7.SP.C.8Common CoreMathStatistics and ProbabilityGrade 7
7.SP.C.8: Probability of Compound Events
In plain English: 7.SP.C.8 is the Common Core grade 7 math standard that asks students to find probabilities of compound events, events that combine two or more chance steps, such as rolling two number cubes. Students show the sample space with organized lists, tables and tree diagrams, count the outcomes in an event, and design simulations to estimate probabilities that are hard to count.
Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation.
a.Understand that, just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs.
b.Represent sample spaces for compound events using methods such as organized lists, tables and tree diagrams. For an event described in everyday language (e.g., "rolling double sixes"), identify the outcomes in the sample space which compose the event.
c.Design and use a simulation to generate frequencies for compound events. For example, use random digits as a simulation tool to approximate the answer to the question: If 40% of donors have type A blood, what is the probability that it will take at least 4 donors to find one with type A blood?
Common Core State Standards for Mathematics · Domain: Statistics and Probability (SP) · Cluster: Investigate chance processes and develop, use, and evaluate probability models. Also written as 7.SP.8 · Official standard
A compound event is an event that combines two or more chance steps, such as rolling two number cubes or flipping a coin three times. Students learn that its probability works just like the probability of a simple event (one step): it is the fraction of the equally likely outcomes in the sample space (the list of all possible outcomes) for which the event happens. The work is in finding all the outcomes without missing any.
Students use three tools to show a sample space: an organized list, a table and a tree diagram. They turn everyday events, such as "rolling double sixes" or "a tie", into a set of outcomes. When counting is hard, they design a simulation: they act out the chance process many times with a simple tool, such as random digits (digits from 0 to 9 picked so that each is equally likely), coins or number cubes, and use the relative frequency. The official example uses random digits to estimate how many blood donors it takes to find one with type A blood. Simulation results on this page are invented, made by a computer simulation so that they vary like real results.
Learning Objectives
By the end of this lesson, students will be able to:
Show the sample space of a compound event with an organized list, a table or a tree diagram
Identify the outcomes that make up an event described in everyday words, such as "a tie" or "double sixes"
Find the probability of a compound event as the fraction of equally likely outcomes in which it happens
Design a simulation for a compound event, carry it out, and use it to estimate a probability
Compare a simulation's estimate with the probability found by counting
Prior Knowledge Required
Students should already be comfortable with:
Uniform probability models, where each outcome has the same probability 7.SP.C.7
Estimating a probability from relative frequency 7.SP.C.6
Probability as a number from 0 to 1 7.SP.C.5
Multiplying whole numbers to count equal groups, for example 6 groups of 6 3.OA.A.1
"You roll two number cubes and add the numbers. Which is more likely: a sum of 7 or a sum of 12? Or are they equally likely? Explain."
Many students say every sum from 2 to 12 is equally likely. Ask them to find all the ways to make each sum. A sum of 12 needs (6, 6), only one way. A sum of 7 can be (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) or (6, 1), six ways. So a sum of 7 is six times as likely: 6/36 against 1/36. Point out that (1, 6) and (6, 1) are different outcomes, because the two cubes are different. This is the idea of the whole lesson: count the equally likely outcomes, not the results.
Direct Instruction20 minutes
Organized list: write the outcomes in a fixed order, so none is missed or repeated. Hold the first step fixed and run through every choice for the second step.
Table: for two steps, put the first step's outcomes down the side and the second step's across the top. Each cell is one outcome (Diagram 1).
Tree diagram: draw one set of branches for each step. Each path from the start to the end of the tree is one outcome (Diagram 2). Trees work for three or more steps.
Find the probability: count the outcomes in the event and divide by the number of outcomes in the sample space, as long as the outcomes are equally likely.
Simulation: choose a tool whose outcomes match the probabilities, decide what one trial (one run of the whole compound event) is and what counts as a success, run many trials, and find the relative frequency of success.
