7th Grade · 7.SP.C.8

Compound probability in 7th grade

Compound probability is the chance that two or more things happen together, such as flipping heads and rolling a 6. In 7th grade, your child finds these chances by listing every possible combination and counting the ones that fit. Organized lists, tables and tree diagrams help make sure no combination is missed.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation.
7.SP.C.8.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.
7.SP.C.8.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.
7.SP.C.8.c Design and use a simulation to generate frequencies for compound events.

What it means, in plain words

A compound event is made of two or more simple events, for example flipping a coin and then rolling a number cube. The sample space is the list of every possible outcome. Here each outcome is a pair, such as heads and 3. Compound probability uses the same rule as a single event: the number of outcomes that fit divided by the total number of outcomes.

Listing every outcome is the hard part, so your child learns three ways to do it. An organized list writes the outcomes in a fixed order. A table puts one event across the top and the other down the side, which works well for two number cubes. Tree diagrams start with one branch for each result of the first event. Each of those branches then splits into one branch for each result of the second event.

Some compound events have too many outcomes to list. Then your child designs a simulation, a way to imitate the events with coins, spinners or a computer. Running it many times gives results that estimate the probability.

Your child should be able to:

List the sample space for compound events with an organized list, a table or a tree diagram
Find the outcomes that match an event described in words, such as rolling doubles with two number cubes
Find the probability of a compound event as a fraction
Design a simple simulation to estimate a probability

Example

A coin is flipped and a number cube is rolled. What is the probability of getting heads and an even number?

Draw a tree diagram: 2 branches for the coin, then 6 branches for the cube at the end of each coin branch.
That gives $2 \times 6 = 12$ outcomes.
The outcomes that fit are H2, H4 and H6 (H2 means heads and 2), so 3 outcomes fit.
$P = \frac{3}{12} = \frac{1}{4}$ . The probability of heads and an even number is $\frac{1}{4}$ .

Example

Two number cubes are rolled. What is the probability that the two numbers add up to 7?

Make a table with one cube across the top and the other down the side. It has $6 \times 6 = 36$ boxes, one for each outcome.
The pairs that add up to 7 are $(1, 6)$ , $(2, 5)$ , $(3, 4)$ , $(4, 3)$ , $(5, 2)$ and $(6, 1)$ : 6 outcomes.
$P = \frac{6}{36} = \frac{1}{6}$ . The probability that the numbers add up to 7 is $\frac{1}{6}$ .

Where children usually get stuck

For heads and an even number, your child adds the two chances, $\frac{1}{2} + \frac{1}{2} = 1$ , and says the result is certain.
Adding is the first thing that comes to mind for "and". Both things must happen at once, which is less likely than either one alone.
What helps. Draw the tree diagram together and count. Only 3 of the 12 outcomes fit, so the answer has to be smaller than each single chance.
Your child says a sum of 7 with two number cubes has a probability of $\frac{1}{11}$ , because the sums go from 2 to 12.
The child treats the 11 possible sums as equally likely. A sum of 7 can happen 6 ways, but a sum of 2 happens only one way.
What helps. Fill in a 6-by-6 table of sums together. Count the 7s and the 2s and compare.
Your child lists outfits from 3 shirts and 2 pairs of pants in no particular order and finds only 5 outfits.
Without an order, one pairing gets skipped and nothing shows that it is missing.
What helps. Have your child write the 3 shirts in a column and draw a branch from each one to each pair of pants. 3 shirts with 2 branches each give 6 outfits.

