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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 1Questions 1–10 of 60
  1. Q1Past Paper · PPSC/FPSC/CSSeasy

    The sample space of a random experiment is

    1. Athe set of all possible outcomes
    2. Bonly the most likely outcome
    3. Conly outcomes with probability 1
    4. Dthe set of impossible events only
    💡 Explanation:

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

  2. Q2Past Paper · PPSC/FPSC/CSSeasy

    A probability model must satisfy P(S) equal to

    1. A1
    2. B0
    3. C0.5
    4. Dany positive number
    💡 Explanation:

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

  3. Q3Past Paper · PPSC/FPSC/CSSeasy

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

    1. Anegative if rare
    2. Bbetween 0 and 1 inclusive
    3. Cgreater than 1 for sure events
    4. Dalways exactly 0.5
    💡 Explanation:

    Probabilities are non-negative and at most 1.

  4. Q4Past Paper · PPSC/FPSC/CSSeasy

    Two events A and B are mutually exclusive if

    1. AP(A) = P(B)
    2. BA and B are independent
    3. CP(A ∪ B) = 1 always
    4. DP(A ∩ B) = 0
    💡 Explanation:

    Disjoint events cannot occur together.

  5. Q5Past Paper · PPSC/FPSC/CSSeasy

    The complement rule states that P(A') equals

    1. AP(A) − 1
    2. B1 − P(A)
    3. C1 + P(A)
    4. DP(A) × P(A')
    💡 Explanation:

    The complement accounts for all outcomes not in A.

  6. Q6Past Paper · PPSC/FPSC/CSSeasy

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

    1. AP(A) + P(B)
    2. BP(A) × P(B)
    3. CP(A) − P(B)
    4. DP(A) / P(B)
    💡 Explanation:

    For disjoint events, addition rule has no overlap term.

  7. Q7Past Paper · PPSC/FPSC/CSSeasy

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

    1. AP(A) + P(B)
    2. BP(A) + P(B) − P(A ∩ B)
    3. CP(A) × P(B)
    4. DP(A) − P(B)
    💡 Explanation:

    Subtracting the intersection avoids double counting.

  8. Q8Past Paper · PPSC/FPSC/CSSmedium

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

    1. A0.7
    2. B0.9
    3. C0.3
    4. D0.1
    💡 Explanation:

    P(A∪B) = 0.4 + 0.5 − 0.2 = 0.7.

  9. Q9Past Paper · PPSC/FPSC/CSSeasy

    Conditional probability P(A|B) is defined as

    1. AP(A ∩ B) / P(B), provided P(B) > 0
    2. BP(A) / P(B) always
    3. CP(B) / P(A)
    4. DP(A) + P(B)
    💡 Explanation:

    Conditioning restricts the sample space to B.

  10. Q10Past Paper · PPSC/FPSC/CSSmedium

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

    1. A0.036
    2. B0.4
    3. C0.42
    4. D0.15
    💡 Explanation:

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

  11. Q11Past Paper · PPSC/FPSC/CSSeasy

    Events A and B are independent if and only if

    1. AP(A) = P(B)
    2. BP(A ∪ B) = 1
    3. CP(A|B) = 1
    4. DP(A ∩ B) = P(A)P(B)
    💡 Explanation:

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

  12. Q12Past Paper · PPSC/FPSC/CSSmedium

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

    1. A0.7
    2. B0.1
    3. C0.12
    4. D0.58
    💡 Explanation:

    P(A∩B) = 0.3 × 0.4 = 0.12.

  13. Q13Past Paper · PPSC/FPSC/CSSeasy

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

    1. AP(A) × P(B)
    2. BP(A) + P(B)
    3. CP(A) − P(B)
    4. DP(A) / P(B)
    💡 Explanation:

    Independence converts joint probability to a product.

  14. Q14Past Paper · PPSC/FPSC/CSSmedium

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

    1. AP(A) + P(B|A)
    2. BP(A) × P(B|A)
    3. CP(B) only
    4. DP(A|B) − P(B)
    💡 Explanation:

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

  15. Q15Past Paper · PPSC/FPSC/CSSmedium

    Bayes' theorem expresses P(A|B) as

    1. AP(A)P(B)
    2. BP(B|A)P(A) / P(B)
    3. CP(B|A) / P(A)
    4. DP(A) + P(B|A)
    💡 Explanation:

    Bayes reverses conditioning using prior and likelihood.

  16. Q16hard

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

    1. Athe law of total probability over a partition
    2. Balways setting it to 1
    3. Cignoring P(A)
    4. Dmultiplying P(A) and P(B|A) only without summing
    💡 Explanation:

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

  17. Q17Past Paper · PPSC/FPSC/CSSeasy

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

    1. A1/3
    2. B2/3
    3. C1/2
    4. D1/6
    💡 Explanation:

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

  18. Q18Past Paper · PPSC/FPSC/CSSeasy

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

    1. A1/4
    2. B3/4
    3. C1/8
    4. D1/2
    💡 Explanation:

    Outcomes HT, TH: 2 of 4 → 1/2.