Organized list
Two friends play one round of rock-paper-scissors, and each picks rock (R), paper (P) or scissors (S) at random. Find the probability of a tie.
Equation: List with the first player's choice fixed: RR, RP, RS, PR, PP, PS, SR, SP, SS, so 9 outcomes. A tie means both pick the same: RR, PP, SS. P(tie) = 3/9 = 1/3.
Table, the official event "rolling double sixes"
Two number cubes are rolled. Find the probability of rolling double sixes and of a sum of 8.
Equation: The table has 6 × 6 = 36 equally likely outcomes. Double sixes is one outcome, (6, 6): P = 1/36. A sum of 8 is (2, 6), (3, 5), (4, 4), (5, 3), (6, 2): P = 5/36 (Diagram 1).
Tree diagram
A coin is flipped 3 times. Find the probability of exactly two heads and of at least one tail.
Equation: The tree has 2 × 2 × 2 = 8 paths. Exactly two heads: HHT, HTH, THH, so P = 3/8. At least one tail: every outcome except HHH, so P = 7/8 (Diagram 2).
Simulation with random digits, the official example
40% of donors have type A blood. What is the probability that it takes at least 4 donors to find one with type A? Use random digits: 0, 1, 2, 3 stand for a type A donor (4 of 10 digits is 40%) and 4 to 9 for another type.
Equation: Read digits until one is 0-3; that is one trial. The row 91740 02975 50473 68813 98491 651 gives 10 trials: 91, 740, 0, 2, 97550, 473, 6881, 3, 98491, 651. Three of them (97550, 6881, 98491) need 4 or more donors: 3/10. A class ran 100 trials and got 23 (invented results), so P ≈ 0.23.
Design a simulation, then check by counting
Jo guesses on a quiz of 4 true-false questions. Estimate the probability that she gets at least 3 right.
Equation: Tool: flip 4 coins for one trial; heads means a right answer. Success: 3 or 4 heads. A class ran 40 trials and got 11 successes (invented results): about 0.28. Check with an organized list: 16 outcomes, and 5 have 3 or more right (4 with exactly 3, 1 with all 4), so P = 5/16 ≈ 0.31.
For the donor problem, show why counting is possible but slow: the first three donors have 10 × 10 × 10 = 1,000 equally likely digit patterns, and 6 × 6 × 6 = 216 of them have no 0-3. So the probability is 216/1000 = 0.216, close to the simulation's 0.23. Ask: "Why is the simulation not exactly 0.216?" (Chance results vary; more trials would come closer.)
Guided Practice15 minutes
Pairs solve these problems. For each one, they choose a list, a table or a tree, and say why.
Guided practice problems with answers
Problem
Answer
A spinner with 4 equal sections, A, B, C and D, is spun twice. Find P(the same letter both times) and P(B, then C).
16 outcomes: P(same) = 4/16 = 1/4; P(B then C) = 1/16
A coin is flipped and a number cube is rolled. Find P(tails and an even number).
A spinner with 3 equal sections numbered 1, 2, 3 is spun, then a coin is flipped. Find P(an odd number and heads).
Tree: 6 outcomes; 1H and 3H, so 2/6 = 1/3
Two number cubes are rolled. List the outcomes in the event "the two numbers differ by 4" and find its probability.
(1, 5), (5, 1), (2, 6), (6, 2): 4/36 = 1/9
Watch for students who count (1, 5) and (5, 1) as one outcome, and for lists that skip outcomes because they were not written in a fixed order.
Independent Practice10 minutes
Independent practice problems with answers
Problem
Answer
A spinner with 3 equal sections (red, blue, yellow) is spun twice. Find P(no red).
9 outcomes; BB, BY, YB, YY: 4/9
Two number cubes are rolled. Find P(both numbers are even).
3 × 3 = 9 of 36 outcomes: 9/36 = 1/4
A coin is flipped, then a spinner with 4 equal sections numbered 1 to 4 is spun. Find P(tails or a 4).
