A binomial experiment requires

fixed n independent trials, two outcomes, constant p each trial

the number of successes in n independent Bernoulli trials

Discrete Probability Distributions MCQs 2026

59 questions with detailed answers · 39 from past papers · 6 quiz batches available

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Page 1 of 1 Questions 1–10 of 59

Q1 Past Paper · PPSC/FPSC/CSS easy

For X ~ Bin(4, 0.5), P(X=2) equals
1. A 1/2
2. B 1/4
3. C 5/8
4. D 3/8
💡 Explanation:
C(4,2)(0.5)⁴ = 6/16 = 3/8.
Q2 Past Paper · PPSC/FPSC/CSS medium

A quality inspector: 10% defect rate, 5 items. X ~ Bin(5,0.1). P(X=0) equals
1. A 0.5
2. B 0.59049
3. C 0.1
4. D 0.9
💡 Explanation:
(0.9)⁵ = 0.59049.
Q3 Past Paper · PPSC/FPSC/CSS medium

Same inspector: P(X=1) equals
1. A 0.1
2. B 0.5
3. C 0.32805
4. D 0.081
💡 Explanation:
C(5,1)(0.1)(0.9)⁴ = 5×0.1×0.6561 = 0.32805.
Q4 Past Paper · PPSC/FPSC/CSS easy

A binomial experiment requires
1. A three or more outcomes per trial
2. B changing p each trial
3. C fixed n independent trials, two outcomes, constant p each trial
4. D dependent trials
💡 Explanation:
Binomial: n iid Bernoulli(p) trials.
Q5 Past Paper · PPSC/FPSC/CSS easy

X ~ Bin(n,p) counts
1. A waiting time to first success
2. B the number of successes in n independent Bernoulli trials
3. C events per fixed interval only
4. D items until r successes without limit
💡 Explanation:
Binomial counts successes in fixed n.
Q6 Past Paper · PPSC/FPSC/CSS easy

The binomial PMF is P(X=k) equals
1. A n! p^k
2. B p^k only
3. C C(n,k) p^n
4. D C(n,k) p^k (1−p)^{n−k}
💡 Explanation:
Standard binomial formula for k successes.
Q7 Past Paper · PPSC/FPSC/CSS easy

For X ~ Bin(10, 0.3), E(X) equals
1. A 0.3
2. B 10
3. C 7
4. D 3
💡 Explanation:
E(X) = np = 10×0.3 = 3.
Q8 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(10, 0.3), Var(X) equals
1. A 3
2. B 2.1
3. C 0.21
4. D 7
💡 Explanation:
Var(X) = np(1−p) = 10×0.3×0.7 = 2.1.
Q9 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(5, 0.4), P(X=2) equals
1. A 0.4
2. B 0.3456
3. C 0.0768
4. D 0.64
💡 Explanation:
C(5,2)(0.4)²(0.6)³ = 10×0.16×0.216 = 0.3456.
Q10 Past Paper · PPSC/FPSC/CSS easy

The Poisson distribution models
1. A fixed n trials with two outcomes
2. B the number of events in a fixed interval with constant average rate λ
3. C sampling without replacement from finite population
4. D continuous waiting times only
💡 Explanation:
Poisson counts rare events in time/space with rate λ.
Q11 Past Paper · PPSC/FPSC/CSS easy

X ~ Poisson(λ) has PMF P(X=k) equals
1. A λ^k only
2. B e^{−λ} only
3. C e^{−λ} λ^k / k!
4. D k! / λ^k
💡 Explanation:
Poisson PMF: e^{−λ}λ^k/k!.
Q12 Past Paper · PPSC/FPSC/CSS easy

For X ~ Poisson(λ), E(X) and Var(X) both equal
1. A λ
2. B λ²
3. C √λ
4. D 2λ
💡 Explanation:
Poisson mean and variance are both λ.
Q13 Past Paper · PPSC/FPSC/CSS medium

For X ~ Poisson(3), P(X=0) equals
1. A 3
2. B 0.5
3. C 1/3
4. D e^{−3} ≈ 0.0498
💡 Explanation:
P(0) = e^{−3}3⁰/0! = e^{−3}.
Q14 Past Paper · PPSC/FPSC/CSS medium

For X ~ Poisson(4), P(X=2) equals
1. A 4e^{−4}
2. B 2e^{−4}
3. C 8e^{−4} ≈ 0.1465
4. D 16e^{−4}
💡 Explanation:
P(2) = e^{−4}4²/2! = 16e^{−4}/2 = 8e^{−4}.
Q15 Past Paper · PPSC/FPSC/CSS medium

