← Back to the Probability Rules worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: S-CP.A.1, S-CP.A.2, S-CP.A.3, S-CP.A.4, S-CP.A.5, S-CP.B.6, S-CP.B.7 · Level: Grade 11 Algebra 2
Problem 1 Sample space for two stages
There are 2 × 6 = 12 equally likely outcomes. Heads with 2, 4 or 6 gives 3 outcomes. 12 outcomes; P = 3/12 = 1/4.
Problem 2 Complement of an event
P(not on time) = 1 − 0.88. 0.12.
Problem 3 Events as subsets
A = {3, 6, 9, 12} and B = {2, 4, 6, 8, 10, 12}. A ∩ B = {6, 12}. The union has 4 + 6 − 2 = 8 outcomes. A ∩ B = {6, 12}; P(A ∪ B) = 8/12 = 2/3.
Problem 4 Addition rule
0.62 + 0.41 − 0.27 = 0.76. 0.76.
Problem 5 Mutually exclusive events
The events cannot overlap, so add: 0.2 + 0.5 = 0.7. Then P(neither) = 1 − 0.7 = 0.3. 0.7 and 0.3.
Problem 6 Check independence
Multiply: 0.35 × 0.20 = 0.07, which equals the probability of both. Yes, they are independent.
Problem 7 Two free throws
Both: 0.7 × 0.7 = 0.49. Exactly one: 0.7 × 0.3 + 0.3 × 0.7 = 0.42. 0.49 and 0.42.
Problem 8 Venn diagram of language classes
Spanish only = 19 − 7 = 12 and French only = 13 − 7 = 6, so 25 students take a language and 35 − 25 = 10 take neither. P(French | Spanish) = 7/19. 10 neither; 7/19.
Problem 9 Two-way table probabilities
P(iPhone) = 64/120 = 8/15. Among the 64 iPhone users, 26 are juniors, so 26/64 = 13/32. Among 60 seniors, 38 have an iPhone, so 38/60 = 19/30. 8/15, 13/32, 19/30.
Problem 10 Conditional probability formula
P(helmet | skateboard) = 0.14 ÷ 0.35 = 0.4. 0.4.
Problem 11 Draws without replacement
Multiply along the mint-mint path: 7/12 × 6/11 = 42/132 = 7/22. 7/22.
Problem 12 Test for independence in a table
P(Android) = 56/120 ≈ 0.467. P(Android | Junior) = 34/60 ≈ 0.567. The values differ, so the events are not independent.
Problem 13 Conditional from percentages
P(bicycle | pet) = 0.25 ÷ 0.60 = 5/12. 5/12 ≈ 0.417.
Problem 14 Addition rule with a table
P(Junior) = 60/120, P(Android) = 56/120 and P(both) = 34/120. The sum is (60 + 56 − 34)/120 = 82/120. 41/60.
Problem 15 Union with cards
Face cards: 12. Spades: 13. Face spades: 3. Union: 12 + 13 − 3 = 22. 22/52 = 11/26.
Problem 16 Two alarms
The complement is that neither rings: 0.08 × 0.08 = 0.0064. So P(at least one) = 1 − 0.0064. 0.9936.
Problem 17 Defective parts from three factories
P(defective) = 0.40(0.01) + 0.35(0.04) + 0.25(0.02) = 0.004 + 0.014 + 0.005 = 0.023. P(B | defective) = 0.014/0.023 ≈ 0.6087. 0.023 and ≈ 0.609.
Problem 18 Two dice, conditional
Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2), so 5 outcomes. Those with a 6: (2,6) and (6,2). 2/5.
Problem 19 Allergy screening
P(allergy and reaction) = 0.03 × 0.90 = 0.027. P(reaction) = 0.027 + 0.97 × 0.05 = 0.027 + 0.0485 = 0.0755. P(allergy | reaction) = 0.027/0.0755 = 0.3576. ≈ 0.36. Most reactions are false alarms because the allergy is rare.
Problem 20 Independence of a complement
P(A ∩ B′) = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A) · P(B′). Therefore A and B′ are independent.