8 outcomes; T1, T2, T3, T4, H4: 5/8
Two number cubes are rolled. Find P(at least one cube shows a 6).
11 of the 36 cells have a 6: 11/36
Closure5 minutes
Exit ticket: (1) Two friends each pick a whole number from 1 to 3 at random. Make an organized list and find P(the two numbers add to 4). (Answer: 11, 12, 13, 21, 22, 23, 31, 32, 33; the sum is 4 for 13, 22 and 31, so 3/9 = 1/3.) (2) An event has a 30% chance each time. Describe how random digits could stand for it. (Answer: for example, 0, 1, 2 mean "yes" and 3 to 9 mean "no".) (3) In one sentence: when would you use a simulation instead of a list?
Differentiation Strategies
For Struggling Students
Give partly filled tables and trees, so students complete the sample space before they count
Use color: highlight the outcomes in the event, as in Diagram 1, before writing the fraction
Run simulations with physical tools (coins, number cubes) before moving to random digits
For Advanced Students
Ask students to find the exact probability for the donor problem that it takes exactly 2 donors, by counting digit patterns, and to check it with the simulation data
Ask students to explain why a tree for 4 coin flips has 16 paths without drawing it
Have students design a simulation for an event with probability 1/3 using a number cube, and one using random digits (hint: skip some digits)
Assessment Guidance
What to Look For
Check that every sample space is complete and that ordered outcomes, such as (1, 5) and (5, 1), are counted separately. Students should circle or highlight the outcomes in the event before writing a probability. For simulations, a complete design names the tool, how its outcomes match the probabilities, what one trial is, what counts as a success, and how many trials to run. When the simulation and the counted probability differ a little, students should say this is normal for chance results.
02
Classroom Activities
3 Activities
1
Sum Race
15 minPairs
Eleven racers, numbered 2 to 12, line up on a track with 8 spaces. Pairs roll two number cubes, and the racer whose number equals the sum moves one space. Students test their warm-up ideas and then explain the results with a table.
Procedure
Before the race, each student writes which racer they think will win and why
Roll two number cubes, add, and move that racer one space. Stop when a racer reaches space 8
Race three times and record the winner of each race
Make a 6 × 6 table of sums, like Diagram 1, and find the probability for each racer's sum
Discussion Questions
Which racer has the best chance to win? Use the table to explain.
Racer 2 and racer 12 have the same probability of moving. What is it?
Would racer 1 ever move? Why is there no racer 1 on the track?
Modification for Distance Learning
Use a two-dice roller app or the random number key of a calculator (two numbers from 1 to 6 each turn) and a shared racetrack slide.
2
Two Bags
15 minGroups of 3
Bag 1 holds a red, a blue and a green cube. Bag 2 holds a red and a blue cube. A player draws one cube from each bag without looking and wins when the colors match. Groups decide whether it is a good game to play.
Procedure
Each student guesses the probability of a match
Draw a tree diagram: 3 branches for bag 1, then 2 for bag 2. List the 6 outcomes and circle the matches, RR and BB. P(match) = 2/6 = 1/3
Play 30 rounds, putting the cubes back each time, and record the number of matches. The model predicts about (1/3)(30) = 10
Organize the same 6 outcomes in a table and in an organized list, and compare the three tools
Discussion Questions
Was your number of matches close to 10? What would you expect if the whole class pooled its rounds?
Why does green never match?
Which change would make a match more likely: adding a green cube to bag 2 or taking the green cube out of bag 1? Use a tree to check.
3
Simulate It
20 minGroups of 3-4
Groups run the official blood donor simulation with random digits, then design their own simulation for a new compound event.