Poisson is often used to approximate binomial when
1. A n is large, p is small, and λ = np is moderate
2. B n and p are both large
3. C p is near 1 only
4. D n is small
💡 Explanation:
Rare events: large n, small p, λ=np fixed.
Q16 Past Paper · PPSC/FPSC/CSS easy

Hypergeometric distribution applies when
1. A sampling without replacement from a finite population of N with K successes
2. B sampling with replacement
3. C infinite population
4. D Poisson rate λ
💡 Explanation:
Hypergeometric: finite population, no replacement.
Q17 Past Paper · PPSC/FPSC/CSS medium

X ~ Hypergeometric(N,K,n) counts successes in sample size n. P(X=k) uses
1. A p^k (1−p)^{n−k}
2. B C(K,k) C(N−K, n−k) / C(N,n)
3. C e^{−λ}λ^k/k!
4. D n! p^k
💡 Explanation:
Hypergeometric PMF uses combinations from finite urn.
Q18 Past Paper · PPSC/FPSC/CSS medium

An urn has 10 items, 4 defective. Sample 3 without replacement. X = defectives. X is
1. A Binomial(3,0.4) exactly always
2. B Poisson(1.2)
3. C Hypergeometric(10,4,3)
4. D Normal(1.2,0.7)
💡 Explanation:
Finite population without replacement → hypergeometric.
Q19 Past Paper · PPSC/FPSC/CSS hard

An urn has 10 items, 4 defective. Sample 3 without replacement. P(exactly 1 defective) equals
1. A 0.3
2. B 0.112
3. C 0.6
4. D 0.5
💡 Explanation:
C(4,1)C(6,2)/C(10,3) = 4×15/120 = 60/120 = 0.5.
Q20 hard

For hypergeometric, when N is large relative to n, it approximates
1. A Poisson with λ=n
2. B binomial with p = K/N
3. C normal with μ=0
4. D uniform
💡 Explanation:
Large N: sampling without replacement ≈ with replacement.
Q21 Past Paper · PPSC/FPSC/CSS medium

Geometric distribution counts
1. A successes in fixed n
2. B the trial number of the first success in independent Bernoulli(p) trials
3. C events per interval
4. D defectives in sample n
💡 Explanation:
Geometric: waiting time to first success.
Q22 Past Paper · PPSC/FPSC/CSS medium

For Geometric(p), P(X=k) (first success on trial k) equals
1. A (1−p)^{k−1} p
2. B p^k
3. C (1−p)^k p
4. D C(n,k)p^k
💡 Explanation:
Need k−1 failures then one success.
Q23 Past Paper · PPSC/FPSC/CSS medium

For Geometric(p), E(X) equals
1. A 1/p
2. B p
3. C 1−p
4. D p²
💡 Explanation:
Mean waiting time to first success is 1/p.
Q24 hard

Negative binomial generalizes geometric by counting
1. A events in fixed interval
2. B successes in fixed n only
3. C items in finite urn only
4. D trials until r successes occur
💡 Explanation:
Negative binomial: waiting for r-th success.
Q25 hard

Multinomial distribution extends binomial to
1. A more than two categories with fixed n trials
2. B continuous outcomes
3. C Poisson rates only
4. D hypergeometric sampling
💡 Explanation:
Multinomial counts outcomes in k categories.
Q26 Past Paper · PPSC/FPSC/CSS easy

For X ~ Bin(20, 0.1), E(X) equals
1. A 2
2. B 0.1
3. C 20
4. D 18
💡 Explanation:
E(X) = np = 20×0.1 = 2.
Q27 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(20, 0.1), Var(X) equals
1. A 2
2. B 0.09
3. C 1.8
4. D 18
💡 Explanation:
Var = 20×0.1×0.9 = 1.8.
Q28 Past Paper · PPSC/FPSC/CSS medium

P(at least one success) for X ~ Bin(n,p) equals
1. A p^n
2. B np
3. C 1 − (1−p)^n
4. D 1 − p
💡 Explanation:
Complement of all failures: 1−(1−p)^n.
Q29 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(6, 0.5), P(X ≥ 1) equals
1. A 63/64
2. B 1/2
3. C 1/64
4. D 7/8
💡 Explanation:
1 − P(0) = 1 − (1/2)⁶ = 1 − 1/64 = 63/64.
Q30 medium

Poisson process assumption includes
1. A events always occur in pairs
2. B rate changes every second always
3. C exactly one event per trial
4. D events occur independently at constant average rate
💡 Explanation:
Homogeneous Poisson: independent increments, constant λ.
Q31 Past Paper · PPSC/FPSC/CSS easy

If accidents occur at rate λ=2 per week, expected accidents in 3 weeks is
1. A 6
2. B 2
3. C 3
4. D 4
💡 Explanation:
E(total) = λt = 2×3 = 6 for Poisson process.
Q32 hard

