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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
Problem 1 Complement Rule
(a) 1 − 0.08 = 0.92. (b) P(neither late) = 0.92 × 0.92 = 0.8464, so P(at least one late) = 1 − 0.8464 = 0.1536.
(a) 0.92; (b) 0.1536
Problem 2 Marginal Relative Frequencies
(a) 105/250 = 0.42. (b) 80/250 = 0.32. (c) 90/250 = 0.36.
(a) 0.42, (b) 0.32, (c) 0.36
Problem 3 Joint and Conditional Frequencies
(a) 35/250 = 0.14. (b) Among the 80 11th graders, 42 are members: 42/80 = 0.525. (c) Among the 105 members, 42 are in Grade 11: 42/105 = 0.40.
(a) 0.14, (b) 0.525, (c) 0.40
Problem 4 Exclusive Events
The events are mutually exclusive, so P(pizza or sandwich) = 0.35 + 0.25 = 0.60. If they were independent, P(both) would be 0.35 × 0.25 = 0.0875, but it is 0.
0.60; not independent
Problem 5 Addition Rule with Overlap
P(A or B) = 0.52 + 0.31 − 0.14 = 0.69. Neither: 1 − 0.69 = 0.31. The regions are 0.38 (A only), 0.14, 0.17 (B only) and 0.31.
P(A or B) = 0.69; P(neither) = 0.31
Problem 6 Conditional Probability from Counts
Given an electric bike owner, the probability of owning a car is 90/120 = 0.75. For a random adult from all 300, P(electric and car) = 90/300 = 0.30.
0.75; 0.30
Problem 7 Testing Independence with a Table
P(glasses | left) = 12/40 = 0.30. P(glasses | right) = 48/160 = 0.30. Overall P(glasses) = 60/200 = 0.30. All three are equal, so knowing handedness does not change the probability of wearing glasses.
Yes, independent
Problem 8 Multiplication Rule for Free Throws
(a) 0.72 × 0.72 × 0.72 = 0.3732. (b) 0.28 × 0.72 × 0.72 = 0.1452.
(a) 0.373; (b) 0.145
Problem 9 Tree Diagram for Defects
P(defective) = 0.60(0.03) + 0.40(0.05) = 0.018 + 0.020 = 0.038.
0.038
Problem 10 Reverse Conditional Probability
P(Machine 1 | defective) = P(Machine 1 and defective)/P(defective) = 0.018/0.038 = 0.4737.
≈ 0.474
Problem 11 Designing a Simulation
Each guess is right with probability 1/5, so let digits 0 and 1 mean a correct answer and digits 2 through 9 an incorrect answer. Four digits make one quiz. The numbers correct in the 8 quizzes are 0, 1, 0, 1, 1, 0, 2, 0. At least 2 correct occurred in 1 of 8 quizzes, so the estimate is 1/8 = 0.125. (More than 8 trials would be used in real studies.)
Digits 0-1 = correct; estimate = 1/8 = 0.125
Problem 12 Using Simulation Results
At least 4: (60 + 31)/200 = 0.455. At most 2: (2 + 11 + 34)/200 = 0.235.
≈ 0.455; ≈ 0.235
Problem 13 Finding a Missing Probability
The probabilities sum to 1: k = 1 − (0.18 + 0.34 + 0.27 + 0.06) = 0.15. Then P(X ≥ 2) = 0.27 + 0.15 + 0.06 = 0.48.
k = 0.15; P(X ≥ 2) = 0.48
Problem 14 Mean of a Random Variable
μ = 0(0.18) + 1(0.34) + 2(0.27) + 3(0.15) + 4(0.06) = 1.57. Over many households, the average number of pets is about 1.57.
μ = 1.57 pets per household
Problem 15 Standard Deviation of a Random Variable
Using μ = 1.57: variance = (0 − 1.57)²(0.18) + (1 − 1.57)²(0.34) + (2 − 1.57)²(0.27) + (3 − 1.57)²(0.15) + (4 − 1.57)²(0.06) = 1.2651. So σ = √1.2651 ≈ 1.125. A typical household differs from the mean by about 1.12 pets.
σ ≈ 1.12 pets
Problem 16 Probability Histogram
P(X ≥ 4) = 0.20 + 0.15 = 0.35. Mean = 1(0.10) + 2(0.25) + 3(0.30) + 4(0.20) + 5(0.15) = 3.05.
0.35; mean = 3.05 tickets
Problem 17 Expected Value of a Game
Expected prize = 0(0.7) + 4(0.2) + 10(0.1) = 1.80. Expected net gain = 1.80 − 3 = −1.20, that is a loss of 1.20. In the long run a player loses an average of $1.20 per play.
Expected net gain = −$1.20 per play
Problem 18 Sampling without Replacement
(a) (105/250)(104/249) = 0.1754. (b) senior then non-senior plus non-senior then senior: 2 × (105/250)(145/249) = 0.4892.
(a) 0.175; (b) 0.489
Problem 19 At Least One Success
(a) P(none buy) = 0.855 = 0.4437, so P(at least one) = 1 − 0.4437 = 0.5563. (b) The first two do not buy and the third does: 0.85 × 0.85 × 0.15 = 0.1084.
(a) 0.556; (b) 0.108
Problem 20 Warranty Cost Distribution
μ = 0(0.85) + 120(0.09) + 300(0.05) + 600(0.01) = 31.80. Variance = (0 − 31.8)²(0.85) + (120 − 31.8)²(0.09) + (300 − 31.8)²(0.05) + (600 − 31.8)²(0.01) = 8384.76, so σ ≈ 91.57. Expected profit = 40 − 31.80 = 8.20.
μ = $31.80, σ ≈ $91.57; expected profit $8.20
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