equal. Would you call breakfast and first-period PE independent? What would help you decide?
Why can two events be independent even when one row total is much larger than the other?
3
From Raw Records to a Table
20 minPairs
Pairs tally 24 raw survey records (invented) about housing type and pet ownership, build the table, and use it as a sample space to answer conditional probability questions in both directions.
Records
1. Apartment, no pet; 2. House, no pet; 3. Apartment, no pet; 4. Apartment, pet; 5. House, no pet; 6. House, pet; 7. Apartment, pet; 8. House, pet; 9. Apartment, no pet; 10. House, no pet; 11. Apartment, no pet; 12. Apartment, no pet; 13. Apartment, no pet; 14. House, pet; 15. House, pet; 16. House, pet; 17. Apartment, no pet; 18. House, pet; 19. House, pet; 20. Apartment, pet; 21. House, no pet; 22. House, pet; 23. House, no pet; 24. House, pet.
Procedure
Partner A reads each record aloud while Partner B makes tally marks in a blank 2-by-2 grid; they swap roles halfway
Pairs add row and column totals and check that the grand total is 24 (House: 9 pet, 5 no pet; Apartment: 3 pet, 7 no pet)
Pairs estimate P(pet | house) ≈ 0.64, P(pet | apartment) = 0.30 and P(house | pet) = 0.75, writing one sentence of interpretation for each
Challenge Variation
Pairs write their own 24 records for a new pair of yes-or-no questions so that the two variables are exactly independent, then trade with another pair to check.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Official Example as a Two-Way Table
Invented data for the official example: 60 randomly chosen students from each grade named a favorite among math, science and English. To estimate P(science | 10th grade), only the tenth-grade row counts, so the denominator is 60, not 240.
Diagram 2: Favorite Subject Within Each Grade
The same data as Diagram 1, drawn to scale as segmented bars: each bar is one grade, split by favorite subject. The dashed line marks where the math segment ends for the whole sample. If grade and favorite subject were independent, every bar would have roughly the same split as the bottom bar; the tenth-grade bar has a clearly larger science segment.
04
Homework Assignment
~30 min
HSS.CP.A.4 Homework: Two-Way Tables and Probability
Directions: Show every table with row and column totals. Write each probability as a fraction and as a decimal rounded to two places, and name the sample space you used ("out of the ..."). All data are invented for practice.
Part 1: Constructing Tables (Problems 1-2)
A random sample of 250 adults were asked whether they are under 30 and whether they pay for a music streaming subscription. In the sample, 110 are under 30, 160 have a subscription, and 88 of the adults under 30 have a subscription. Construct the two-way frequency table with all totals.
A smoothie shop records the size (S = small, L = large) and whether the customer bought an add-on (Y = yes, N = no) for its first 20 orders: S-N, L-Y, S-N, S-Y, L-Y, L-Y, S-N, S-Y, L-N, S-N, L-Y, S-N, S-N, L-Y, S-Y, S-N, L-N, L-N, L-N, L-Y. Construct the two-way table, then estimate the probability that a large smoothie order included an add-on.
Part 2: Reading and Estimating (Problems 3-4)
A sample of 200 students (100 ninth graders and 100 twelfth graders) reported how they usually get lunch. Ninth graders: cafeteria 60, bring lunch 35, off campus 5. Twelfth graders: cafeteria 38, bring lunch 30, off campus 32. (a) Build the table with totals. (b) Estimate P(off campus | 12th grade) and P(12th grade | off campus). (c) Explain in one sentence why the two answers in (b) are different.
A teacher surveyed 150 students about whether they use a planner and whether they turned in every homework assignment last week. Of the 60 planner users, 48 turned in everything; of the 90 students without a planner, 45 turned in everything. Estimate P(turned in everything | planner), P(turned in everything | no planner) and P(turned in everything). Do the two variables appear independent in this sample? Justify with your numbers.
Part 3: Independence (Problems 5-6)
A sample of 300 phone users recorded operating system (System A or System B) and whether they use a phone case. System A: 126 with a case, 54 without. System B: 84 with a case, 36 without. Show that "uses a case" and "System A" appear independent in this sample by comparing two probabilities, then check the other direction by comparing P(System A | case) with P(System A).
