Partition of a sample space. 20 mcqs on it with explained answers
Here are 20 MCQs on Partition of a Sample Space with detailed explained answers:
1. A sample space $$S$$ is partitioned into events $$A$$ and $$B$$. If $$P(A) = 0.4$$, what is $$P(B)$$?
- a) 0.4
- b) 0.6
- c) 0.5
- d) 0.3
Answer: Since $$A$$ and $$B$$ partition $$S$$, $$P(B) = 1 - P(A) = 0.6$$.
2. If $$A_1, A_2, A_3$$ form a partition of sample space with $$P(A_1) = 0.2$$, $$P(A_2) = 0.5$$, find $$P(A_3)$$?
- a) 0.5
- b) 0.3
- c) 0.7
- d) 0.8
Answer: Sum equals 1, so $$P(A_3) = 1 - 0.2 - 0.5 = 0.3$$.
3. Events $$A_1, A_2, A_3$$ partition $$S$$. If $$P(B|A_1)=0.1$$, $$P(B|A_2)=0.4$$, $$P(B|A_3)=0.5$$, and $$P(A_1)=0.3$$, $$P(A_2)=0.4$$, $$P(A_3)=0.3$$, find $$P(B)$$.
- a) 0.33
- b) 0.36
- c) 0.39
- d) 0.41
Answer: Total Probability: $$0.3 \times 0.1 + 0.4 \times 0.4 + 0.3 \times 0.5 = 0.36$$.
4. Partition of a sample space means:
- a) Events are mutually exclusive and collectively exhaustive
- b) Events are independent
- c) Events are disjoint only
- d) Events are dependent
Answer: Partition requires events to be mutually exclusive and collectively exhaustive.
5. If $$\{A_i\}$$ is a partition, then $$\sum P(A_i) = ?$$
- a) 0
- b) 1
- c) Less than 1
- d) More than 1
Answer: Sum of probabilities in a partition equals 1.
6. If events $$A$$ and $$B$$ are partitions, which relation holds?
- a) $$A \cap B = \emptyset$$
- b) $$A \cup B = S$$
- c) Both a and b
- d) None
Answer: For partitions, both mutually exclusive and covering entire space; so (c).
7. Events $$A_1, A_2, A_3$$ partition sample space with equal probabilities. What is $$P(A_1)$$?
- a) 1/3
- b) 1/2
- c) 1/4
- d) 1/5
Answer: Equal parts implies $$P(A_1) = \frac{1}{3}$$.
8. If $$A$$ and $$B$$ partition $$S$$ and $$P(A) = 0.7$$, what is $$P(B)$$?
- a) 0.3
- b) 0.7
- c) 0.5
- d) 0.9
Answer: $$P(B) = 1 - 0.7 = 0.3$$.
9. Which of the following is NOT true about partition of sample space?
- a) Events are mutually exclusive
- b) Events cover entire sample space
- c) Events must be independent
- d) Probabilities sum to 1
Answer: Partitions do NOT require independence.
10. $$P(A_1)=0.2$$, $$P(A_2)=0.3$$, $$P(A_3) = 0.5$$, $$P(B|A_1) = 0.2$$, $$P(B|A_2) = 0.5$$, $$P(B|A_3) = 0.6$$. Find $$P(B)$$.
- a) 0.48
- b) 0.52
- c) 0.46
- d) 0.50
Answer: $$0.2 \times 0.2 + 0.3 \times 0.5 + 0.5 \times 0.6 = 0.48$$.
11. Partitioning a sample space helps to:
- a) Simplify probability calculations
- b) Define impossible events
- c) Combine probabilities of independent events
- d) None
Answer: Helps simplify calculations using total probability.
12. If $$A_1, A_2, A_3$$ form a partition, then the events are:
- a) Exhaustive but not mutually exclusive
- b) Mutually exclusive but not exhaustive
- c) Both mutually exclusive and exhaustive
- d) Neither exhaustive nor exclusive
Answer: Partition events are both mutually exclusive and exhaustive.
13. If $$A_1, A_2$$ partition $$S$$ and $$P(B|A_1) = 0.3$$, $$P(B|A_2) = 0.9$$, $$P(A_1) = 0.6$$, find $$P(B)$$.
- a) 0.54
- b) 0.72
- c) 0.60
- d) 0.36
Answer: $$0.6 \times 0.3 + 0.4 \times 0.9 = 0.54$$.
14. If events $$A, B, C$$ form a partition, then $$P(A \cup B \cup C) = $$ ?
- a) 1
- b) 0
- c) sum of individual probabilities
- d) product of individual probabilities
Answer: 1.
15. The total probability theorem is based on:
- a) Partition of sample space
- b) Independence of events
- c) Union of events
- d) Intersection of events
Answer: Partition of sample space.
16. If $$P(A_1) = 0.25$$, $$P(A_2) = 0.35$$, partitioning sample space, find $$P(A_3)$$.
- a) 0.35
- b) 0.4
- c) 0.3
- d) 0.25
Answer: $$1 - 0.25 - 0.35 = 0.4$$.
17. To form a partition, events must be:
- a) Disjoint and their union is the sample space
- b) Independent and exhaustive
- c) Complementary
- d) Overlapping and exhaustive
Answer: (a).
18. If $$A_1, A_2, A_3$$ form a partition and $$P(B|A_i)$$ known, how to find $$P(B)$$?
- a) Average of $$P(B|A_i)$$
- b) Sum of $$P(A_i)P(B|A_i)$$
- c) Product of $$P(A_i)$$ and $$P(B|A_i)$$
- d) $$P(B)$$ is independent
Answer: Use total probability formula (b).
19. If $$P(A_1) = 0.4$$, $$P(B|A_1) = 0.5$$, $$P(A_2) = 0.6$$, $$P(B|A_2) = 0.2$$, find $$P(B)$$.
- a) 0.32
- b) 0.38
- c) 0.40
- d) 0.44
Answer: $$0.4 \times 0.5 + 0.6 \times 0.2 = 0.38$$.
20. If events $$A_1, A_2$$ partition $$S$$, then $$P(A_1 \cap A_2) = ?$$
- a) 0
- b) 1
- c) $$P(A_1)P(A_2)$$
- d) Cannot tell
Answer: 0 because partitions are mutually exclusive.
