Math 2 Quiz: Addition Rules For Probability
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Addition Rules For ProbabilityQuestion 1 of 16

A survey of 200 smartphone users found that 120 use social media apps, 80 use gaming apps, and 140 use either social media or gaming apps (or both). If a user is selected at random from this survey, what is the probability that they use both social media and gaming apps?

310\frac{3}{10}
710\frac{7}{10}
15\frac{1}{5}
25\frac{2}{5}
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Math 2 Quiz

Math 2 Quiz: Addition Rules For Probability

Practice Addition Rules For Probability in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Addition Rules For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A survey of 200 smartphone users found that 120 use social media apps, 80 use gaming apps, and 140 use either social media or gaming apps (or both). If a user is selected at random from this survey, what is the probability that they use both social media and gaming apps?

  1. 310\frac{3}{10} (correct answer)
  2. 710\frac{7}{10}
  3. 15\frac{1}{5}
  4. 25\frac{2}{5}
Explanation: Using the addition rule: P(S or G) = P(S) + P(G) - P(S and G). We know P(S) = 120/200 = 3/5, P(G) = 80/200 = 2/5, and P(S or G) = 140/200 = 7/10. Solving: 7/10 = 3/5 + 2/5 - P(S and G), so P(S and G) = 3/5 + 2/5 - 7/10 = 6/10 + 4/10 - 7/10 = 3/10. Choice B is P(S or G). Choice C incorrectly uses 40/200. Choice D is P(G).

Question 2

A medical test is given to 1000 patients. The results show that 300 patients test positive for condition X, 250 patients test positive for condition Y, and 400 patients test positive for at least one of the two conditions. What is the probability that a randomly selected patient from this group tests positive for condition Y but negative for condition X?

  1. 0.100.10
  2. 0.150.15 (correct answer)
  3. 0.250.25
  4. 0.400.40
Explanation: Using the addition rule: P(X or Y) = P(X) + P(Y) - P(X and Y). We have 0.4 = 0.3 + 0.25 - P(X and Y), so P(X and Y) = 0.15. The probability of testing positive for Y but negative for X is P(Y) - P(X and Y) = 0.25 - 0.15 = 0.15. Choice A represents P(X and Y) - 0.05. Choice C is P(Y). Choice D is P(X or Y).

Question 3

In a group of 150 students, 90 study French, 75 study Spanish, and 45 study both languages. If we define event D as "a student studies French or Spanish but not both languages," what is the probability of event D?

  1. 45\frac{4}{5}
  2. 35\frac{3}{5}
  3. 25\frac{2}{5}
  4. 12\frac{1}{2} (correct answer)
Explanation: Students who study French only: 90 - 45 = 45. Students who study Spanish only: 75 - 45 = 30. Event D includes students who study exactly one language: P(D) = (45 + 30)/150 = 75/150 = 1/2. Choice A represents P(French or Spanish) = 120/150. Choice B is P(French or Spanish but not both) calculated incorrectly as 90/150. Choice C represents 60/150.

Question 4

At a college, 60%60\% of students are enrolled in mathematics courses, 45%45\% are enrolled in science courses, and 30%30\% are enrolled in both mathematics and science courses. If a student is selected at random, what is the probability that the student is enrolled in mathematics or science, but not both?

  1. 0.450.45 (correct answer)
  2. 0.500.50
  3. 0.750.75
  4. 1.051.05
Explanation: Using the addition rule: P(M or S) = P(M) + P(S) - P(M and S) = 0.60 + 0.45 - 0.30 = 0.75. However, the question asks for students enrolled in mathematics OR science but NOT both. This means P(M only) + P(S only) = (0.60 - 0.30) + (0.45 - 0.30) = 0.30 + 0.15 = 0.45. Choice B represents half of P(M or S). Choice C is P(M or S) including both. Choice D incorrectly adds all probabilities without subtracting the overlap.

