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Academic Support AP Statistics Unit 2: Probability, Random Variables, and Probability Distributions

Probability Rules: Addition, Multiplication, and Complements

Translate verbal events into unions, intersections, and complements before applying the correct probability rule.

Statistics guide Ethical learning support SPSS/R/Python/Excel friendly
AP Statistics Topic Guide

Probability Rules: Addition, Multiplication, and Complements

Translate verbal events into unions, intersections, and complements before applying the correct probability rule.

StatusCurrent Unit 2 core
Main keywordprobability rules
Worked cases12
Practice36 MCQs + 14 FRQs
Study progress0 completed

Probability Rules: direct answer

Probability rules connect complements, unions, intersections, and conditional events. Translate the event first, then use the complement rule, general addition rule, or conditional multiplication rule that matches that event.

This page keeps the lesson centered on probability rules. Practice is included only after the method, assumptions, interpretation, and common decision points are explained.

Quick reference: Probability Rules: Addition, Multiplication, and Complements

Addition ruleP(A∪B)=P(A)+P(B)−P(A∩B)
ComplementP(Aᶜ)=1−P(A)

Concept mastery: Probability Rules: Addition, Multiplication, and Complements

Translate words into set operations first

“A or B” means the union A∪B, “A and B” means the intersection A∩B, and “not A” means the complement Aᶜ. This translation should happen before arithmetic. Many probability errors come from recognizing familiar numbers and applying a memorized operation without first identifying the event.

The complement rule is often the shortest route

P(Aᶜ)=1−P(A). It is especially effective for “at least one” events because calculating every possible successful case separately can be cumbersome. For independent trials, the probability of at least one success equals 1 minus the probability of no successes.

The general addition rule prevents double counting

P(A∪B)=P(A)+P(B)−P(A∩B). The intersection is subtracted because outcomes that belong to both events were counted twice. The simplified sum P(A)+P(B) is valid only when A and B are mutually exclusive, meaning their intersection has probability zero.

The multiplication rule is conditional unless independence is established

P(A∩B)=P(A)P(B|A). If A and B are independent, then P(B|A)=P(B), producing the familiar product P(A)P(B). Using the independent product without evidence of independence is one of the most common probability mistakes.

Probability values provide automatic error checks

Every probability must lie from 0 through 1. A union cannot be smaller than either component event. An intersection cannot exceed either component probability. A conditional probability uses the conditioning event as its reference set. These bounds catch many arithmetic and event-translation errors immediately.

Mutual exclusivity and independence describe opposite kinds of relationships

If two events with positive probability are mutually exclusive, occurrence of one makes the other impossible, so they cannot be independent. Independent events do not change each other’s probability and can occur together. The terms should never be treated as synonyms.

12 worked probability rules cases

Worked case 1: Complement

Scenario. A machine succeeds with probability 0.93 on one job.

Reasoning. The probability of failure is 1−0.93=0.07.

Case 1: Complement check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 1; do not replace it with a memorized generic sentence.

Worked case 2: At least one

Scenario. Three independent alerts each fail with probability 0.10.

Reasoning. P(at least one works)=1−P(all fail)=1−0.10^3=0.999.

Case 2: At least one check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 2; do not replace it with a memorized generic sentence.

Worked case 3: General addition

Scenario. P(A)=0.55, P(B)=0.40, and P(A∩B)=0.20.

Reasoning. P(A∪B)=0.55+0.40−0.20=0.75.

Case 3: General addition check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 3; do not replace it with a memorized generic sentence.

Worked case 4: Mutually exclusive

Scenario. Two prize categories cannot both occur and have probabilities 0.18 and 0.27.

Reasoning. Their union probability is 0.45 because the intersection is 0.

Case 4: Mutually exclusive check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 4; do not replace it with a memorized generic sentence.

Worked case 5: Conditional multiplication

Scenario. P(A)=0.60 and P(B|A)=0.25.

Reasoning. P(A∩B)=0.60×0.25=0.15.

Case 5: Conditional multiplication check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 5; do not replace it with a memorized generic sentence.

Worked case 6: Independent multiplication

Scenario. Independent events have probabilities 0.7 and 0.4.