In a sample of 200 people, 50 describe themselves as night owls and 150 as early birds; 120 of the 200 prefer coffee and 80 prefer tea. Complete the two-way table so that "night owl" and "prefers coffee" are exactly independent, and explain how you found the first cell.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Table Construction
All cells and totals correct and consistent
One cell or total wrong
Table missing or several errors
Sample Space
Correct denominator named for every probability
Denominator wrong once
Grand total used for conditional probabilities
Independence
Compares a conditional and an overall probability and concludes correctly
Correct numbers, conclusion unclear
Compares counts or no comparison
Interpretation
Every answer explained in context, results called estimates
Some answers explained
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Answer each question, then open the explanation. All data are invented. Your score updates as you go, and Reset quiz clears your answers so you 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
A random sample of 200 adults (invented data) gives this table. City residents: 42 take transit, 28 drive. Suburb residents: 18 take transit, 112 drive. Estimate the probability that a randomly chosen city resident from this sample takes transit.
Answer: C
Restrict to the 70 city residents: 42/70 = 0.60. Choice A divides by the grand total (42/200), which is the probability of being a city resident who takes transit. Choice D reverses the condition: 42/60 is the share of transit riders who live in the city.
Question 2 of 20 · Multiple Choice
A random sample of 200 adults (invented data) gives this table. City residents: 42 take transit, 28 drive. Suburb residents: 18 take transit, 112 drive. What does the number 18 represent?
Answer: B
Each cell counts the people who belong to both categories: here, suburb and transit. Choice A treats the count as a percent of the suburb row, but the suburb row has 130 people, not 100. Choice D puts the count in the wrong row.
Question 3 of 20 · Multiple Choice
A random sample of 200 adults (invented data) gives this table. City residents: 42 take transit, 28 drive. Suburb residents: 18 take transit, 112 drive. Do "takes transit" and "lives in the city" appear independent in this sample?
Answer: C
P(transit | suburb) = 18/130 ≈ 0.14, while P(transit) = 60/200 = 0.30, so knowing where someone lives changes the chance they take transit. Choice D compares the two column totals, which says nothing about independence: a table can have far more drivers and still show independence.
Question 4 of 20 · Multiple Choice
In a survey of 90 students, 50 are ninth graders and the rest are tenth graders. In all, 36 students own a bike, and 20 of the ninth graders own a bike. How many tenth graders do not own a bike?
Answer: D
There are 90 - 50 = 40 tenth graders, and 36 - 20 = 16 of them own a bike, so 40 - 16 = 24 do not. Choice A is the number of ninth graders without a bike (50 - 20). Choice C is everyone without a bike (90 - 36).
Question 5 of 20 · Multiple Choice
Each option gives a 2-by-2 table as (row 1: cell, cell; row 2: cell, cell). In which table are the row variable and the column variable independent?
Answer: B
In B, the first column is 12/30 = 0.4 of row 1 and 30/75 = 0.4 of row 2, which matches the overall 42/105 = 0.4. In A the shares are 0.25 and 1/3; in C, 0.5 and 0.25; in D, about 0.83 and 0.36, so those rows differ.
Question 6 of 20 · Multiple Choice
A station inspected a sample of 500 cars (invented data). Under 10 years old: 270 passed, 30 failed. 10 years or older: 150 passed, 50 failed. Estimate P(fail | 10 years or older).
Answer: A
Restrict to the 200 older cars: 50/200 = 0.25. Choice B is 50/80, the share of failed cars that were older, which reverses the condition. Choice C divides by all 500 cars, and choice D is the overall failure rate 80/500.
Question 7 of 20 · Multiple Choice
In the car inspection data, which probability does the phrase "the fraction of passing cars that are under 10 years old" describe?
Answer: C
The phrase "of passing cars" names the group being restricted to, so passing is the condition: P(under 10 years | pass). Choice A reverses the roles: it describes the fraction of newer cars that passed. Choice B is a joint probability out of all 500 cars.
Question 8 of 20 · Multiple Choice
In a random sample of 180 students, 45 are seniors, and 18 of the seniors chose art as their favorite elective. Estimate the probability that a senior at this school favors art.
Answer: D
Given that the student is a senior, the sample space is the 45 seniors: 18/45 = 0.40. Choice A divides by all 180 students, choice B is the share of the sample who are seniors, and choice C is the share of seniors who did not choose art.