Question 5

A bag contains colored marbles where P(red or blue)=34P(\text{red or blue}) = \frac{3}{4}, P(red)=25P(\text{red}) = \frac{2}{5}, and P(blue)=12P(\text{blue}) = \frac{1}{2}. If these are the only marble colors that satisfy this relationship, what is P(red and blue)P(\text{red and blue})?

  1. 320\frac{3}{20} (correct answer)
  2. 720\frac{7}{20}
  3. 920\frac{9}{20}
  4. 1120\frac{11}{20}
Explanation: Using the addition rule: P(red or blue) = P(red) + P(blue) - P(red and blue). Substituting: 3/4 = 2/5 + 1/2 - P(red and blue). Converting to common denominators: 15/20 = 8/20 + 10/20 - P(red and blue), so 15/20 = 18/20 - P(red and blue). Therefore, P(red and blue) = 18/20 - 15/20 = 3/20. Choice B is P(red) + P(red and blue). Choice C is P(red) + P(blue). Choice D is P(red) + P(blue) + P(red and blue).

Question 6

A student takes two independent probability quizzes. The probability of passing the first quiz is 23\frac{2}{3}, and the probability of passing the second quiz is 34\frac{3}{4}. What is the probability that the student passes at least one of the two quizzes?

  1. 56\frac{5}{6}
  2. 1712\frac{17}{12}
  3. 1112\frac{11}{12} (correct answer)
  4. 712\frac{7}{12}
Explanation: When you encounter probability questions asking about "at least one" event occurring, you have two main approaches: calculate the probability directly by adding all favorable outcomes, or use the complement rule. The complement approach is usually easier. The complement of "passing at least one quiz" is "failing both quizzes." Since the quizzes are independent, you can multiply their individual failure probabilities. The probability of failing the first quiz is 1−23=131 - \frac{2}{3} = \frac{1}{3}. The probability of failing the second quiz is 1−34=141 - \frac{3}{4} = \frac{1}{4}. Therefore, the probability of failing both quizzes is 13×14=112\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}. The probability of passing at least one quiz is 1−112=11121 - \frac{1}{12} = \frac{11}{12}, which is answer C. Let's examine why the other answers are wrong. Answer A (56\frac{5}{6}) might result from incorrectly adding the passing probabilities without accounting for overlap: 23+34−23×34\frac{2}{3} + \frac{3}{4} - \frac{2}{3} \times \frac{3}{4} calculated incorrectly. Answer B (1712\frac{17}{12}) comes from simply adding 23+34=812+912=1712\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}, which exceeds 1 and is impossible for a probability. Answer D (712\frac{7}{12}) might result from subtracting probabilities incorrectly. Remember: for "at least one" probability questions, try the complement rule first. Calculate the probability that none of the events occur, then subtract from 1. This approach often involves simpler calculations and fewer errors.

Question 7

In a survey of 800 people about their exercise habits, it was found that 320 people jog, 480 people swim, and 600 people do at least one of these activities. What is the probability that a person selected at random from this survey both jogs and swims?

  1. 18\frac{1}{8}
  2. 12\frac{1}{2}
  3. 38\frac{3}{8}
  4. 14\frac{1}{4} (correct answer)
Explanation: When you encounter problems about overlapping groups, you're dealing with set theory and the principle of inclusion-exclusion. The key insight is that when you add the number of people in each activity, you're double-counting those who do both activities. Let's use the inclusion-exclusion principle: |J ∪ S| = |J| + |S| - |J ∩ S|, where J represents joggers, S represents swimmers, and the intersection represents people who do both. We know 600 people do at least one activity, 320 jog, and 480 swim. Substituting: 600 = 320 + 480 - |J ∩ S| Solving: 600 = 800 - |J ∩ S| Therefore: |J ∩ S| = 200 The probability that a randomly selected person both jogs and swims is 200800=14\frac{200}{800} = \frac{1}{4}. Looking at the wrong answers: Choice A (18\frac{1}{8}) might result from incorrectly calculating 2001600\frac{200}{1600} by using the sum of joggers and swimmers as the denominator. Choice B (12\frac{1}{2}) could come from using 400800\frac{400}{800}, perhaps by miscalculating the overlap as 400. Choice C (38\frac{3}{8}) equals 300800\frac{300}{800}, suggesting an error in finding the intersection, possibly confusing it with those who do only one activity. Remember: when dealing with overlapping sets, always account for double-counting. Draw a Venn diagram if it helps visualize the relationships, and use the inclusion-exclusion principle to find intersections systematically.