Reasoning. P(A∩B)=0.7×0.4=0.28 because independence makes the conditional probability equal the marginal probability.

Case 6: Independent multiplication check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 6; do not replace it with a memorized generic sentence.

Worked case 7: Union bound check

Scenario. A calculation gives P(A∪B)=0.31 while P(A)=0.48.

Reasoning. The result is impossible because a union must be at least as probable as either event it contains.

Case 7: Union bound check check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 7; do not replace it with a memorized generic sentence.

Worked case 8: Intersection bound check

Scenario. A calculation gives P(A∩B)=0.62 while P(A)=0.51.

Reasoning. The result is impossible because an intersection cannot exceed the probability of A.

Case 8: Intersection bound check check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 8; do not replace it with a memorized generic sentence.

Worked case 9: Exclusive versus independent

Scenario. Two positive-probability events cannot occur together.

Reasoning. They are dependent: observing one forces the probability of the other to zero.

Case 9: Exclusive versus independent check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 9; do not replace it with a memorized generic sentence.

Worked case 10: Two-stage path

Scenario. A product is selected from plant 1 with probability 0.6 and is defective given plant 1 with probability 0.02.

Reasoning. The joint path probability is 0.6×0.02=0.012.

Case 10: Two-stage path check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 10; do not replace it with a memorized generic sentence.

Worked case 11: Disjoint paths

Scenario. A defect can come from plant 1 or plant 2, and the source events are disjoint.

Reasoning. Compute each source-and-defect path, then add the disjoint path probabilities.

Case 11: Disjoint paths check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 11; do not replace it with a memorized generic sentence.

Worked case 12: Event translation

Scenario. The phrase “neither A nor B” appears.

Reasoning. Translate it as (A∪B)ᶜ, so its probability is 1−P(A∪B).

Case 12: Event translation check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 12; do not replace it with a memorized generic sentence.

Deep-dive notes for Probability Rules: Addition, Multiplication, and Complements

Use Venn logic before arithmetic

Sketching or mentally partitioning the sample space can prevent formula errors. If events overlap, the common region belongs to both marginal probabilities and must be corrected in a union calculation. If the desired event is everything outside a region, a complement is often simpler. Event structure should dictate the arithmetic, not the other way around.

Addition and multiplication answer different questions

Addition combines alternative paths to an event, usually after ensuring the paths are disjoint or correcting overlap. Multiplication follows a single joint path through successive conditions. In a tree diagram, multiply down branches and add across different terminal paths. This simple rule organizes many complicated probability problems without memorizing separate formulas for each story.

Conditional probability is part of the multiplication rule

The general multiplication rule is P(A and B)=P(A)P(B|A). Independence is a special case that lets P(B|A) become P(B). Starting from the conditional form is safer because it forces you to ask whether the second event changes after the first has occurred rather than assuming independence from convenience.

Complements simplify repeated-trial language

Phrases such as “at least one,” “not all,” or “one or more” often describe a large collection of cases. The complement can collapse them into one event. For example, at least one success is the complement of zero successes. This is especially efficient in independent repeated trials, where the no-success probability is often a simple product.

Probability bounds diagnose impossible outputs

A probability below zero or above one is invalid, but stronger event-specific bounds are useful too. P(A∩B) cannot exceed either P(A) or P(B). P(A∪B) cannot be smaller than either event. If A is contained in B, P(A) cannot exceed P(B). These relationships catch setup mistakes before interpretation.

Mutually exclusive events usually become dependent

When two positive-probability events cannot happen together, observing one drives the probability of the other to zero. That is the opposite of independence. Students sometimes see “separate events” and assume independence, but mutual exclusivity is a strong form of dependence unless one of the events itself has probability zero.

Sampling without replacement creates dependence

Selecting one item without replacement changes the composition available for the next draw. The second probability therefore depends on the first outcome. The correct multiplication uses conditional probabilities. For very large populations relative to sample size, the change may be small, but exact finite-population calculations should still reflect the updated denominator.

Total probability can be built from disjoint routes

When an outcome can occur through several source groups, partition by source. Compute each source-and-outcome joint probability, then add those disjoint paths. This strategy is the foundation for screening-test, manufacturing-source, and classification problems and naturally prepares the denominator needed for reverse conditional probabilities.