Part 1: The Donor Simulation
Use a printed table of random digits or a calculator's random number key (digits 0 to 9, each equally likely)
0, 1, 2, 3 mean a type A donor; 4 to 9 mean another type. Read digits until a 0-3 appears. That is one trial; record how many digits (donors) it took
Each group runs 10 trials, and the class pools its trials to estimate P(at least 4 donors)
Part 2: Design Your Own
A cereal brand puts one of 3 stickers, each equally likely, in every box. Estimate the probability that 4 boxes give all 3 stickers
Design the simulation: for example, roll a number cube, where 1-2 is sticker A, 3-4 is sticker B and 5-6 is sticker C. One trial is 4 rolls, and a success is a trial with all three stickers
Run 10 trials per group and pool the class results. In one sample class (invented results), 24 of 60 trials were successes: about 0.40
Discussion Questions
Why do 0, 1, 2, 3 stand for type A, and not 0 to 4?
Why must a trial in Part 2 be 4 rolls and not 1 roll?
For teachers: counting all 3 × 3 × 3 × 3 = 81 equally likely outcomes shows that 36 contain all three stickers, so the exact probability is 36/81 = 4/9 ≈ 0.44. How close did the class come?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Table for Two Number Cubes
The table shows all 36 equally likely outcomes of rolling two number cubes, with the first cube down the side and the second across the top. Each cell shows the sum. The five shaded cells make up the event "a sum of 8", so P(sum of 8) = 5/36. The event "double sixes" is the single cell (6, 6), so its probability is 1/36.
Diagram 2: A Tree Diagram for Three Coin Flips
Each flip splits every branch into H (heads) and T (tails), so there are 2 × 2 × 2 = 8 paths, one for each equally likely outcome. The highlighted outcomes HHT, HTH and THH make up the event "exactly two heads", so its probability is 3/8.
04
Homework Assignment
~30 min
7.SP.C.8 Homework: Compound Events, Sample Spaces and Simulations
Directions: Show the sample space for every problem with an organized list, a table or a tree diagram. Circle the outcomes in each event before you write a probability. For simulations, name the tool, what one trial is and what counts as a success.
Part 1: Lists and Tables (Problems 1-2)
A sandwich shop's "surprise lunch" picks one bread (white or wheat) and one filling (turkey, cheese or egg) at random. (a) Make an organized list of the outcomes. (b) Find P(wheat bread with egg). (c) Find P(no turkey).
A number cube is rolled and a spinner with 4 equal sections numbered 1 to 4 is spun, and the two numbers are added. (a) Make a table of the sums. (b) Find P(a sum of 6). (c) Find P(the number on the cube is larger than the number on the spinner). (d) Describe in everyday words an event with probability 1/8.
Part 2: Tree Diagrams (Problems 3-4)
Kim picks an outfit by flipping a coin three times: the first flip chooses the shirt (red or white), the second the pants (jeans or khakis), the third the shoes (sneakers or sandals). (a) Draw a tree diagram. (b) Find P(white shirt, jeans and sneakers). (c) Find P(the outfit has sandals or jeans).
Spinner A has 4 equal sections numbered 1 to 4, and spinner B has 2 equal sections numbered 1 and 2. Both are spun and the numbers are added. (a) Show the sample space with a tree or a table. (b) Find P(the sum is 4). (c) Find P(the sum is odd).
Part 3: Simulations (Problems 5-6)
A cereal brand puts a prize code in 20% of its boxes. You want to estimate the probability that you must buy at least 3 boxes to get your first code. (a) Explain why the digits 0 and 1 can stand for a box with a code. (b) Explain why only the first 2 digits of each trial matter. (c) Use these 20 pairs of random digits, one pair per trial, to estimate the probability: 65 32 69 48 46 90 30 72 25 24 99 01 90 64 31 57 28 01 12 20. (d) Counting all 100 equally likely pairs gives 64/100. Compare.