Question 8

A quality control inspector examines products from two assembly lines. Line A produces 60% of the total output, while Line B produces 40%. Historical data shows that 8% of products from Line A are defective, and 12% of products from Line B are defective.

Based on the information above, what is the probability that a randomly selected product is either from Line A or defective (or both)?

  1. 0.6280.628
  2. 0.6680.668
  3. 0.6480.648 (correct answer)
  4. 0.6880.688
Explanation: When you encounter probability questions involving "or" conditions, you're dealing with the addition rule for probability. The key insight is recognizing that "either from Line A or defective (or both)" means you need to find P(Line A ∪ Defective). Using the addition rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Here, you need P(Line A) + P(Defective) - P(Line A and Defective). First, find each probability:
  • P(Line A) = 0.60
  • P(Defective) = P(Defective from A) + P(Defective from B) = (0.60 × 0.08) + (0.40 × 0.12) = 0.048 + 0.048 = 0.096
  • P(Line A and Defective) = 0.60 × 0.08 = 0.048
Therefore: P(Line A or Defective) = 0.60 + 0.096 - 0.048 = 0.648 Answer C (0.648) is correct. Answer A (0.628) likely results from incorrectly calculating the total defective rate or making an arithmetic error. Answer B (0.668) probably comes from adding P(Line A) and P(Defective) without subtracting the overlap, violating the addition rule. Answer D (0.688) suggests a more significant computational error, possibly in calculating the defective probabilities. Remember: whenever you see "or" in probability, check if events overlap. If they do, you must subtract the intersection to avoid double-counting. Always use the complete addition rule formula, not just simple addition.

Question 9

A medical study tracks three risk factors for a disease: smoking (S), obesity (O), and high blood pressure (H). In a population of 1000 people, 320 have factor S, 280 have factor O, 250 have factor H, 120 have both S and O, 90 have both S and H, 110 have both O and H, and 40 have all three factors. If a person is selected randomly from this population, what is the probability that they have at least one of these risk factors?

  1. 0.85
  2. 0.70 (correct answer)
  3. 0.75
  4. 0.80
Explanation: Using the inclusion-exclusion principle: P(S ∪ O ∪ H) = P(S) + P(O) + P(H) - P(S ∩ O) - P(S ∩ H) - P(O ∩ H) + P(S ∩ O ∩ H). Converting to probabilities: 320/1000 + 280/1000 + 250/1000 - 120/1000 - 90/1000 - 110/1000 + 40/1000 = (320 + 280 + 250 - 120 - 90 - 110 + 40)/1000 = 570/1000 = 0.70. Choice A incorrectly adds all individual probabilities without proper subtraction. Choice C forgets to add back the triple intersection. Choice D makes an arithmetic error in the inclusion-exclusion calculation.

Question 10

Two events A and B satisfy P(A) = 0.6, P(B) = 0.7, and P(A∩B) = 0.5. Event C is defined such that P(A∩C) = 0.2, P(B∩C) = 0.3, and P(A∩B∩C) = 0.1. What is P(A∪B∪C)?

