7th Grade · 7.SP.C.7

Theoretical probability in 7th grade

Theoretical probability is the chance of an event found by reasoning, before anything is flipped, rolled or spun. In 7th grade, your child lists every possible outcome, gives each one a probability and uses that list to answer questions. Your child also compares these predictions with real results and explains why they do not always match.

What the standard says

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

7.SP.C.7 Develop a probability model and use it to find probabilities of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy.
7.SP.C.7.a Develop a uniform probability model by assigning equal probability to all outcomes, and use the model to determine probabilities of events.
7.SP.C.7.b Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process.

What it means, in plain words

An outcome is one possible result, such as landing on 4 when a number cube is rolled. A probability model is a list of every possible outcome with a probability for each one. The probabilities in a probability model always add up to 1.

When all the outcomes are equally likely, the model is called uniform. A number cube has 6 outcomes, each with a probability of $\frac{1}{6}$ . For an event that includes several outcomes, the theoretical probability is the number of outcomes in the event divided by the total number of outcomes. Rolling an even number includes 2, 4 and 6, so its probability is $\frac{3}{6} = \frac{1}{2}$ .

Some probability models are not uniform. A thumbtack can land point up or point down, but those two outcomes are not equally likely. For objects like this, your child builds the model from data: drop the tack many times and use the results as the probabilities. When real results are far from what a model predicts, your child looks for a reason, such as too few trials or an object that is not fair.

Your child should be able to:

List all the outcomes of a chance process
Build a uniform probability model and use it to find the probability of an event
Build a model from data when the outcomes are not equally likely
Compare what a model predicts with real results and explain a big difference

Example

A bag holds 5 tiles with the letters A, B, C, D and E. One tile is drawn without looking. What is the probability of drawing a vowel?

There are 5 equally likely outcomes, so each letter has a probability of $\frac{1}{5}$ .
The vowels are A and E, so the event includes 2 outcomes.
$P(\text{vowel}) = \frac{2}{5}$ , which is 0.4.
The probability of drawing a vowel is $\frac{2}{5}$ .

Example

A thumbtack is dropped 100 times. It lands point up 62 times and point down 38 times. What probability model fits these results?

The two outcomes are not equally likely, so the model comes from the data.
$P(\text{up}) = \frac{62}{100} = 0.62$ and $P(\text{down}) = \frac{38}{100} = 0.38$ .
Check that the model is complete: $0.62 + 0.38 = 1$ .
The model gives point up a probability of 0.62 and point down a probability of 0.38.

Where children usually get stuck

On a spinner that is half red, with the other half split into a blue quarter and a green quarter, your child says the chance of red is $\frac{1}{3}$ .
There are 3 colors, so the child counts 3 outcomes. The colors are not equally likely, because red covers half the spinner.
What helps. Draw a line that splits the red half into two quarters. Now there are 4 equal parts and 2 of them are red, so the chance of red is $\frac{2}{4} = \frac{1}{2}$ .
After a number cube lands on 6 three times in a row, your child says a 6 is now less likely on the next roll.
It feels like the cube should even itself out. The cube has no memory, so each roll still has a probability of $\frac{1}{6}$ for a 6.
What helps. Ask what has changed about the cube itself. Nothing has, so its chances have not changed either.
For a bag of red, blue and green marbles, your child writes a model of red 0.5 and blue 0.3, and stops.
The child lists the outcomes the question mentions and forgets the rest. Every outcome needs a place in the model.
What helps. Have your child add up the probabilities. They must total 1, so the missing 0.2 belongs to green.

