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Basic Probability MCQs 2026

60 questions with detailed answers · 42 from past papers · 6 quiz batches available

📚 Statistics 📄 42 Past-Paper Qs ✓ Free · No Login Needed
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Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1 Questions 1–10 of 60
  1. Q1 Past Paper · PPSC/FPSC/CSS easy

    The sample space of a random experiment is

    1. A the set of all possible outcomes
    2. B only the most likely outcome
    3. C only outcomes with probability 1
    4. D the set of impossible events only
    💡 Explanation:

    The sample space S lists every elementary outcome of the experiment.

  2. Q2 Past Paper · PPSC/FPSC/CSS easy

    A probability model must satisfy P(S) equal to

    1. A 1
    2. B 0
    3. C 0.5
    4. D any positive number
    💡 Explanation:

    One of Kolmogorov's axioms: the probability of the entire sample space is 1.

  3. Q3 Past Paper · PPSC/FPSC/CSS easy

    For any event A, the axiom of probability requires P(A) to be

    1. A negative if rare
    2. B between 0 and 1 inclusive
    3. C greater than 1 for sure events
    4. D always exactly 0.5
    💡 Explanation:

    Probabilities are non-negative and at most 1.

  4. Q4 Past Paper · PPSC/FPSC/CSS easy

    Two events A and B are mutually exclusive if

    1. A P(A) = P(B)
    2. B A and B are independent
    3. C P(A ∪ B) = 1 always
    4. D P(A ∩ B) = 0
    💡 Explanation:

    Disjoint events cannot occur together.

  5. Q5 Past Paper · PPSC/FPSC/CSS easy

    The complement rule states that P(A') equals

    1. A P(A) − 1
    2. B 1 − P(A)
    3. C 1 + P(A)
    4. D P(A) × P(A')
    💡 Explanation:

    The complement accounts for all outcomes not in A.

  6. Q6 Past Paper · PPSC/FPSC/CSS easy

    If A and B are mutually exclusive, then P(A ∪ B) equals

    1. A P(A) + P(B)
    2. B P(A) × P(B)
    3. C P(A) − P(B)
    4. D P(A) / P(B)
    💡 Explanation:

    For disjoint events, addition rule has no overlap term.

  7. Q7 Past Paper · PPSC/FPSC/CSS easy

    The general addition rule is P(A ∪ B) equals

    1. A P(A) + P(B)
    2. B P(A) + P(B) − P(A ∩ B)
    3. C P(A) × P(B)
    4. D P(A) − P(B)
    💡 Explanation:

    Subtracting the intersection avoids double counting.

  8. Q8 Past Paper · PPSC/FPSC/CSS medium

    Given P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, P(A ∪ B) is

    1. A 0.7
    2. B 0.9
    3. C 0.3
    4. D 0.1
    💡 Explanation:

    P(A∪B) = 0.4 + 0.5 − 0.2 = 0.7.

  9. Q9 Past Paper · PPSC/FPSC/CSS easy

    Conditional probability P(A|B) is defined as

    1. A P(A ∩ B) / P(B), provided P(B) > 0
    2. B P(A) / P(B) always
    3. C P(B) / P(A)
    4. D P(A) + P(B)
    💡 Explanation:

    Conditioning restricts the sample space to B.

  10. Q10 Past Paper · PPSC/FPSC/CSS medium

    If P(A ∩ B) = 0.12 and P(B) = 0.3, then P(A|B) equals

    1. A 0.036
    2. B 0.4
    3. C 0.42
    4. D 0.15
    💡 Explanation:

    P(A|B) = 0.12/0.3 = 0.4.

  11. Q11 Past Paper · PPSC/FPSC/CSS easy

    Events A and B are independent if and only if

    1. A P(A) = P(B)
    2. B P(A ∪ B) = 1
    3. C P(A|B) = 1
    4. D P(A ∩ B) = P(A)P(B)
    💡 Explanation:

    Independence means occurrence of one does not change probability of the other.

  12. Q12 Past Paper · PPSC/FPSC/CSS medium

    If A and B are independent with P(A) = 0.3 and P(B) = 0.4, then P(A ∩ B) is

    1. A 0.7
    2. B 0.1
    3. C 0.12
    4. D 0.58
    💡 Explanation:

    P(A∩B) = 0.3 × 0.4 = 0.12.

  13. Q13 Past Paper · PPSC/FPSC/CSS easy

    The multiplication rule for independent events gives P(A ∩ B) as

    1. A P(A) × P(B)
    2. B P(A) + P(B)
    3. C P(A) − P(B)
    4. D P(A) / P(B)
    💡 Explanation:

    Independence converts joint probability to a product.