  1. 0.9 - P(C) + P(B∩C) + P(A∩C)
  2. 0.8 + P(C) - P(A∩C) - P(B∩C)
  3. 0.8 + P(C) - 0.4 (correct answer)
  4. 0.9 + P(C) - P(A∩C) - P(B∩C)
Explanation: Using inclusion-exclusion: P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C). Substituting known values: P(A∪B∪C) = 0.6 + 0.7 + P(C) - 0.5 - 0.2 - 0.3 + 0.1 = 0.8 + P(C) - 0.4. Choice A incorrectly starts with 0.9 instead of 0.8 and adds instead of subtracts some terms. Choice B has the wrong sign on P(C). Choice D starts with 0.9 and doesn't simplify the numerical terms correctly.

Question 11

Events X and Y are such that P(X∪Y) = 0.75, P(X) = 0.45, and P(YcY^c) = 0.4. Event Z is independent of both X and Y, with P(Z) = 0.3. What is P(X∪Y∪Z)?

  1. 0.775
  2. 0.90
  3. 0.875
  4. 0.825 (correct answer)
Explanation: When you encounter probability questions involving unions and independent events, you need to systematically work through the given information and apply the appropriate formulas. First, let's find P(Y)P(Y). Since P(Yc)=0.4P(Y^c) = 0.4, we know P(Y)=1−0.4=0.6P(Y) = 1 - 0.4 = 0.6. Next, we'll find P(X∩Y)P(X \cap Y) using the union formula: P(X∪Y)=P(X)+P(Y)−P(X∩Y)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y). Substituting: 0.75=0.45+0.6−P(X∩Y)0.75 = 0.45 + 0.6 - P(X \cap Y), so P(X∩Y)=0.3P(X \cap Y) = 0.3. Now for P(X∪Y∪Z)P(X \cup Y \cup Z). Since Z is independent of both X and Y, it's also independent of their union. We can use: P(X∪Y∪Z)=P(X∪Y)+P(Z)−P((X∪Y)∩Z)P(X \cup Y \cup Z) = P(X \cup Y) + P(Z) - P((X \cup Y) \cap Z). Since Z is independent of (X∪Y)(X \cup Y): P((X∪Y)∩Z)=P(X∪Y)×P(Z)=0.75×0.3=0.225P((X \cup Y) \cap Z) = P(X \cup Y) \times P(Z) = 0.75 \times 0.3 = 0.225. Therefore: P(X∪Y∪Z)=0.75+0.3−0.225=0.825P(X \cup Y \cup Z) = 0.75 + 0.3 - 0.225 = 0.825. Answer choice A (0.775) likely comes from forgetting to add P(Z)P(Z) properly. Choice B (0.90) probably results from simply adding P(X∪Y)+P(Z)P(X \cup Y) + P(Z) without subtracting the intersection. Choice C (0.875) might stem from incorrectly calculating the intersection term. Strategy tip: Always identify what information you have versus what you need, work step-by-step through probability formulas, and remember that independence means P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

Question 12

A software company tracks three types of user errors: input errors (I), logic errors (L), and syntax errors (S). Analysis of user sessions shows P(I) = 0.3, P(L) = 0.4, P(S) = 0.2, P(I∩L) = 0.15, P(I∩S) = 0.08, P(L∩S) = 0.12, and P(I∩L∩S) = 0.05. The company wants to send targeted help messages. What is the probability that a randomly selected user session has exactly two types of errors?

  1. 0.20 (correct answer)
  2. 0.18
  3. 0.22
  4. 0.15
Explanation: Exactly two types means: (I∩L∩ScI∩L∩S^c) ∪ (I∩LcI∩L^c∩S) ∪ (IcI^c∩L∩S). These are mutually exclusive, so we add: P(I∩L∩ScI∩L∩S^c) = P(I∩L) - P(I∩L∩S) = 0.15 - 0.05 = 0.10. P(I∩LcI∩L^c∩S) = P(I∩S) - P(I∩L∩S) = 0.08 - 0.05 = 0.03. P(IcI^c∩L∩S) = P(L∩S) - P(I∩L∩S) = 0.12 - 0.05 = 0.07. Total = 0.10 + 0.03 + 0.07 = 0.20. Choice B miscalculates one of the pairwise differences. Choice C adds the original pairwise intersections without subtracting the triple intersection. Choice D represents some other calculation error.