In a video game, each treasure chest has a 1/4 chance of holding a key. (a) Design a simulation, with a spinner or with random digits, to estimate the probability of finding at least one key in 2 chests. Name the tool, one trial and a success. (b) Find the exact probability by listing the 16 equally likely outcomes of two spins of a spinner with 4 equal sections.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Sample Space
Complete list, table or tree with every outcome once
One or two outcomes missing or repeated
Sample space missing or mostly wrong
Identifying the Event
All outcomes in the event marked correctly
Most outcomes marked
Event outcomes not identified
Probability
Fraction of outcomes in the event, simplified
Correct count, wrong total or not simplified
Missing or wrong
Simulation
Tool, trial and success clearly defined; estimate found and compared
Design has one unclear part
No working design
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
A coin is flipped twice. Which list is the sample space?
Answer: C
Each flip can be H or T, and order matters, so there are 2 × 2 = 4 outcomes: HH, HT, TH, TT. Choice A treats HT and TH as the same outcome. Choice B lists only the outcomes where both flips match. Choice D is the sample space for one flip.
Question 2 of 20 · Multiple Choice
A number cube is rolled and a spinner with 5 equal sections numbered 1 to 5 is spun. How many outcomes are in the sample space?
Answer: D
Each of the 6 numbers on the cube pairs with each of the 5 numbers on the spinner: 6 × 5 = 30. Choice A adds 6 + 5 instead of multiplying. Choice B is 6 × 6, the sample space for two number cubes. Choice C is the sample space for the spinner alone.
Question 3 of 20 · Multiple Choice
Two number cubes are rolled. What is the probability of rolling doubles (both cubes show the same number)?
Answer: A
There are 6 doubles, (1, 1) through (6, 6), in the 36 outcomes: 6/36 = 1/6. Choice B counts only double sixes. Choice C divides the 6 doubles by 12, adding 6 + 6 instead of multiplying. Choice D divides by 21, as if (1, 2) and (2, 1) were the same outcome: 6/21 = 2/7.
Question 4 of 20 · Multiple Choice
Two number cubes are rolled. Which outcomes make up the event "the sum is 5"?
Answer: D
The sum is 5 for (1, 4), (4, 1), (2, 3) and (3, 2), 4 outcomes, so P = 4/36 = 1/9. Choice A counts each pair of numbers once and ignores order. Choice B misses (4, 1). Choice C uses a 0, which is not on a number cube.
Question 5 of 20 · Multiple Choice
A coin is flipped, and then a spinner with 5 equal sections (A, B, C, D, E) is spun. How many paths does the tree diagram have?
Answer: B
Each of the 2 coin branches splits into 5 spinner branches: 2 × 5 = 10 paths. Choice A adds 2 + 5 instead of multiplying. Choice C is 2 × 2 × 2, the tree for three coin flips. Choice D is 5 × 5, as if the spinner were spun twice.
Question 6 of 20 · Multiple Choice
Two number cubes are rolled. What is the probability that the sum is 4?
Answer: C
A sum of 4 is (1, 3), (2, 2) or (3, 1): 3/36 = 1/12. Choice A treats the 11 sums as equally likely. Choice B forgets (3, 1) and gets 2/36. Choice D counts the sums of 4 or less, 6/36.
Question 7 of 20 · Multiple Choice
A bag has 2 red and 3 blue cubes. You draw a cube, put it back, and draw again. What is P(both draws are blue)?
Answer: D
Label the cubes R1, R2, B1, B2, B3. Two draws give 5 × 5 = 25 equally likely outcomes, and 3 × 3 = 9 of them are blue both times: 9/25. Choice A is the probability for one draw. Choice B treats RR, RB, BR and BB as equally likely. Choice C counts only draws of two different blue cubes (3 × 2 = 6) and forgets that the same blue cube can be drawn twice.
Question 8 of 20 · Multiple Choice
A team has a 1/3 chance of winning each game. Which simulation estimates the probability that it wins at least 2 of its next 3 games?
Answer: C
One trial must act out all 3 games, and 2 of the 6 faces (1 or 2) match the 1/3 chance of a win, so roll 3 number cubes per trial and count the trials with at least 2 wins. Choice A uses coins, which give a 1/2 chance, not 1/3. Choice B acts out only one game. Choice D runs just one trial, which cannot estimate a probability.