  14. Q14 Past Paper · PPSC/FPSC/CSS medium

    The general multiplication rule states P(A ∩ B) equals

    1. A P(A) + P(B|A)
    2. B P(A) × P(B|A)
    3. C P(B) only
    4. D P(A|B) − P(B)
    💡 Explanation:

    Chain rule: P(A∩B) = P(A)P(B|A) = P(B)P(A|B).

  15. Q15 Past Paper · PPSC/FPSC/CSS medium

    Bayes' theorem expresses P(A|B) as

    1. A P(A)P(B)
    2. B P(B|A)P(A) / P(B)
    3. C P(B|A) / P(A)
    4. D P(A) + P(B|A)
    💡 Explanation:

    Bayes reverses conditioning using prior and likelihood.

  16. Q16 hard

    In Bayes' theorem, P(B) in the denominator is found by

    1. A the law of total probability over a partition
    2. B always setting it to 1
    3. C ignoring P(A)
    4. D multiplying P(A) and P(B|A) only without summing
    💡 Explanation:

    P(B) = Σ P(B|Ai)P(Ai) over a partition {Ai}.

  17. Q17 Past Paper · PPSC/FPSC/CSS easy

    A fair die is rolled once. P(getting an even number) equals

    1. A 1/3
    2. B 2/3
    3. C 1/2
    4. D 1/6
    💡 Explanation:

    Even faces {2,4,6}: 3 outcomes out of 6 → 3/6 = 1/2.

  18. Q18 Past Paper · PPSC/FPSC/CSS easy

    Two fair coins are tossed. P(exactly one head) equals

    1. A 1/4
    2. B 3/4
    3. C 1/8
    4. D 1/2
    💡 Explanation:

    Outcomes HT, TH: 2 of 4 → 1/2.

  19. Q19 Past Paper · PPSC/FPSC/CSS easy

    A card is drawn from a standard 52-card deck. P(red card) equals

    1. A 1/4
    2. B 1/13
    3. C 1/2
    4. D 26/52 only if deck is biased
    💡 Explanation:

    26 red cards out of 52 → 1/2.

  20. Q20 Past Paper · PPSC/FPSC/CSS easy

    The classical definition of probability applies when

    1. A all outcomes are equally likely
    2. B outcomes have unequal weights always
    3. C sample space is infinite only
    4. D no sample space exists
    💡 Explanation:

    Classical probability = favourable outcomes / total equally likely outcomes.

  21. Q21 easy

    Empirical (relative frequency) probability is based on

    1. A logical symmetry only
    2. B axioms without data
    3. C proportion of times an event occurs in repeated trials
    4. D subjective belief only
    💡 Explanation:

    Long-run relative frequency estimates probability from data.

  22. Q22 Past Paper · PPSC/FPSC/CSS medium

    The number of permutations of n distinct objects taken r at a time is

    1. A n! / r!
    2. B n! × r!
    3. C r! / n!
    4. D n! / (n − r)!
    💡 Explanation:

    nPr = n!/(n−r)! counts ordered arrangements.

  23. Q23 Past Paper · PPSC/FPSC/CSS medium

    The number of combinations of n objects taken r at a time is

    1. A n! / (n − r)!
    2. B n × r
    3. C r! × n!
    4. D n! / [r!(n − r)!]
    💡 Explanation:

    nCr = n!/[r!(n−r)!] counts unordered selections.

  24. Q24 Past Paper · PPSC/FPSC/CSS easy

    The value of 5C2 equals

    1. A 20
    2. B 5
    3. C 10
    4. D 25
    💡 Explanation:

    5C2 = 5!/(2!3!) = 120/12 = 10.

  25. Q25 Past Paper · PPSC/FPSC/CSS easy

    The value of 5P2 equals

    1. A 10
    2. B 5
    3. C 20
    4. D 25
    💡 Explanation:

    5P2 = 5!/3! = 120/6 = 20.

  26. Q26 easy

    How many ways can 4 people sit in 4 chairs in a row

    1. A 12
    2. B 24
    3. C 16
    4. D 4
    💡 Explanation:

    4! = 24 permutations.

  27. Q27 Past Paper · PPSC/FPSC/CSS medium

    A committee of 3 is chosen from 7 people. The number of ways is

    1. A 21
    2. B 210
    3. C 35
    4. D 7
    💡 Explanation:

    7C3 = 7!/(3!4!) = 35.

  28. Q28 medium

    If order matters, 3-digit codes from digits 1–5 without repetition: total codes equal

    1. A 125
    2. B 60
    3. C 10
    4. D 15
    💡 Explanation:

    5P3 = 5!/2! = 60.

  29. Q29 Past Paper · PPSC/FPSC/CSS medium

    Drawing 2 aces in succession without replacement from a deck: use

    1. A independence always
    2. B addition rule only
    3. C permutations of 52 only
    4. D multiplication rule with changing conditional probabilities
    💡 Explanation:

    Without replacement, P(second ace|first ace) changes.

