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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: AP Statistics Unit 2 (2026-27) · Level: Grade 12 AP Statistics
Question 1 Two-Way Table (2 points)
(a) 36/180 = 0.20; 96/180 = 0.533. (b) 48/80 = 0.60. Overall P(hot) = 84/180 = 0.467, which is different from 0.60, so the variables are not independent.
(a) 0.20 and 0.533; (b) 0.60, not independent
Scoring: 1 point each part; part (b) requires comparing conditional and marginal probabilities.
Question 2 Venn Diagram Probabilities (2 points)
(a) 0.30 + 0.15 + 0.25 = 0.70. (b) P(B) = 0.15 + 0.25 = 0.40, so P(A | B) = 0.15/0.40 = 0.375.
(a) 0.70; (b) 0.375
Scoring: 1 point each part.
Question 3 Independent Events (2 points)
P(A and B) = 0.6 + 0.5 − 0.8 = 0.3. Also P(A)P(B) = 0.6 × 0.5 = 0.3. The values are equal.
Yes, independent
Scoring: 1 point for finding P(A and B), 1 point for the comparison.
Question 4 Tree Diagram (3 points)
(a) 0.30(0.10) + 0.45(0.20) + 0.25(0.04) = 0.03 + 0.09 + 0.01 = 0.13. (b) P(bus | late) = 0.09/0.13 = 0.692. (c) P(late | bus) = 0.20 is not equal to P(late) = 0.13, so they are not independent.
(a) 0.13; (b) 0.692; (c) not independent
Scoring: 1 point each part.
Question 5 Simulation Design (2 points)
Let digits 0 through 5 represent a made shot and 6 through 9 a miss, so P(make) = 6/10. Read 4 digits for one trial and count the made shots. Repeat for many trials (for example 100). The estimate is the number of trials with at least 3 made shots divided by the total number of trials.
Digits 0-5 = make; estimate = (trials with at least 3 makes)/(number of trials)
Scoring: 1 point for correct digit assignment, 1 point for the trial and estimate.
Question 6 Discrete Random Variable (3 points)
(a) k = 1 − 0.60 = 0.40. (b) μ = 0(0.30) + 1(0.40) + 2(0.20) + 3(0.10) = 1.1. (c) Variance = (0 − 1.1)²(0.30) + (1 − 1.1)²(0.40) + (2 − 1.1)²(0.20) + (3 − 1.1)²(0.10) = 0.89, so σ ≈ 0.943.
k = 0.40; μ = 1.1; σ ≈ 0.94
Scoring: 1 point each part.
Question 7 Expected Value (2 points)
Net gains: 6 gives 6 − 2 = 4 with probability 1/6; 4 or 5 gives 0 with probability 1/3; other rolls give −2 with probability 1/2. Mean = 4(1/6) + 0(1/3) − 2(1/2) = 2/3 − 1 = −1/3 ≈ −0.33. In the long run, a player loses about 33 cents per play.
Expected net gain ≈ −$0.33 per play
Scoring: 1 point for the calculation, 1 point for the interpretation.
Question 8 Without Replacement (2 points)
(a) (5/12)(4/11) = 20/132 = 0.1515. (b) 2 × (5/12)(7/11) = 70/132 = 0.5303.
(a) 0.152; (b) 0.530
Scoring: 1 point each part.
Question 9 At Least One (2 points)
P(no green) = 0.6 × 0.6 × 0.6 = 0.216. P(at least one green) = 1 − 0.216 = 0.784.
0.784
Scoring: 1 point for complement setup, 1 point for the answer.
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