Probability Rules: Addition, Multiplication, and Complements: 36 multiple-choice questions

These questions stay within this page’s topic. Work them after the concept and worked-case sections so practice reinforces the method rather than replacing instruction.

Question 1. Probability Rules

For two events A and B in a county election office in Cedar Grove during a yearly program evaluation, P(A)=0.467, P(B)=0.285, and P(A∩B)=0.235. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.752, and they are mutually exclusive.
  2. B. 0.483, and mutual exclusivity cannot be assessed.
  3. C. 0.517, and the events are not mutually exclusive.
  4. D. 0.235, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.467+0.285−0.235=0.517. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 2. Probability Rules

For two events A and B in a public high school in Riverbend during a monthly quality review, P(A)=0.409, P(B)=0.397, and P(A∩B)=0.374. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.568, and mutual exclusivity cannot be assessed.
  2. B. 0.806, and they are mutually exclusive.
  3. C. 0.374, because union and intersection are the same.
  4. D. 0.432, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.409+0.397−0.374=0.432. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 3. Probability Rules

For two events A and B in a university advising center in Central County during a baseline measurement week, P(A)=0.586, P(B)=0.46, and P(A∩B)=0.141. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.905, and the events are not mutually exclusive.
  2. B. 1.046, and they are mutually exclusive.
  3. C. 0.141, because union and intersection are the same.
  4. D. 0.095, and mutual exclusivity cannot be assessed.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.586+0.46−0.141=0.905. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 4. Probability Rules

For two events A and B in a food safety laboratory in North Valley during a community outreach cycle, P(A)=0.375, P(B)=0.364, and P(A∩B)=0.116. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.116, because union and intersection are the same.
  2. B. 0.623, and the events are not mutually exclusive.
  3. C. 0.377, and mutual exclusivity cannot be assessed.
  4. D. 0.739, and they are mutually exclusive.

Answer: B

P(A∪B)=P(A)+P(B)−P(A∩B)=0.375+0.364−0.116=0.623. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 5. Probability Rules

For two events A and B in a county library in Metro East during a spring 2027 pilot, P(A)=0.583, P(B)=0.543, and P(A∩B)=0.204. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.078, and mutual exclusivity cannot be assessed.
  2. B. 0.204, because union and intersection are the same.
  3. C. 1.126, and they are mutually exclusive.
  4. D. 0.922, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.583+0.543−0.204=0.922. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 6. Probability Rules

For two events A and B in a public health department in Metro East during a six-week field trial, P(A)=0.389, P(B)=0.365, and P(A∩B)=0.046. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.292, and mutual exclusivity cannot be assessed.
  2. B. 0.754, and they are mutually exclusive.
  3. C. 0.708, and the events are not mutually exclusive.
  4. D. 0.046, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.389+0.365−0.046=0.708. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 7. Probability Rules

For two events A and B in a grocery cooperative in South Harbor during a service-improvement study, P(A)=0.641, P(B)=0.426, and P(A∩B)=0.289. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 1.067, and they are mutually exclusive.
  2. B. 0.222, and mutual exclusivity cannot be assessed.
  3. C. 0.289, because union and intersection are the same.
  4. D. 0.778, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.641+0.426−0.289=0.778. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 8. Probability Rules

For two events A and B in a wildlife clinic in Sunbelt district during a quarterly performance study, P(A)=0.597, P(B)=0.252, and P(A∩B)=0.176. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.849, and they are mutually exclusive.
  2. B. 0.176, because union and intersection are the same.
  3. C. 0.673, and the events are not mutually exclusive.
  4. D. 0.327, and mutual exclusivity cannot be assessed.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.597+0.252−0.176=0.673. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 9. Probability Rules

For two events A and B in a community college in South Harbor during a winter readiness review, P(A)=0.387, P(B)=0.532, and P(A∩B)=0.288. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.369, and mutual exclusivity cannot be assessed.
  2. B. 0.288, because union and intersection are the same.
  3. C. 0.919, and they are mutually exclusive.
  4. D. 0.631, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.387+0.532−0.288=0.631. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 10. Probability Rules