Try four practice questions

Real 7th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Spin Spinner A and Spinner B once. Complete the setup for landing on C on Spinner A and landing on 3 on Spinner B.
Spinner A
Spinner B
Favorable outcomes
$\displaystyle \times$
$\displaystyle =$
Total outcomes
$\displaystyle \times$
$\displaystyle =$
Probability
$\displaystyle /$
Show the answer and how to solve it
The answer is $\frac{1}{16}$ because there is 1 favorable pair out of 16 total pairs.
1. 1On Spinner A, exactly 1 sector shows C. On Spinner B, exactly 1 sector shows 3. Each spinner contributes 1 favorable choice.
2. 2Each of Spinner A's 4 sectors can pair with each of Spinner B's 4 sectors, giving $4 \times 4 = 16$ total pairs. The matching sector on A pairs with the matching sector on B, giving one favorable pair.
3. 3Write the probability as favorable outcomes over total outcomes.
Common mistake. A common mistake is writing 4 in the numerator, using the total number of sectors on one spinner instead of the count of favorable sectors. Only the sector labeled C on Spinner A and the sector labeled 3 on Spinner B qualify, giving $1 \times 1 = 1$ favorable pair.
The idea underneath. For a compound event with separate stages, multiply the matching choices at each stage to count favorable outcomes, and multiply all choices at each stage to count total outcomes.
2On level
The compound event is getting red on a spinner, then flipping heads on a coin. Is the simulated estimate consistent with the theoretical probability?
Quantity Decimal
Theoretical probability 0.30
Simulated estimate 0.34
Allowed margin 0.05
A
Yes
B
No
Show the answer and how to solve it
The answer is Yes because 0.34 is 0.04 away from 0.30, and 0.04 is within the allowed margin of 0.05.
1. 1Look at the table. The theoretical probability is 0.30, the simulated estimate is 0.34, and the allowed margin is 0.05.
2. 2Find the difference between the simulated estimate and the theoretical probability.
3. 3The allowed margin is 0.05. Since 0.04 is less than 0.05, the estimate is within the allowed margin.
4. 4So the answer is Yes.
Common mistake. A common mistake is saying No just because 0.34 is greater than 0.30. Consistency depends on how far apart the numbers are, not on whether the estimate is above or below the theoretical probability.
The idea underneath. A simulated estimate is consistent when its distance from the theoretical probability is at most the allowed margin; only the size of the difference matters.
3On level
A bag has 3 green tiles and 2 yellow tiles. Two tiles are drawn one at a time without replacement. You want the theoretical probability that both tiles are green. Sofia says, "Use a table, because there are 2 draws so a 5-by-5 grid will show all the outcomes." Noah says, "Use a tree diagram, because the color choices for the second draw change after the first tile is taken out." Whose method makes the changing chances after each draw easier to track: Sofia, Noah, or Neither?
A
Sofia
B
Noah
C
Neither
Show the answer and how to solve it
Noah is the answer for the stated goal. The tree makes the changing chances easier to track; the other method can also be made correct.
1. 1The question asks which method makes the changing chances easier to track.
2. 2After a green tile is drawn, 2 green and 2 yellow remain. After a yellow, 3 green and 1 yellow remain.
3. 3Noah's tree branches after each draw. Labeling the branches with the remaining counts shows the updated chances beside each possible path.
4. 4A table can also work. Label the individual objects and cross out pairs that repeat an object already removed. For more stages, add entries for the later results.
5. 5Sofia's proposed table organizes outcomes, but does not describe the changing chances at each stage. Noah's tree puts those updated chances beside the matching branches, making them easier to track.
Common mistake. Do not call the grid wrong just because the draws are without replacement. Crossing out the 5 pairs that repeat a tile makes it work, but the tree shows the changing chances after each draw more directly.
The idea underneath. For draws without replacement, a tree makes updated chances easy to follow. A correctly marked table or a list with the right chances can also work.
4Stretch
Two fair spinners each have sections labeled 1, 2, and 3. Complete the sample-space table with rows in the order 1, 2, 3 (first spinner) and columns in the order 1, 2, 3 (second spinner). Then check every outcome where both spinners land on the same number, and enter the favorable and total counts.
Favorable outcomes
Total outcomes
Show the answer and how to solve it
Enter 3 for favorable outcomes and 9 for total outcomes because 3 of the 9 equally likely outcomes have both spinners showing the same number.
1. 1Fill in the headers using the order given. The row labels are 1, 2, 3 (first spinner) and the column labels are 1, 2, 3 (second spinner).
2. 2There are 3 sections on each spinner, so the sample space has 9 equally likely outcomes.
3. 3The event is "both spinners show the same number." Check along the main diagonal: (1, 1), (2, 2), and (3, 3) each satisfy the condition. Every other cell pairs two different numbers and does not qualify. That gives 3 favorable outcomes.
4. 4Enter 3 for favorable outcomes and 9 for total outcomes. The probability that both spinners match is $\frac{3}{9} = \frac{1}{3}$ .
Common mistake. A common mistake is finding only one matching outcome instead of all three. Remember to scan every row: (1, 1) in row 1, (2, 2) in row 2, and (3, 3) in row 3 all qualify.
The idea underneath. A sample-space table makes each equally likely outcome one box, so probability comes from counting favorable boxes over total boxes; for an "or" event, the shared box still counts only once.

Favorable outcomes	$\displaystyle \times$ $\displaystyle =$
Total outcomes	$\displaystyle \times$ $\displaystyle =$
Probability	$\displaystyle /$

Quantity	Decimal
Theoretical probability	0.30
Simulated estimate	0.34
Allowed margin	0.05