7.SP.7: Develop and Test Probability Models
Also written as 7.SP.C.7 and CCSS.MATH.CONTENT.7.SP.C.7 · Cluster: Investigate chance processes and develop, use, and evaluate probability models.
The Official Wording
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.
Common Core State Standards, Grade 7, Statistics & Probability · Read it at the source · Checked October 14, 2026
What 7.SP.7 Means in Plain English
Seventh graders build probability models two ways. In part a every outcome is equally likely: pick one of 28 students and P(Jane) = 1/28, P(girl) = 16/28. In part b the model comes from data: a paper cup lands on its side 34 times in 50, so about 0.68.
Then they hold models up against results. If 70 name draws give a girl only 31 times when the model predicts 40, they look for a cause, like slips folded differently. High school judges models against data by simulation (HSS-IC.2).
7.SP.7 I Can Statements
- I can build a model where every outcome is equally likely.
- I can build a model from frequencies I observe.
- I can explain why observed results differ from a model.
7.SP.7 Worked Example
Two models side by side: one from equal chances (part a), one from 50 tosses of a cup (part b), and a test of the first against real draws.
(a) A class has 28 students, 16 girls and 12 boys, and one name is drawn at random. Find P(Jane) and P(a girl). (b) A paper cup tossed 50 times landed on its side 34 times, open end down 10 times and open end up 6 times. Build a model for the cup. Then compare: 70 draws of names, with replacement, gave a girl 31 times.
- (a) Uniform model: each of the 28 names is equally likely. Jane is one outcome.P(Jane) = 1/28
- (a) The event “a girl” is 16 of the 28 outcomes.P(girl) = 16/28 = 4/7 ≈ 0.57
- (b) The cup has three ways to land, but nothing says they are equally likely, so build the model from the tosses.P(side) = 34/50 = 0.68, P(down) = 10/50 = 0.2, P(up) = 6/50 = 0.12
- (b) Test equal chances against the data: a uniform model would give each landing 1/3, about 0.33. The side came up about twice that often.0.68 against 0.33: the cup’s outcomes are not equally likely
- Compare model (a) with real draws. The model predicts 4/7 of 70, but the draws gave 31 girls. A gap that size calls for a look at how the names were mixed.Expected 4/7 × 70 = 40; observed 31/70 ≈ 0.44
Check it: Each model’s probabilities must add to 1: 16/28 + 12/28 = 28/28 = 1, and 0.68 + 0.2 + 0.12 = 1.
Common Mistakes with 7.SP.7
Assuming every outcome is equally likely
The cup can land three ways, so each has probability 1/3
Try this: Toss the object 50 times before writing a model; equal chances need a reason, like symmetry, and a cup has none.
Dividing by the wrong total
P(girl) = 16/12, girls over boys
Try this: Put the whole sample space in the denominator, all 28 students, so P(girl) = 16/28, and check that the answer is below 1.
Calling every gap bad luck
31 girls in 70 draws instead of 40 is just luck
Try this: Compare the gap with the expected count and inspect the process; when results stay far off, check the slips, the mixing and the tally.
Where 7.SP.7 Fits
Practice 7.SP.7
Quick Check
- A bag has 5 red, 3 blue and 2 yellow tiles. Using a uniform model, what is P(blue)?
- A spinning penny landed heads 18 times in 60 spins. Build a model for one spin.
- A model says P(red) = 1/2, but 100 spins gave red 72 times. Name one possible reason.