For two events A and B in a community bank in South Harbor during a community outreach cycle, P(A)=0.591, P(B)=0.517, and P(A∩B)=0.136. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 1.108, and they are mutually exclusive.
  2. B. 0.028, and mutual exclusivity cannot be assessed.
  3. C. 0.136, because union and intersection are the same.
  4. D. 0.972, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.591+0.517−0.136=0.972. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 11. Probability Rules

For two events A and B in a county library in Cedar Grove during a pre-exam training cycle, P(A)=0.556, P(B)=0.406, and P(A∩B)=0.337. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.375, and mutual exclusivity cannot be assessed.
  2. B. 0.962, and they are mutually exclusive.
  3. C. 0.337, because union and intersection are the same.
  4. D. 0.625, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.556+0.406−0.337=0.625. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 12. Probability Rules

For two events A and B in a university advising center in Lakeside district during a regional benchmarking study, P(A)=0.417, P(B)=0.278, and P(A∩B)=0.256. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.256, because union and intersection are the same.
  2. B. 0.561, and mutual exclusivity cannot be assessed.
  3. C. 0.439, and the events are not mutually exclusive.
  4. D. 0.695, and they are mutually exclusive.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.417+0.278−0.256=0.439. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 13. Probability Rules

For two events A and B in a digital learning platform in Capital Region during a regional benchmarking study, P(A)=0.427, P(B)=0.317, and P(A∩B)=0.131. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.613, and the events are not mutually exclusive.
  2. B. 0.131, because union and intersection are the same.
  3. C. 0.387, and mutual exclusivity cannot be assessed.
  4. D. 0.744, and they are mutually exclusive.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.427+0.317−0.131=0.613. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 14. Probability Rules

For two events A and B in a regional manufacturer in Mountain Region during a semester-long cohort study, P(A)=0.344, P(B)=0.45, and P(A∩B)=0.115. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.794, and they are mutually exclusive.
  2. B. 0.321, and mutual exclusivity cannot be assessed.
  3. C. 0.679, and the events are not mutually exclusive.
  4. D. 0.115, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.344+0.45−0.115=0.679. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 15. Probability Rules

For two events A and B in a public health department in Atlantic Corridor during a two-month observation window, P(A)=0.531, P(B)=0.457, and P(A∩B)=0.214. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.226, and mutual exclusivity cannot be assessed.
  2. B. 0.774, and the events are not mutually exclusive.
  3. C. 0.988, and they are mutually exclusive.
  4. D. 0.214, because union and intersection are the same.

Answer: B

P(A∪B)=P(A)+P(B)−P(A∩B)=0.531+0.457−0.214=0.774. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 16. Probability Rules

For two events A and B in a university advising center in Mountain Region during a two-month observation window, P(A)=0.314, P(B)=0.417, and P(A∩B)=0.288. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.557, and mutual exclusivity cannot be assessed.
  2. B. 0.731, and they are mutually exclusive.
  3. C. 0.443, and the events are not mutually exclusive.
  4. D. 0.288, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.314+0.417−0.288=0.443. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 17. Probability Rules

For two events A and B in a wildlife clinic in New England network during a randomized pilot period, P(A)=0.637, P(B)=0.265, and P(A∩B)=0.138. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.236, and mutual exclusivity cannot be assessed.
  2. B. 0.902, and they are mutually exclusive.
  3. C. 0.764, and the events are not mutually exclusive.
  4. D. 0.138, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.637+0.265−0.138=0.764. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 18. Probability Rules

For two events A and B in a regional hospital in Cedar Grove during a weekday operations study, P(A)=0.456, P(B)=0.341, and P(A∩B)=0.309. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.512, and mutual exclusivity cannot be assessed.
  2. B. 0.488, and the events are not mutually exclusive.
  3. C. 0.797, and they are mutually exclusive.
  4. D. 0.309, because union and intersection are the same.

Answer: B

P(A∪B)=P(A)+P(B)−P(A∩B)=0.456+0.341−0.309=0.488. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 19. Probability Rules

For two events A and B in a municipal water office in Great Lakes during a two-month observation window, P(A)=0.547, P(B)=0.468, and P(A∩B)=0.158. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 1.015, and they are mutually exclusive.
  2. B. 0.857, and the events are not mutually exclusive.
  3. C. 0.158, because union and intersection are the same.
  4. D. 0.143, and mutual exclusivity cannot be assessed.