Question 9 of 20 · Multiple Choice
In a sample of 400 shoppers, 240 used a coupon. Of the 100 shoppers who came on a weekend, 60 used a coupon. Do "weekend shopper" and "used a coupon" appear independent in this sample?
Answer: A
P(coupon | weekend) = 60/100 = 0.60 and P(coupon) = 240/400 = 0.60, so knowing the shopper came on a weekend does not change the proportion. Choice B compares counts from groups of different sizes, and choice C is about the size of the weekend group, which does not matter for independence.
Question 10 of 20 · Multiple Choice
A two-way table of grade level by favorite sport shows row percentages, so each row adds to 100%. What do the percentages in the "Juniors" row show?
Answer: D
A row percentage divides each cell by its row total, so the Juniors row describes only juniors: it shows P(sport | junior) for each sport. Choice B describes column percentages, which divide by each sport's total instead.
Question 11 of 20 · Multiple Choice
When you use a two-way table to estimate P(A | B), what goes in the denominator?
Answer: A
The condition "given B" restricts the sample space to B, so divide the count of objects in both A and B by the total for B. Choice C gives the joint probability P(A and B), and choice B gives P(B | A).
Question 12 of 20 · Multiple Choice
Twelve students gave their grade and whether they are in a club: (9, yes), (10, no), (9, no), (9, yes), (10, yes), (10, no), (9, yes), (10, no), (9, no), (10, yes), (9, yes), (10, no). Which is the correct grade-10 row of the two-way table?
Answer: A
The grade-10 records are (10, no) four times and (10, yes) twice, for 6 tenth graders. Choice B swaps the two cells, and choice D counts the ninth-grade club members (4) into the grade-10 row.
Question 13 of 20 · Multiple Choice
Suppose "left-handed" and "owns a dog" are independent in a large sample of adults. What should be true?
Answer: D
Independence means that knowing someone is left-handed does not change the chance that they own a dog, so P(dog | left-handed) ≈ P(dog). Choice B compares counts: there are far more right-handed adults, so their count of dog owners would be larger even under independence. Choice C describes events that cannot happen together, which are not independent.
Question 14 of 20 · Multiple Choice
In a sample, 35 of 50 freshmen and 35 of 70 sophomores said they usually eat school lunch. Which statement does the sample support?
Answer: B
Compare proportions within each grade: 35/50 = 0.70 and 35/70 = 0.50. Choice A compares counts from groups of different sizes. Choice C is wrong because 70/120 ≈ 0.58, and choice D is wrong because freshmen are 35 of the 70 school-lunch eaters, which is 50%.
Question 15 of 20 · Short Answer
A town surveyed 120 residents (invented data): 45 own a bicycle, 70 live within 2 miles of work, and 30 of the bicycle owners live within 2 miles of work. Construct the two-way table with all totals.
Bicycle: 30 within 2 miles, 15 farther (total 45). No bicycle: 40 within 2 miles, 35 farther (total 75). Column totals: 70 within 2 miles and 50 farther; grand total 120. Start with the cell you know (30), then subtract from the row and column totals.
Question 16 of 20 · Short Answer
A theater sampled 300 customers (invented data). 3D showing: 63 teens, 27 adults. 2D showing: 57 teens, 153 adults. Estimate P(teen | 3D) and P(3D | teen), and say in words what each one means.
P(teen | 3D) = 63/90 = 0.70: of the customers at the 3D showing, 70% were teens. P(3D | teen) = 63/120 = 0.525: of the 120 teens, about 53% chose the 3D showing. The numerator is the same, but each uses a different group as the sample space.
Question 17 of 20 · Short Answer
A company surveyed 500 employees (invented data). Remote workers: 96 satisfied, 24 not satisfied. Office workers: 304 satisfied, 76 not satisfied. Decide whether "works remotely" and "satisfied" appear independent. Show your comparison.
P(satisfied | remote) = 96/120 = 0.80, P(satisfied | office) = 304/380 = 0.80, and P(satisfied) = 400/500 = 0.80. They appear independent: knowing where someone works does not change the proportion who are satisfied in this sample.
Question 18 of 20 · Short Answer
Like the official example, a school sampled 150 students (invented data) about their favorite of math, science and English. The 50 eleventh graders chose math 14 times, science 12 times and English 24 times. In the whole sample, 45 students chose science. Estimate P(science | 11th grade), compare it with P(science), and do the same comparison for English given 11th grade if 42 students in the whole sample chose English.