  19. Q19Past Paper · PPSC/FPSC/CSSeasy

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

    1. A1/4
    2. B1/13
    3. C1/2
    4. D26/52 only if deck is biased
    💡 Explanation:

    26 red cards out of 52 → 1/2.

  20. Q20Past Paper · PPSC/FPSC/CSSeasy

    The classical definition of probability applies when

    1. Aall outcomes are equally likely
    2. Boutcomes have unequal weights always
    3. Csample space is infinite only
    4. Dno sample space exists
    💡 Explanation:

    Classical probability = favourable outcomes / total equally likely outcomes.

  21. Q21easy

    Empirical (relative frequency) probability is based on

    1. Alogical symmetry only
    2. Baxioms without data
    3. Cproportion of times an event occurs in repeated trials
    4. Dsubjective belief only
    💡 Explanation:

    Long-run relative frequency estimates probability from data.

  22. Q22Past Paper · PPSC/FPSC/CSSmedium

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

    1. An! / r!
    2. Bn! × r!
    3. Cr! / n!
    4. Dn! / (n − r)!
    💡 Explanation:

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

  23. Q23Past Paper · PPSC/FPSC/CSSmedium

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

    1. An! / (n − r)!
    2. Bn × r
    3. Cr! × n!
    4. Dn! / [r!(n − r)!]
    💡 Explanation:

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

  24. Q24Past Paper · PPSC/FPSC/CSSeasy

    The value of 5C2 equals

    1. A20
    2. B5
    3. C10
    4. D25
    💡 Explanation:

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

  25. Q25Past Paper · PPSC/FPSC/CSSeasy

    The value of 5P2 equals

    1. A10
    2. B5
    3. C20
    4. D25
    💡 Explanation:

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

  26. Q26easy

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

    1. A12
    2. B24
    3. C16
    4. D4
    💡 Explanation:

    4! = 24 permutations.

  27. Q27Past Paper · PPSC/FPSC/CSSmedium

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

    1. A21
    2. B210
    3. C35
    4. D7
    💡 Explanation:

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

  28. Q28medium

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

    1. A125
    2. B60
    3. C10
    4. D15
    💡 Explanation:

    5P3 = 5!/2! = 60.

  29. Q29Past Paper · PPSC/FPSC/CSSmedium

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

    1. Aindependence always
    2. Baddition rule only
    3. Cpermutations of 52 only
    4. Dmultiplication rule with changing conditional probabilities
    💡 Explanation:

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

  30. Q30medium

    Drawing with replacement makes successive draws

    1. Aindependent (if each draw has same probabilities)
    2. Balways mutually exclusive
    3. Calways dependent
    4. Dimpossible to model
    💡 Explanation:

    Replacement restores the same population each draw.

  31. Q31Past Paper · PPSC/FPSC/CSSmedium

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

    1. A1/6
    2. B1/36
    3. C5/36
    4. D11/36
    💡 Explanation:

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

  32. Q32Past Paper · PPSC/FPSC/CSSeasy

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

    1. A1/36
    2. B1/6
    3. C2/6
    4. D1/18
    💡 Explanation:

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

  33. Q33hard

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

    1. Amutually exclusive
    2. Bindependent with P(A∩B)=0.3 confirmed independent
    3. Cimpossible
    4. Dnot 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. Q34Past Paper · PPSC/FPSC/CSSmedium

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

    1. A0.9
    2. B0.2
    3. C0.6
    4. D0.7
    💡 Explanation:

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

  35. Q35Past Paper · PPSC/FPSC/CSSeasy

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

    1. A2/5
    2. B1/5
    3. C3/5
    4. D3/10
    💡 Explanation:

    3 red out of 5 total → 3/5.

  36. Q36Past Paper · PPSC/FPSC/CSSmedium

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

    1. A3/10
    2. B9/25
    3. C6/25
    4. D1/5
    💡 Explanation:

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

  37. Q37Past Paper · PPSC/FPSC/CSSmedium

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

    1. A3/10
    2. B9/25
    3. C6/25
    4. D3/5
    💡 Explanation:

    P = (3/5)² = 9/25.

  38. Q38Past Paper · PPSC/FPSC/CSSeasy

    The probability of an impossible event is

    1. A0
    2. B1
    3. Cundefined
    4. D1/2
    💡 Explanation:

    Empty set has probability zero by axioms.