Question 9 of 20 · Multiple Choice
A student ran 40 trials of the donor simulation and found that 9 trials needed at least 4 donors. What is the estimate of P(at least 4 donors)?
Answer: B
9/40 = 0.225. Choice A divides 9 by 100 instead of by 40. Choice C divides 9 by 10. Choice D divides 9 by the 31 trials that needed fewer donors: 9/31 ≈ 0.29.
Question 10 of 20 · Multiple Choice
In the donor simulation, the digits 0-3 mean a type A donor. Which string of digits is one trial that needed at least 4 donors?
Answer: D
Reading 7, 4, 9 (not type A) and then 1 (type A) means the fourth donor was the first with type A. Choice A finds type A at the third donor (2). Choice B finds it at the first donor. Choice C finds it at the second donor (0).
Question 11 of 20 · Multiple Choice
A family picks a snack at random (popcorn or pretzels) and a drink at random (water, juice or lemonade). What is P(pretzels and not water)?
Answer: B
There are 2 × 3 = 6 snack-drink pairs, and 2 of them are pretzels with juice or lemonade: 2/6 = 1/3. Choice A counts only one of the two pairs. Choice C is P(pretzels) alone. Choice D divides 2 by the 5 items, adding snacks and drinks instead of listing pairs.
Question 12 of 20 · Multiple Choice
A coin is flipped 3 times. What is P(all three flips are the same)?
Answer: C
Of the 8 outcomes in the tree, 2 are all the same: HHH and TTT. 2/8 = 1/4. Choice A counts only HHH. Choice B treats 0, 1, 2 and 3 heads as equally likely, with 2 of those 4 all the same. Choice D treats "all heads", "all tails" and "mixed" as equally likely.
Question 13 of 20 · Multiple Choice
How do you find the probability of a compound event when all the outcomes are equally likely?
Answer: A
Just as with simple events, the probability is the fraction of outcomes in the sample space for which the event happens. Choice B compares the event with the rest of the outcomes instead of with all of them. Choices C and D do not count the outcomes in the event against the whole sample space.
Question 14 of 20 · Multiple Choice
Two spinners each have 3 equal sections numbered 1, 2, 3. Both are spun and the numbers are multiplied. What is P(the product is even)?
Answer: A
A product is even when at least one number is 2: (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), 5 of the 9 outcomes. Choice B is the probability of an odd product. Choice C counts only the outcomes where the first spin is 2. Choice D assumes even and odd are equally likely.
Question 15 of 20 · Short Answer
Two of the four students Ana, Ben, Cal and Dee will be picked at random to be class representatives. Make an organized list of the possible pairs, then find P(Ana is picked).
Pairs: Ana-Ben, Ana-Cal, Ana-Dee, Ben-Cal, Ben-Dee, Cal-Dee, so 6 pairs. Ana is in 3 of them: P = 3/6 = 1/2. (Here order does not matter: Ana-Ben is the same pair as Ben-Ana.)
Question 16 of 20 · Short Answer
Two number cubes are rolled. List the outcomes in the event "the two numbers differ by 2" and find its probability.
A coin is flipped twice, and then a spinner with 3 equal sections numbered 1, 2, 3 is spun. (a) How many outcomes does the tree diagram have? (b) Find P(exactly one head and an odd number).
(a) 2 × 2 × 3 = 12 outcomes. (b) Exactly one head: HT or TH. Odd number: 1 or 3. That gives 2 × 2 = 4 outcomes (HT1, HT3, TH1, TH3): P = 4/12 = 1/3.
Question 18 of 20 · Short Answer
For the official donor example (40% of donors have type A blood), explain how random digits act out the problem. Then use these invented results: in 60 trials, 13 needed at least 4 donors. Estimate the probability.
Let 0, 1, 2, 3 stand for type A (4 of the 10 digits, 40%) and 4 to 9 for other types. One trial is reading digits until a 0-3 appears; a success is a trial that needs 4 or more digits. Estimate: 13/60 ≈ 0.22.