Question 13

In a clinical trial, patients receive treatment A, treatment B, both treatments, or neither. The probability that a patient receives treatment A is 0.6, the probability of receiving treatment B is 0.5, and the probability of receiving at least one treatment is 0.8. A patient is selected randomly from the trial. Given that this patient receives at least one treatment, what is the probability that they receive exactly one treatment?

  1. 58\frac{5}{8} (correct answer)
  2. 38\frac{3}{8}
  3. 12\frac{1}{2}
  4. 34\frac{3}{4}
Explanation: First find P(A∩B): P(A∪B) = P(A) + P(B) - P(A∩B), so 0.8 = 0.6 + 0.5 - P(A∩B), giving P(A∩B) = 0.3. The probability of exactly one treatment is P(A∩BcA∩B^c) + P(AcA^c∩B) = [P(A) - P(A∩B)] + [P(B) - P(A∩B)] = (0.6 - 0.3) + (0.5 - 0.3) = 0.3 + 0.2 = 0.5. Given at least one treatment, P(exactly one | at least one) = P(exactly one)/P(at least one) = 0.5/0.8 = 5/8. Choice B gives the probability of both treatments given at least one. Choice C is the unconditional probability of exactly one treatment. Choice D incorrectly calculates the conditional probability.

Question 14

Events A and B are such that P(A)=0.4P(A) = 0.4, P(B)=0.7P(B) = 0.7, and P(A or B)=0.9P(A \text{ or } B) = 0.9. If event C is defined as "A occurs but B does not occur," what is P(C)P(C)?

  1. 0.10.1
  2. 0.20.2 (correct answer)
  3. 0.30.3
  4. 0.40.4
Explanation: First find P(A and B) using the addition rule: P(A or B) = P(A) + P(B) - P(A and B), so 0.9 = 0.4 + 0.7 - P(A and B), giving P(A and B) = 0.2. Event C is "A but not B," which means P(C) = P(A) - P(A and B) = 0.4 - 0.2 = 0.2. Choice A is P(A and B) - P(A). Choice C is P(A) - P(A or B) + P(B). Choice D is simply P(A).

Question 15

Events M and N are mutually exclusive with P(M)=0.3P(M) = 0.3 and P(N)=0.4P(N) = 0.4. Event R is defined as the complement of (M or N)(M \text{ or } N). What is P(R)P(R)?

  1. 0.10.1
  2. 0.30.3 (correct answer)
  3. 0.40.4
  4. 0.70.7
Explanation: Since M and N are mutually exclusive, P(M and N) = 0. Therefore, P(M or N) = P(M) + P(N) - P(M and N) = 0.3 + 0.4 - 0 = 0.7. Event R is the complement of (M or N), so P(R) = 1 - P(M or N) = 1 - 0.7 = 0.3. Choice A would result from 1 - 0.9. Choice C equals P(N). Choice D equals P(M or N).

Question 16

Two dice are rolled simultaneously. What is the probability that the sum is either 7 or 11, or that at least one die shows a 6?

  1. 49\frac{4}{9}
  2. 512\frac{5}{12}
  3. 1736\frac{17}{36} (correct answer)
  4. 12\frac{1}{2}
Explanation: P(sum is 7) = 6/36, P(sum is 11) = 2/36, P(at least one 6) = 11/36. However, we must account for overlaps: (6,1), (1,6), (6,5), (5,6) give sums of 7 and include a 6. So P(sum is 7 or 11 or at least one 6) = 6/36 + 2/36 + 11/36 - 4/36 = 15/36. Wait, let me recalculate: outcomes with at least one 6: 11 total. Outcomes with sum 7: 6 total. Outcomes with sum 11: 2 total. Overlaps: (1,6) and (6,1) for sum 7. Total: 6 + 2 + 11 - 2 = 17. So 17/36. Choice A is 16/36. Choice B is 15/36. Choice D is 18/36.