Answer: B

P(A∪B)=P(A)+P(B)−P(A∩B)=0.547+0.468−0.158=0.857. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 20. Probability Rules

For two events A and B in a public health department in Great Lakes during a two-month observation window, P(A)=0.324, P(B)=0.273, and P(A∩B)=0.132. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.132, because union and intersection are the same.
  2. B. 0.597, and they are mutually exclusive.
  3. C. 0.465, and the events are not mutually exclusive.
  4. D. 0.535, and mutual exclusivity cannot be assessed.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.324+0.273−0.132=0.465. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 21. Probability Rules

For two events A and B in a state park in Lakeside district during a quarterly performance study, P(A)=0.397, P(B)=0.494, and P(A∩B)=0.068. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.823, and the events are not mutually exclusive.
  2. B. 0.891, and they are mutually exclusive.
  3. C. 0.177, and mutual exclusivity cannot be assessed.
  4. D. 0.068, because union and intersection are the same.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.397+0.494−0.068=0.823. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 22. Probability Rules

For two events A and B in a recycling program in Cedar Grove during a multiweek validation study, P(A)=0.39, P(B)=0.429, and P(A∩B)=0.061. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.242, and mutual exclusivity cannot be assessed.
  2. B. 0.819, and they are mutually exclusive.
  3. C. 0.061, because union and intersection are the same.
  4. D. 0.758, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.39+0.429−0.061=0.758. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 23. Probability Rules

For two events A and B in a community bank in New England network during a multiweek validation study, P(A)=0.37, P(B)=0.464, and P(A∩B)=0.185. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.834, and they are mutually exclusive.
  2. B. 0.185, because union and intersection are the same.
  3. C. 0.649, and the events are not mutually exclusive.
  4. D. 0.351, and mutual exclusivity cannot be assessed.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.37+0.464−0.185=0.649. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 24. Probability Rules

For two events A and B in a regional manufacturer in Sunbelt district during a school-year data collection, P(A)=0.337, P(B)=0.451, and P(A∩B)=0.145. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.643, and the events are not mutually exclusive.
  2. B. 0.145, because union and intersection are the same.
  3. C. 0.357, and mutual exclusivity cannot be assessed.
  4. D. 0.788, and they are mutually exclusive.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.337+0.451−0.145=0.643. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 25. Probability Rules

For two events A and B in a housing authority in Great Lakes during a weekday operations study, P(A)=0.625, P(B)=0.54, and P(A∩B)=0.514. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.514, because union and intersection are the same.
  2. B. 0.651, and the events are not mutually exclusive.
  3. C. 1.165, and they are mutually exclusive.
  4. D. 0.349, and mutual exclusivity cannot be assessed.

Answer: B

P(A∪B)=P(A)+P(B)−P(A∩B)=0.625+0.54−0.514=0.651. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 26. Probability Rules

For two events A and B in a municipal emergency dispatch center in Westview during a school-year data collection, P(A)=0.265, P(B)=0.494, and P(A∩B)=0.05. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.759, and they are mutually exclusive.
  2. B. 0.05, because union and intersection are the same.
  3. C. 0.709, and the events are not mutually exclusive.
  4. D. 0.291, and mutual exclusivity cannot be assessed.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.265+0.494−0.05=0.709. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 27. Probability Rules

For two events A and B in a digital learning platform in Desert County during a baseline measurement week, P(A)=0.251, P(B)=0.355, and P(A∩B)=0.097. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.491, and mutual exclusivity cannot be assessed.
  2. B. 0.097, because union and intersection are the same.
  3. C. 0.509, and the events are not mutually exclusive.
  4. D. 0.606, and they are mutually exclusive.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.251+0.355−0.097=0.509. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 28. Probability Rules

For two events A and B in a housing authority in Mountain Region during a fall 2026 audit, P(A)=0.553, P(B)=0.545, and P(A∩B)=0.281. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 1.098, and they are mutually exclusive.
  2. B. 0.281, because union and intersection are the same.
  3. C. 0.817, and the events are not mutually exclusive.
  4. D. 0.183, and mutual exclusivity cannot be assessed.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.553+0.545−0.281=0.817. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 29. Probability Rules