  39. Q39Past Paper · PPSC/FPSC/CSSeasy

    The probability of a sure (certain) event is

    1. A0
    2. B1
    3. C0.5
    4. Ddepends on sample space size
    💡 Explanation:

    The whole sample space has probability 1.

  40. Q40Past Paper · PPSC/FPSC/CSSeasy

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

    1. A0.35
    2. B0.65
    3. C1.35
    4. D−0.35
    💡 Explanation:

    P(A') = 1 − 0.35 = 0.65.

  41. Q41Past Paper · PPSC/FPSC/CSSmedium

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

    1. AP(A)
    2. BP(B)
    3. CP(A ∩ B)
    4. D1 − P(A)
    💡 Explanation:

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

  42. Q42Past Paper · PPSC/FPSC/CSSmedium

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

    1. A0.8
    2. B1.1
    3. C0.3
    4. D0.5
    💡 Explanation:

    P(A∪B) = 0.6 + 0.5 − 0.3 = 0.8.

  43. Q43medium

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

    1. A0.6
    2. B0.3
    3. C0.5
    4. D0.2
    💡 Explanation:

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

  44. Q44hard

    Law of total probability requires

    1. Adependent events only
    2. Ba single event
    3. Cprobabilities summing to 0
    4. Da 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. Q45Past Paper · PPSC/FPSC/CSShard

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

    1. A0.95 only
    2. B0.02 only
    3. C0.90 only
    4. D0.02×0.95 + 0.98×0.10 = 0.117
    💡 Explanation:

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

  46. Q46Past Paper · PPSC/FPSC/CSShard

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

    1. A0.95
    2. B0.162
    3. C0.02
    4. D0.90
    💡 Explanation:

    P(D|+) = 0.019/0.117 ≈ 0.162.

  47. Q47hard

    Odds in favour of an event with probability p equal

    1. A1 − p
    2. Bp × (1 − p)
    3. Cp / (1 − p)
    4. D1 / p
    💡 Explanation:

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

  48. Q48hard

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

    1. A3/4
    2. B1/4
    3. C1/3
    4. D3/1
    💡 Explanation:

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

  49. Q49medium

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

    1. A15600
    2. B26
    3. C78
    4. D17576
    💡 Explanation:

    26³ = 17576.

  50. Q50medium

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

    1. A45
    2. B100
    3. C90 ways
    4. D20
    💡 Explanation:

    10P2 = 10×9 = 90 ordered selections.

  51. Q51Past Paper · PPSC/FPSC/CSSmedium

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

    1. A1/2
    2. B3/8
    3. C7/8
    4. D1/8
    💡 Explanation:

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

  52. Q52medium

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

    1. AP(A) + P(B)
    2. B1 always
    3. Cmin(P(A), P(B))
    4. Dmax(P(A), P(B))
    💡 Explanation:

    Joint probability is bounded by each marginal probability.

  53. Q53Past Paper · PPSC/FPSC/CSSeasy

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

    1. A0
    2. BP(A)²
    3. C2P(A)
    4. D1
    💡 Explanation:

    An event and its complement partition the sample space.

  54. Q54hard

    The birthday problem illustrates that

    1. Abirthdays are independent of probability
    2. BP(shared) is always zero below 365 people
    3. Cshared birthdays are more likely than intuition suggests in moderate groups
    4. Donly leap years matter
    💡 Explanation:

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

  55. Q55medium

    A Venn diagram for three events can represent

    1. Aonly two regions
    2. Ball eight regions of elementary outcome groupings for A, B, C
    3. Conly independent events
    4. Donly continuous outcomes
    💡 Explanation:

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

  56. Q56hard

    Inclusion–exclusion for three events adds back

    1. Aonly P(A)+P(B)+P(C)
    2. Bno intersection terms
    3. CP(A∩B), P(A∩C), P(B∩C) and subtracts P(A∩B∩C)
    4. DP(A)×P(B)×P(C)
    💡 Explanation:

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

  57. Q57Past Paper · PPSC/FPSC/CSSeasy

    From a deck, P(king) equals

    1. A1/52
    2. B4/52 only if stated wrong
    3. C1/4
    4. D1/13
    💡 Explanation:

    4 kings out of 52 → 4/52 = 1/13.

  58. Q58Past Paper · PPSC/FPSC/CSSmedium

    Exactly two heads in three tosses: P equals

    1. A3/8
    2. B1/8
    3. C1/2
    4. D3/4
    💡 Explanation:

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

  59. Q59hard

    A lottery picks 6 numbers from 49. Total combinations equal

    1. A49
    2. B294
    3. C720
    4. D13983816
    💡 Explanation:

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

  60. Q60hard

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

    1. Amutually exclusive with B
    2. Ba subset of B
    3. Cindependent of B
    4. Dequal to B
    💡 Explanation:

    Same conditional probability given B or B' implies independence.

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