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Quiz Chapter 9: Probability of Combined Events

10 questions · Form 4 Mathematics Bab 9: Probability of Combined Events

Question 1 of 10Score: 0

A single fair six-sided die is rolled. What is the probability of getting an even number OR a number greater than 4?

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. A single fair six-sided die is rolled. What is the probability of getting an even number OR a number greater than 4?

  1. A. 12
  2. B. 23
  3. C. 56
  4. D. 13
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Answer: B

Even = {2, 4, 6} → P(Even) = 36. Greater than 4 = {5, 6} → P(>4) = 26. Both = {6} → P(Both) = 16. P(Union) = 36 + 26 - 16 = 46 = 23.

2. Which formula calculates the probability of union of two non-mutually exclusive events A and B?

  1. A. P(A ∪ B) = P(A) + P(B)
  2. B. P(A ∪ B) = P(A) × P(B)
  3. C. P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
  4. D. P(A ∪ B) = 1 - P(A ∩ B)
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Answer: C

For non-mutually exclusive events, overlapping outcomes counted twice must be subtracted: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

3. If P(A ∪ B) = 0.75, what is the probability of the complement event P((A ∪ B)')?

  1. A. 0.25
  2. B. 0.50
  3. C. 0.75
  4. D. 1.00
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Answer: A

P((A ∪ B)') = 1 - P(A ∪ B) = 1 - 0.75 = 0.25.

4. Events X and Y are mutually exclusive. Given P(X) = 13 and P(Y) = 14, find P(X ∪ Y).

  1. A. 112
  2. B. 712
  3. C. 17
  4. D. 512
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Answer: B

Since X and Y are mutually exclusive: P(X ∪ Y) = P(X) + P(Y) = 13 + 14 = 412 + 312 = 712.

5. In a class of 30 students, 18 study Physics, 12 study Chemistry, and 5 study both. A student is picked at random. What is the probability that the student studies Physics OR Chemistry?

  1. A. 2530
  2. B. 3030
  3. C. 1830
  4. D. 2030
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Answer: A

P(P ∪ C) = P(P) + P(C) - P(P ∩ C) = 1830 + 1230 - 530 = 2530 = 56.

6. A bag contains 5 red marbles and 3 green marbles. A marble is drawn, its color recorded, and REPLACED. A second marble is drawn. What is the probability of drawing two red marbles?

  1. A. 2564
  2. B. 514
  3. C. 1556
  4. D. 964
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Answer: A

P(Red) = 58. Since drawing is with replacement, events are independent. P(Red and Red) = (58) × (58) = 2564.

7. A target is shot by Ali and Ahmad independently. Probability of Ali hitting is 0.6 and Ahmad hitting is 0.7. Find the probability that BOTH hit the target.

  1. A. 0.42
  2. B. 0.88
  3. C. 0.13
  4. D. 0.26
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Answer: A

Independent events: P(Both hit) = P(Ali) × P(Ahmad) = 0.6 × 0.7 = 0.42.

8. Using the target scenario above (P(Ali)=0.6, P(Ahmad)=0.7), find the probability that AT LEAST ONE hits the target.

  1. A. 0.42
  2. B. 0.88
  3. C. 0.12
  4. D. 0.98
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Answer: B

P(At least one) = 1 - P(Neither hits) = 1 - (1 - 0.6)(1 - 0.7) = 1 - (0.4 × 0.3) = 1 - 0.12 = 0.88.

9. When selecting two items from a container WITHOUT replacement, the sample space size for the second selection is:

  1. A. Unchanged relative to the first selection
  2. B. Decreased by 1 relative to the first selection
  3. C. Increased by 1 relative to the first selection
  4. D. Doubled
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Answer: B

Without replacement means 1 item is removed, so the total number of items available for the second draw decreases by 1.

10. A box contains cards labeled with numbers 1 to 10. A card is drawn at random. Let A = prime numbers, B = multiples of 3. Find P(A ∪ B).

  1. A. 710
  2. B. 310
  3. C. 410
  4. D. 610
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Answer: A

A = {2, 3, 5, 7} → 4 items. B = {3, 6, 9} → 3 items. A ∩ B = {3} → 1 item. A ∪ B = {2, 3, 5, 6, 7, 9} → 6 items? Wait: Prime numbers in 1-10 are 2, 3, 5, 7 (4 items). Multiples of 3 are 3, 6, 9 (3 items). Union = {2, 3, 5, 6, 7, 9} which is 6 items. P(A ∪ B) = 610 or 35. Option D is 610.

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