  30. Q30 medium

    Drawing with replacement makes successive draws

    1. A independent (if each draw has same probabilities)
    2. B always mutually exclusive
    3. C always dependent
    4. D impossible to model
    💡 Explanation:

    Replacement restores the same population each draw.

  31. Q31 Past Paper · PPSC/FPSC/CSS medium

    P(at least one six in two rolls of a fair die) equals

    1. A 1/6
    2. B 1/36
    3. C 5/36
    4. D 11/36
    💡 Explanation:

    P(at least one) = 1 − P(no six) = 1 − (5/6)² = 1 − 25/36 = 11/36.

  32. Q32 Past Paper · PPSC/FPSC/CSS easy

    P(both sixes in two rolls of a fair die) equals

    1. A 1/36
    2. B 1/6
    3. C 2/6
    4. D 1/18
    💡 Explanation:

    P(both) = (1/6)² = 1/36 assuming independence.

  33. Q33 hard

    If P(A) = 0.6 and P(B) = 0.5 with P(A ∩ B) = 0.3, events A and B are

    1. A mutually exclusive
    2. B independent with P(A∩B)=0.3 confirmed independent
    3. C impossible
    4. D not mutually exclusive but may or may not be independent
    💡 Explanation:

    P(A∩B)≠0 so not disjoint; 0.6×0.5=0.3 so they are independent here.

  34. Q34 Past Paper · PPSC/FPSC/CSS medium

    For independent A, B: P(A ∪ B) with P(A)=0.4, P(B)=0.5 equals

    1. A 0.9
    2. B 0.2
    3. C 0.6
    4. D 0.7
    💡 Explanation:

    P(A∪B) = 0.4 + 0.5 − 0.4×0.5 = 0.9 − 0.2 = 0.7.

  35. Q35 Past Paper · PPSC/FPSC/CSS easy

    A box has 3 red and 2 blue balls. One drawn at random. P(red) equals

    1. A 2/5
    2. B 1/5
    3. C 3/5
    4. D 3/10
    💡 Explanation:

    3 red out of 5 total → 3/5.

  36. Q36 Past Paper · PPSC/FPSC/CSS medium

    From the same box, two balls drawn without replacement. P(both red) equals

    1. A 3/10
    2. B 9/25
    3. C 6/25
    4. D 1/5
    💡 Explanation:

    P = (3/5)(2/4) = 6/20 = 3/10.

  37. Q37 Past Paper · PPSC/FPSC/CSS medium

    From the same box, two balls drawn with replacement. P(both red) equals

    1. A 3/10
    2. B 9/25
    3. C 6/25
    4. D 3/5
    💡 Explanation:

    P = (3/5)² = 9/25.

  38. Q38 Past Paper · PPSC/FPSC/CSS easy

    The probability of an impossible event is

    1. A 0
    2. B 1
    3. C undefined
    4. D 1/2
    💡 Explanation:

    Empty set has probability zero by axioms.

  39. Q39 Past Paper · PPSC/FPSC/CSS easy

    The probability of a sure (certain) event is

    1. A 0
    2. B 1
    3. C 0.5
    4. D depends on sample space size
    💡 Explanation:

    The whole sample space has probability 1.

  40. Q40 Past Paper · PPSC/FPSC/CSS easy

    If P(A) = 0.35, then P(A') equals

    1. A 0.35
    2. B 0.65
    3. C 1.35
    4. D −0.35
    💡 Explanation:

    P(A') = 1 − 0.35 = 0.65.

  41. Q41 Past Paper · PPSC/FPSC/CSS medium

    A and B independent implies P(A|B) equals

    1. A P(A)
    2. B P(B)
    3. C P(A ∩ B)
    4. D 1 − P(A)
    💡 Explanation:

    Independence: conditioning on B does not change P(A).

  42. Q42 Past Paper · PPSC/FPSC/CSS medium

    A survey: 60% read newspaper A, 50% read B, 30% read both. P(read A or B) equals

    1. A 0.8
    2. B 1.1
    3. C 0.3
    4. D 0.5
    💡 Explanation:

    P(A∪B) = 0.6 + 0.5 − 0.3 = 0.8.

  43. Q43 medium

    From the same survey, P(read A but not B) equals

    1. A 0.6
    2. B 0.3
    3. C 0.5
    4. D 0.2
    💡 Explanation:

    P(A only) = P(A) − P(A∩B) = 0.6 − 0.3 = 0.3.

  44. Q44 hard

    Law of total probability requires

    1. A dependent events only
    2. B a single event
    3. C probabilities summing to 0
    4. D a partition of the sample space into mutually exclusive events covering S
    💡 Explanation:

    Partition {Bi} with ΣP(Bi)=1 allows P(A)=ΣP(A|Bi)P(Bi).