P(science | 11th) = 12/50 = 0.24, lower than P(science) = 45/150 = 0.30. P(English | 11th) = 24/50 = 0.48, higher than P(English) = 42/150 = 0.28. In this sample, eleventh graders lean toward English more than the sample as a whole. These are estimates from one sample, so another sample would give somewhat different values.
Question 19 of 20 · Short Answer
In a sample of 80 students, 40 are in band, 30 take Spanish, and 24 are in band and take Spanish. Compute 24/80, 24/40 and 24/30, and say what probability each fraction estimates.
24/80 = 0.30 estimates P(band and Spanish), a joint probability out of all 80 students. 24/40 = 0.60 estimates P(Spanish | band): out of the band members. 24/30 = 0.80 estimates P(band | Spanish): out of the Spanish students.
Question 20 of 20 · Short Answer
A table from a random sample of 1,000 adults gives P(A | B) = 0.41 and P(A) = 0.40. A student says A and B are clearly dependent because the numbers are not equal. Do you agree? Explain.
Not necessarily. With sample data, a conditional probability and an overall probability almost never match exactly, even for events that are independent in the population. A difference of 0.01 is small enough to come from sampling variation, so the events appear approximately independent. A much larger difference, or the same pattern in several samples, would be stronger evidence of dependence.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.A.4 mean?
HSS.CP.A.4 means students can organize two-category data in a two-way frequency table and then use that table to answer probability questions. They build the table, explain what its cells and totals mean, estimate conditional probabilities from a row or column, and decide whether two events appear independent.
Is HSS.CP.A.4 taught in Geometry, Algebra 2 or Statistics?
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 instead. The content is the same wherever it is taught.
How do you make a two-way frequency table?
List the categories of one variable as rows and the other as columns, then count how many objects fall into each row-column pair. Add a total for every row and column and a grand total. When you are given summary statements instead of raw data, fill in the cell you know and use subtraction from the totals for the rest. The four cells of a 2-by-2 table must add to the grand total.
How do you find a conditional probability from a two-way table?
Find the row or column for the condition, and divide the cell you want by that row or column total. For example, if 16 of the 40 members of a hiking club are also in a photography club, P(photography | hiking) is 16/40 = 0.4. The word after "given" (or "of the", "among") tells you which total to use.
How do you tell if two events are independent using a two-way table?
Compare a conditional probability with the overall probability. If P(A | B) equals P(A), knowing B does not change the chance of A, and the events are independent. With sample data the two numbers are rarely exactly equal, so students should say "appear independent" when they are very close and "not independent" when they clearly differ.
What is the difference between P(A | B) and P(B | A)?
They use different sample spaces. P(A | B) looks only at the objects in B; P(B | A) looks only at the objects in A. Both have the same numerator, the count in both A and B, but different denominators, so they are usually different numbers. Mixing them up is a common error on tests.
How is HSS.CP.A.4 different from HSS.ID.B.5?
HSS.ID.B.5 is about summarizing categorical data in two-way tables and interpreting joint, marginal and conditional relative frequencies. HSS.CP.A.4 uses the same kind of table as a sample space for probability: it adds the decision about independence and the estimate of conditional probabilities for a randomly chosen member. Many teachers teach the two together.
Why does the standard say to approximate conditional probabilities?
Because the table usually comes from a sample, not from the whole population. A proportion computed from a sample estimates the probability for the larger group, and a different random sample would give a slightly different estimate. Larger random samples give estimates that tend to be closer to the population value.
What mistakes do students make with two-way tables?
Common errors include dividing by the grand total when a conditional probability is asked, reversing the condition, comparing raw counts from groups of different sizes, and forgetting to check that cells add up to the totals. Asking "out of which group?" before every fraction prevents the first two errors.
Does HSS.CP.A.4 appear on the SAT?
Yes, the skill does. The digital SAT's Problem-Solving and Data Analysis domain includes questions that give a two-way table and ask for a probability or a conditional probability. Students who always name the sample space before dividing are well prepared for those items.
07
Related Standards
5 standards
These standards connect to HSS.CP.A.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.SP.A.4Prerequisite
Show patterns of association in categorical data with two-way tables