Question 19 of 20 · Short Answer
25% of cereal boxes contain a free ticket. Design a simulation to estimate the probability that none of 3 boxes has a ticket. Name the tool, one trial and a success. Then find the exact probability with a spinner model of 4 equal sections.
Tool: a spinner with 4 equal sections, where 1 section means a ticket (or random digits 1-4, skipping 0 and 5-9, with 1 meaning a ticket). One trial: 3 spins. Success: no ticket in the 3 spins. Exact: 4 × 4 × 4 = 64 equally likely outcomes, and 3 × 3 × 3 = 27 have no ticket: P = 27/64 ≈ 0.42.
Question 20 of 20 · Short Answer
Two number cubes are rolled. Describe in everyday words the event made of the outcomes (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6), and give its probability.
The event is "the sum is 10 or more" (a sum of at least 10). It has 6 of the 36 outcomes: P = 6/36 = 1/6.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does 7.SP.C.8 mean?
7.SP.C.8 means students find probabilities of compound events, events made of two or more chance steps. They show all the outcomes with an organized list, a table or a tree diagram, count the outcomes in the event, and divide by the total. When counting is hard, they design a simulation and use its relative frequency.
What is a compound event in 7th grade math?
A compound event combines two or more chance steps, such as flipping two coins or rolling a number cube and spinning a spinner. "Getting heads on both coins" is a compound event. A simple event has one step, such as rolling a 4 on one number cube.
When should students use a list, a table or a tree diagram?
Any of the three works, and the best choice depends on the number of steps. A table is clear for exactly two steps, such as two number cubes. A tree diagram works for two or more steps and shows the order of the steps. An organized list is quick when there are few outcomes, as in rock-paper-scissors.
What grade is 7.SP.C.8, and what comes after it?
7.SP.C.8 is a grade 7 standard, the last in the grade 7 probability cluster. It builds on probability models from 7.SP.C.7. In high school, students study independent events and conditional probability (HSS.CP.A.2) and use counting methods for compound events (HSS.CP.B.9).
Why are (1, 6) and (6, 1) different outcomes?
They are different because the two number cubes are different objects. Imagine one red cube and one blue cube: red 1 with blue 6 is not the same as red 6 with blue 1. Counting both keeps all 36 outcomes equally likely. Treating them as one outcome is a frequent source of wrong answers.
What is a simulation in probability?
A simulation acts out a chance process with a simpler tool, such as coins, number cubes, spinners or random digits, many times. Each run is a trial. The relative frequency of success in many trials estimates the probability. Simulations are useful when the sample space is too large to list.
How do random digits work in the 7.SP.C.8 blood donor example?
Random digits work because each digit from 0 to 9 is equally likely, so 4 of the 10 digits stand for a 40% chance. In the official example, 0 to 3 mean a donor with type A blood. Students read digits until one is 0 to 3 and count how many donors that took. Many trials give an estimate of the probability that it takes at least 4 donors.
Why doesn't the simulation give the exact answer?
A simulation is a chance process itself, so its results vary from run to run. With more trials, its relative frequency usually gets closer to the exact probability. That is why classes pool their trials, and why a simulation estimate is always stated as "about".
What mistakes should teachers watch for?
A common mistake is counting results instead of outcomes, for example treating the 11 sums of two cubes as equally likely. Other frequent errors are missing outcomes in unorganized lists, adding the number of choices instead of multiplying them, and simulations in which one trial does not act out the whole compound event.
How can parents help with 7.SP.C.8 at home?
Parents can use everyday choices. Ask: "If we pick a random shirt from 3 and random pants from 2, how many outfits are possible?" (6.) Play a few rounds of rock-paper-scissors, and ask your child to list all 9 outcomes and find the chance of a tie.
07
Related Standards
5 standards
These standards connect to 7.SP.C.8: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.SP.C.7Prerequisite
Develop probability models, find probabilities, and compare them with observed data