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
A student wants a probability model for the next flip of a fair coin. The table shows results from 12 recent flips. Lily says, "Use an empirical model because the table shows the actual results of the flips." Jordan says, "Use a uniform model because the coin is fair and heads and tails are equally likely." Which student is correct?
Result Count
Heads 7
Tails 5
A
Lily
B
Jordan
Show the answer and how to solve it
Jordan is correct because the coin is fair, making heads and tails equally likely by design. A uniform model is appropriate.
1. 1Look at the table. The coin landed heads 7 times and tails 5 times in 12 flips.
2. 2A uniform model is used when the situation itself makes outcomes equally likely by design, such as a fair coin. An empirical model is used when outcomes are not equally likely by design and we rely on observed frequencies instead.
3. 3The coin is described as fair, which guarantees that heads and tails are equally likely. Having a table of results does not mean we must use an empirical model. Lily's reasoning is wrong. Jordan is correct.
Common mistake. A common mistake is thinking that any table of observed results requires an empirical model. If the device itself is fair, a uniform model is correct no matter what the counts show.
The idea underneath. Choose the model based on the kind of situation, not the presence of data: fair equal-chance devices call for uniform models, while real performance or real-world data call for empirical models.
2On level
The model for a spinner is shown below, followed by the results of 2 runs. Mia says, 'The spinner is probably biased toward red because red has the highest count in both runs.' Carlos says, 'These results look uneven because 80 spins per run is not enough to be certain.' Whose explanation is better supported by the evidence?
$\displaystyle P(\text{red}) = P(\text{blue}) = P(\text{green}) = P(\text{yellow}) = \frac{1}{4}$
Run Red Blue Green Yellow Total
1 30 18 17 15 80
2 29 19 18 14 80
A
Mia's explanation
B
Carlos's explanation
Show the answer and how to solve it
Mia's explanation is better supported because red is the highest count in both runs by a clear margin, making repeated bias a more likely explanation than random variation from sample size.
1. 1Start with the model. It says all 4 colors are equally likely, each with probability $\frac{1}{4}$ .
2. 2Look at the table. There are 2 runs and each run has 80 spins. All row sums check out: $30 + 18 + 17 + 15 = 80$ and $29 + 19 + 18 + 14 = 80$ .
3. 3Red is the largest count in both runs: 30 in run 1 and 29 in run 2. Red exceeds the next-highest color by 12 in run 1 and 10 in run 2. The same color leading by a wide margin in every run is the key pattern.
4. 4Because red stays far ahead in both 80-spin runs, Mia's explanation that the spinner is biased toward red is better supported than Carlos's claim that the sample is too small.
Common mistake. A common mistake is treating any mismatch with the model as a sample-size problem. Here the clue is that red is the highest count in every run, not just once, which points to the spinner itself rather than chance.
The idea underneath. One uneven run can happen by chance, but the same strong skew in every run is stronger evidence about the device itself.
3On level
Select all situations where an empirical model is the better choice for a future probability question.
A
A standard number cube was rolled 24 times, and a 4 came up 5 times; you want the chance of rolling a 4 on the next roll.
B
A tennis player made 14 of 25 first serves in her last match; you want the chance she makes her next first serve.
C
A fair coin was flipped 30 times, and tails came up 16 times; you want the chance of tails on the next flip.
D
In the last 10 rainy mornings, Priya's school called a delay 4 times; you want the chance of a delay on the next rainy morning.
E
A bag has 6 red chips and 6 blue chips, and one chip is chosen without looking; you want the chance of picking a red chip.
Show the answer and how to solve it
Select the tennis serve record and the rainy-morning delay record because both probabilities must be estimated from observed performance or real-world conditions, not from known equal-likelihood outcomes.
1. 1Read each situation and ask one question: are the outcomes already known to be equally likely, or do you need past results to estimate the chance?
2. 2The tennis serve record needs an empirical model because a player's serve success depends on observed skill, not equally likely outcomes. The 14 makes in 25 first serves are the evidence to use.
3. 3The rainy-morning delay record also needs an empirical model. School delays depend on real conditions, so the 4 delays in 10 rainy mornings are the relevant past results.
4. 4The standard number cube, the equal-chip bag, and the fair coin all use uniform models. Their outcomes are already known to be equally likely, so extra sample data like 5 fours in 24 rolls or 16 tails in 30 flips do not change the model choice.
Common mistake. A common mistake is selecting the die or the coin just because they show past results, or skipping the delay record because 4 of 10 looks close to half. What matters is whether equal likelihood is already known, not whether data are shown or whether results look balanced.
The idea underneath. The key question is not "Is there data?" but "Are the probabilities already known from equally likely outcomes?"

Result	Count
Heads	7
Tails	5

Run	Red	Blue	Green	Yellow	Total
1	30	18	17	15	80
2	29	19	18	14	80

4Stretch

Priya spun the spinner 50 times and landed on a number greater than 7 a total of 12 times. A count is reasonable if the difference from the expected count is at most 3. Which statement best describes the error in Priya's check?

Probability	$\displaystyle P(\text{number} > 7) = \frac{3}{10}$
Expected count	$\displaystyle \text{Expected count} = \frac{3}{10} \times 50 = 15$
Compare	$\displaystyle \|12 - 15\| = 3$
Conclusion	So 12 is not reasonable

The check multiplied to find the expected count incorrectly.

The check used the probability of not getting a number greater than 7.

The check treated a count exactly 3 away from 15 as outside the allowed range.

The check said 12 was unreasonable just because it is less than 15, without checking whether 12 is at most 3 away from 15.

Show the answer and how to solve it

The correct choice is that the check treated a count exactly 3 away from the expected count as outside the allowed range. Since the criterion is at most 3 away, 12 is on the boundary and is reasonable.

1The numbers greater than 7 on this spinner are 8, 9, and 10 — that is 3 favorable outcomes out of 10 equal sectors.
2The expected count in 50 spins is $\frac{3}{10}$ of 50, which equals 15.
3A margin of ±3 means any count from 12 to 18 is reasonable. The boundary values 12 and 18 are included because the criterion says at most 3 away.
4The check correctly computes |12 − 15| = 3, but then concludes 12 is not reasonable. Because a difference of exactly 3 satisfies 'at most 3 away,' 12 sits right on the boundary and is still reasonable. The error is treating the boundary value as outside the allowed range.

Common mistake. A common mistake is reading 'at most 3 away' as 'strictly fewer than 3 away.' A count that is exactly 3 from the expected count still satisfies the criterion and should be judged reasonable.

The idea underneath. Expected count is a center point, not a required exact result; reasonableness depends on how far the observed count is from that point.