← Back to the Probability Models worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: 7.SP.C.5, 7.SP.C.6, 7.SP.C.7 · Level: Grade 7
Problem 1 Certain event
Every face is less than 7, so the event always happens.
certain; probability 1
Problem 2 Describe the probability
0.92 is close to 1; 0 never happens; 1 always happens; 0.08 is close to 0; 0.5 is halfway.
0.92 likely; 0 impossible; 1 certain; 0.08 unlikely; 0.5 equally likely as not
Problem 3 Marble bag
Total: 4 + 6 + 5 = 15. P(yellow) = 6/15 = 2/5 = 0.4.
2/5 = 0.4
Problem 4 Spinner with 8 sections
The primes from 1 to 8 are 2, 3, 5 and 7, which is 4 outcomes out of 8.
P(prime) = 4/8 = 1/2.
1/2
Problem 5 Experimental probability
14 ÷ 40 = 0.35.
0.35
Problem 6 Complement
1 − 0.18 = 0.82.
0.82
Problem 7 Which is more likely
3/8 = 0.375, and 0.4 is greater than 0.375.
Event B
Problem 8 Predict a count
Expected = 3/10 × 250 = 75.
about 75 red tiles
Problem 9 Coin flips
Expected: 1/2 × 60 = 30. Observed 36 is 6 more, which is 10% of the flips.
A difference this size can happen by chance, so the result is reasonable.
30 heads expected; 36 is reasonable
Problem 10 Missing probability
The probabilities add to 1: P(green) = 1 − 0.40 − 0.35 = 0.25.
P(not red) = 1 − 0.40 = 0.60.
P(green) = 0.25; P(not red) = 0.60
Problem 11 Die results
Experimental: 10 ÷ 50 = 0.2. Theoretical: 1/6 ≈ 0.167.
Expected number of 4s: 1/6 × 50 ≈ 8.3.
0.2 compared with about 0.167; about 8 fours expected
Problem 12 Probability model
Outcome 1 covers 2 of 5 sections, 2 covers 1 of 5, and 3 covers 2 of 5.
Model: P(1) = 2/5, P(2) = 1/5, P(3) = 2/5. P(2 or 3) = 1/5 + 2/5 = 3/5.
P(2 or 3) = 3/5
Problem 13 Free throws
Estimate: 140 ÷ 200 = 0.7.
Prediction: 0.7 × 30 = 21.
0.7; about 21 free throws
Problem 14 Mystery bag
Experimental P(red) = 22 ÷ 80 = 0.275.
Estimate: 0.275 × 40 = 11.
about 11 red chips
Problem 15 Is it uniform
If the outcomes were equally likely, each would be expected about 100 ÷ 3 ≈ 33 times.
A was 15 above that and C was 11 below, which is large for 100 spins.
Experimental probabilities: 0.48, 0.30, 0.22. A better model is about P(A) = 0.5, P(B) = 0.3, P(C) = 0.2.
The claim is not supported; the model is non-uniform.
Problem 16 Prize boxes
Experimental: 14/40 = 0.35, 9/40 = 0.225, 8/40 = 0.2, 9/40 = 0.225.
Equally likely would predict 1/4 × 40 = 10 of each. The counts differ from 10 by at most 4, which can happen by chance in 40 boxes.
The claim is reasonable.
Problem 17 Weather model
P(rain) = 1 − 0.45 − 0.30 = 0.25.
Sunny: 0.45 × 30 = 13.5. Cloudy: 0.30 × 30 = 9. Rain: 0.25 × 30 = 7.5.
P(rain) = 0.25; about 13.5 sunny, 9 cloudy and 7.5 rainy days
Problem 18 Defective bulbs
P(defective) = 5 ÷ 250 = 0.02. P(not defective) = 1 − 0.02 = 0.98.
Defective bulbs: 0.02 × 12,000 = 240.
0.98; about 240 defective bulbs
Problem 19 Weighted number cube
The other five faces share 1 − 0.25 = 0.75, so each has 0.75 ÷ 5 = 0.15.
P(even) = P(2) + P(4) + P(6) = 0.15 + 0.15 + 0.25 = 0.55.
Expected 6s: 0.25 × 80 = 20.
0.15 each; P(even) = 0.55; 20 sixes
Problem 20 Carnival game
P(prize) = 4/10 = 0.4. Predicted prizes: 0.4 × 150 = 60.
Predicted grand prizes: 1/10 × 150 = 15.
52 is 8 fewer than 60, which is plausible for 150 plays.
0.4; 60 prizes; 15 grand prizes; 52 is reasonably close