  45. Q45 Past Paper · PPSC/FPSC/CSS hard

    In a diagnostic test: disease prevalence 2%, sensitivity 95%, specificity 90%. P(positive test) approximately uses

    1. A 0.95 only
    2. B 0.02 only
    3. C 0.90 only
    4. D 0.02×0.95 + 0.98×0.10 = 0.117
    💡 Explanation:

    P(+)=P(+|D)P(D)+P(+|D')P(D') = 0.019+0.098 = 0.117.

  46. Q46 Past Paper · PPSC/FPSC/CSS hard

    Using those test figures, P(disease | positive) by Bayes is approximately

    1. A 0.95
    2. B 0.162
    3. C 0.02
    4. D 0.90
    💡 Explanation:

    P(D|+) = 0.019/0.117 ≈ 0.162.

  47. Q47 hard

    Odds in favour of an event with probability p equal

    1. A 1 − p
    2. B p × (1 − p)
    3. C p / (1 − p)
    4. D 1 / p
    💡 Explanation:

    Odds for = P(event)/P(complement).

  48. Q48 hard

    If odds against are 3:1, probability of the event equals

    1. A 3/4
    2. B 1/4
    3. C 1/3
    4. D 3/1
    💡 Explanation:

    Odds against 3:1 → P = 1/(3+1) = 1/4.

  49. Q49 medium

    A password has 3 letters A–Z with repetition allowed. Total passwords equal

    1. A 15600
    2. B 26
    3. C 78
    4. D 17576
    💡 Explanation:

    26³ = 17576.

  50. Q50 medium

    Choosing a president and vice-president from 10 members (distinct roles) gives

    1. A 45
    2. B 100
    3. C 90 ways
    4. D 20
    💡 Explanation:

    10P2 = 10×9 = 90 ordered selections.

  51. Q51 Past Paper · PPSC/FPSC/CSS medium

    At least one head in three fair coin tosses: P equals

    1. A 1/2
    2. B 3/8
    3. C 7/8
    4. D 1/8
    💡 Explanation:

    1 − P(no heads) = 1 − (1/2)³ = 7/8.

  52. Q52 medium

    For any events, P(A ∩ B) cannot exceed

    1. A P(A) + P(B)
    2. B 1 always
    3. C min(P(A), P(B))
    4. D max(P(A), P(B))
    💡 Explanation:

    Joint probability is bounded by each marginal probability.

  53. Q53 Past Paper · PPSC/FPSC/CSS easy

    For any event A, P(A) + P(A') equals

    1. A 0
    2. B P(A)²
    3. C 2P(A)
    4. D 1
    💡 Explanation:

    An event and its complement partition the sample space.

  54. Q54 hard

    The birthday problem illustrates that

    1. A birthdays are independent of probability
    2. B P(shared) is always zero below 365 people
    3. C shared birthdays are more likely than intuition suggests in moderate groups
    4. D only leap years matter
    💡 Explanation:

    Pairwise comparisons grow rapidly (nC2), raising collision probability.

  55. Q55 medium

    A Venn diagram for three events can represent

    1. A only two regions
    2. B all eight regions of elementary outcome groupings for A, B, C
    3. C only independent events
    4. D only continuous outcomes
    💡 Explanation:

    Three-set Venn partitions into 8 regions for inclusion–exclusion.

  56. Q56 hard

    Inclusion–exclusion for three events adds back

    1. A only P(A)+P(B)+P(C)
    2. B no intersection terms
    3. C P(A∩B), P(A∩C), P(B∩C) and subtracts P(A∩B∩C)
    4. D P(A)×P(B)×P(C)
    💡 Explanation:

    General formula: ΣP(Ai) − ΣP(Ai∩Aj) + P(A∩B∩C).

  57. Q57 Past Paper · PPSC/FPSC/CSS easy

    From a deck, P(king) equals

    1. A 1/52
    2. B 4/52 only if stated wrong
    3. C 1/4
    4. D 1/13
    💡 Explanation:

    4 kings out of 52 → 4/52 = 1/13.

  58. Q58 Past Paper · PPSC/FPSC/CSS medium

    Exactly two heads in three tosses: P equals

    1. A 3/8
    2. B 1/8
    3. C 1/2
    4. D 3/4
    💡 Explanation:

    3 outcomes HHT, HTH, THH out of 8 → 3/8.

  59. Q59 hard

    A lottery picks 6 numbers from 49. Total combinations equal

    1. A 49
    2. B 294
    3. C 720
    4. D 13983816
    💡 Explanation:

    49C6 = 49!/(6!43!) = 13983816.

  60. Q60 hard

    If P(A|B) = P(A|B'), then A is

    1. A mutually exclusive with B
    2. B a subset of B
    3. C independent of B
    4. D equal to B
    💡 Explanation:

    Same conditional probability given B or B' implies independence.

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