For two events A and B in a food safety laboratory in Pine Ridge during a semester-long cohort study, P(A)=0.382, P(B)=0.238, and P(A∩B)=0.174. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.174, because union and intersection are the same.
  2. B. 0.62, and they are mutually exclusive.
  3. C. 0.554, and mutual exclusivity cannot be assessed.
  4. D. 0.446, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.382+0.238−0.174=0.446. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 30. Probability Rules

For two events A and B in a grocery cooperative in Riverbend during a summer implementation review, P(A)=0.515, P(B)=0.438, and P(A∩B)=0.248. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.705, and the events are not mutually exclusive.
  2. B. 0.248, because union and intersection are the same.
  3. C. 0.953, and they are mutually exclusive.
  4. D. 0.295, and mutual exclusivity cannot be assessed.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.515+0.438−0.248=0.705. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 31. Probability Rules

For two events A and B in a county library in New England network during a monthly quality review, P(A)=0.277, P(B)=0.258, and P(A∩B)=0.044. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.491, and the events are not mutually exclusive.
  2. B. 0.044, because union and intersection are the same.
  3. C. 0.535, and they are mutually exclusive.
  4. D. 0.509, and mutual exclusivity cannot be assessed.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.277+0.258−0.044=0.491. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 32. Probability Rules

For two events A and B in a public high school in Atlantic Corridor during a monthly quality review, P(A)=0.496, P(B)=0.458, and P(A∩B)=0.097. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.857, and the events are not mutually exclusive.
  2. B. 0.143, and mutual exclusivity cannot be assessed.
  3. C. 0.954, and they are mutually exclusive.
  4. D. 0.097, because union and intersection are the same.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.496+0.458−0.097=0.857. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 33. Probability Rules

For two events A and B in a community college in South Harbor during a randomized pilot period, P(A)=0.277, P(B)=0.324, and P(A∩B)=0.232. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.601, and they are mutually exclusive.
  2. B. 0.631, and mutual exclusivity cannot be assessed.
  3. C. 0.369, and the events are not mutually exclusive.
  4. D. 0.232, because union and intersection are the same.

Answer: C

P(A∪B)=P(A)+P(B)−P(A∩B)=0.277+0.324−0.232=0.369. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 34. Probability Rules

For two events A and B in a city recreation department in Metro East during a weekday operations study, P(A)=0.502, P(B)=0.224, and P(A∩B)=0.169. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.557, and the events are not mutually exclusive.
  2. B. 0.443, and mutual exclusivity cannot be assessed.
  3. C. 0.726, and they are mutually exclusive.
  4. D. 0.169, because union and intersection are the same.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.502+0.224−0.169=0.557. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 35. Probability Rules

For two events A and B in a municipal emergency dispatch center in Prairie District during a winter readiness review, P(A)=0.417, P(B)=0.286, and P(A∩B)=0.071. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.632, and the events are not mutually exclusive.
  2. B. 0.368, and mutual exclusivity cannot be assessed.
  3. C. 0.071, because union and intersection are the same.
  4. D. 0.703, and they are mutually exclusive.

Answer: A

P(A∪B)=P(A)+P(B)−P(A∩B)=0.417+0.286−0.071=0.632. Because P(A∩B) is not zero, the events are not mutually exclusive.

Question 36. Probability Rules

For two events A and B in a county election office in Cedar Grove during a spring 2027 pilot, P(A)=0.281, P(B)=0.471, and P(A∩B)=0.044. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. A. 0.292, and mutual exclusivity cannot be assessed.
  2. B. 0.044, because union and intersection are the same.
  3. C. 0.752, and they are mutually exclusive.
  4. D. 0.708, and the events are not mutually exclusive.

Answer: D

P(A∪B)=P(A)+P(B)−P(A∩B)=0.281+0.471−0.044=0.708. Because P(A∩B) is not zero, the events are not mutually exclusive.

Probability Rules: Addition, Multiplication, and Complements: 14 free-response questions

For each response, show the statistical reasoning, use the scenario’s language, and state only the conclusion supported by the design or probability model.

FRQ set 1: Probability Rules

Scenario. For two events A and B in a municipal water office in Lakeside district during a monthly quality review, P(A)=0.54, P(B)=0.466, and P(A∩B)=0.199. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.54+0.466−0.199=0.807. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 2: Probability Rules

Scenario. For two events A and B in a solar installer in Cedar Grove during a school-year data collection, P(A)=0.418, P(B)=0.503, and P(A∩B)=0.069. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.418+0.503−0.069=0.852. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 3: Probability Rules

Scenario. For two events A and B in a wildlife clinic in New England network during a winter readiness review, P(A)=0.339, P(B)=0.262, and P(A∩B)=0.197. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.339+0.262−0.197=0.404. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 4: Probability Rules

Scenario. For two events A and B in a county election office in Westview during a school-year data collection, P(A)=0.46, P(B)=0.4, and P(A∩B)=0.297. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.46+0.4−0.297=0.563. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 5: Probability Rules

Scenario. For two events A and B in a solar installer in Coastal Plains during a weekday operations study, P(A)=0.544, P(B)=0.221, and P(A∩B)=0.125. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.544+0.221−0.125=0.64. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 6: Probability Rules

Scenario. For two events A and B in a public health department in Sunbelt district during a follow-up evaluation period, P(A)=0.312, P(B)=0.268, and P(A∩B)=0.106. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.312+0.268−0.106=0.474. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 7: Probability Rules

Scenario. For two events A and B in a regional hospital in New England network during a baseline measurement week, P(A)=0.475, P(B)=0.333, and P(A∩B)=0.208. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.475+0.333−0.208=0.6. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 8: Probability Rules

Scenario. For two events A and B in a grocery cooperative in Prairie District during a service-improvement study, P(A)=0.445, P(B)=0.22, and P(A∩B)=0.167. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.445+0.22−0.167=0.498. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 9: Probability Rules

Scenario. For two events A and B in a city recreation department in Lakeside district during a regional benchmarking study, P(A)=0.623, P(B)=0.342, and P(A∩B)=0.078. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.623+0.342−0.078=0.887. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 10: Probability Rules

Scenario. For two events A and B in a farm cooperative in Riverbend during a semester-long cohort study, P(A)=0.595, P(B)=0.408, and P(A∩B)=0.137. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.595+0.408−0.137=0.866. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 11: Probability Rules

Scenario. For two events A and B in a grocery cooperative in Central County during a yearly program evaluation, P(A)=0.414, P(B)=0.344, and P(A∩B)=0.053. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.414+0.344−0.053=0.705. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 12: Probability Rules

Scenario. For two events A and B in a public high school in Mountain Region during a multiweek validation study, P(A)=0.558, P(B)=0.498, and P(A∩B)=0.108. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.558+0.498−0.108=0.948. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 13: Probability Rules

Scenario. For two events A and B in a community college in Mountain Region during a fall 2026 audit, P(A)=0.313, P(B)=0.489, and P(A∩B)=0.084. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.313+0.489−0.084=0.718. Because P(A∩B) is not zero, the events are not mutually exclusive.

FRQ set 14: Probability Rules

Scenario. For two events A and B in a city transit agency in Midwest consortium during a regional benchmarking study, P(A)=0.502, P(B)=0.484, and P(A∩B)=0.023. Find P(A∪B) and determine whether the events are mutually exclusive.

  1. Define the event or random variable and state the requested probability or long-run quantity.
  2. Verify the relevant model conditions or probability-distribution requirements.
  3. Show the probability, expected-value, or distribution calculation with labeled terms.
  4. Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.

Model response

P(A∪B)=P(A)+P(B)−P(A∩B)=0.502+0.484−0.023=0.963. Because P(A∩B) is not zero, the events are not mutually exclusive.

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Engr. Muhammad Yar Saqib author profile photo

Engr. Muhammad Yar Saqib

Engr. Muhammad Yar Saqib is an electrical engineer educated at the University of Bradford, United Kingdom, a writer and poet, and an Assistant Education Officer in the School Education Department, Punjab, serving since July 2017. He writes practical guides on statistics, SPSS, data analysis, mathematics and educational technology, with an emphasis on transparent methods, reproducible calculations and ethical learning support.