Sum of independent Poisson(λ1) and Poisson(λ2) is
1. A Binomial
2. B Poisson(λ1 + λ2)
3. C Normal always
4. D Hypergeometric
💡 Explanation:
Poisson sums add rates.
Q33 Past Paper · PPSC/FPSC/CSS easy

For X ~ Poisson(5), standard deviation equals
1. A 5
2. B √5
3. C 25
4. D 2.5
💡 Explanation:
σ = √λ = √5.
Q34 medium

Binomial is symmetric about np when
1. A p = 0.1
2. B p = 0.9
3. C p = 0.5
4. D always for any p
💡 Explanation:
p=0.5 gives symmetric binomial.
Q35 hard

Most likely value (mode) of Bin(n,p) is near
1. A np/2 always
2. B n always
3. C 0 always
4. D floor((n+1)p) or (n+1)p depending on integer
💡 Explanation:
Mode is approximately np for binomial.
Q36 Past Paper · PPSC/FPSC/CSS medium

For n=100, p=0.05, Poisson approximation uses λ equal to
1. A 5
2. B 0.05
3. C 100
4. D 95
💡 Explanation:
λ = np = 100×0.05 = 5.
Q37 hard

Using Poisson(5) to approximate Bin(100,0.05), P(X=3) ≈
1. A 0.1404
2. B 0.5
3. C 0.01
4. D 0.25
💡 Explanation:
P(3) = e^{−5}5³/6 ≈ 0.1404.
Q38 Past Paper · PPSC/FPSC/CSS medium

Hypergeometric(20,8,5): population N=20, K=8 successes, n=5 drawn. E(X) equals
1. A 2
2. B 8
3. C 5
4. D 1.6
💡 Explanation:
E(X) = n(K/N) = 5×(8/20) = 2.
Q39 hard

For X ~ Bin(8, 0.25), P(X=2) equals
1. A 0.25
2. B 0.5
3. C 0.125
4. D 0.311
💡 Explanation:
C(8,2)(0.25)²(0.75)⁶ = 28×0.0625×0.1779785 ≈ 0.311.
Q40 Past Paper · PPSC/FPSC/CSS medium

For Bin(3, 1/3), P(X=1) equals
1. A 1/3
2. B 4/9
3. C 2/9
4. D 8/27
💡 Explanation:
C(3,1)(1/3)(2/3)² = 3×1/3×4/9 = 4/9.
Q41 medium

Poisson(λ=0.5) might model
1. A exactly 50% success rate
2. B hypergeometric with N=2
3. C rare defects when n is large and p is tiny with np=0.5
4. D always binomial n=1
💡 Explanation:
Small λ from rare events.
Q42 hard

Hypergeometric: N=50, K=10, n=5. P(X=0) equals
1. A 0.5
2. B C(40,5)/C(50,5) ≈ 0.418
3. C 0.1
4. D 0.9
💡 Explanation:
No successes: choose all 5 from 40 non-successes.
Q43 easy

Binomial coefficient C(n,k) equals 0 when
1. A k = n/2
2. B p = 0.5
3. C λ = 0
4. D k > n or k < 0
💡 Explanation:
Cannot choose more than n items.
Q44 hard

If Y ~ Bin(n,p) and Z ~ Bin(m,p) independent, Y+Z ~
1. A Bin(nm,p)
2. B Poisson(np)
3. C Hypergeometric
4. D Bin(n+m, p)
💡 Explanation:
Sum of independent binomials with same p adds n.
Q45 Past Paper · PPSC/FPSC/CSS easy

Poisson parameter λ represents
1. A probability of success
2. B sample size n
3. C expected number of events in the specified interval
4. D variance squared
💡 Explanation:
λ is mean count per interval.
Q46 Past Paper · PPSC/FPSC/CSS medium

For telephone calls: avg 4 per hour. Calls in 15 min ~ Poisson(1). P(0 calls) equals
1. A 0.25
2. B 0.5
3. C e^{−1} ≈ 0.368
4. D 0
💡 Explanation:
λ = 4×(15/60) = 1; P(0)=e^{−1}.
Q47 hard

Lottery: match 3 of 6 drawn from 49. Count follows
1. A Poisson(3)
2. B Binomial with p=3
3. C normal Z only
4. D hypergeometric-type counting (combinatorial selection)
💡 Explanation:
Selecting matched numbers from finite pool.
Q48 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(12, 0.4), E(X) and Var(X) are
1. A 4.8 and 4.8
2. B 4.8 and 2.88
3. C 12 and 0.4
4. D 2.4 and 1.44
💡 Explanation:
E=np=4.8; Var=12×0.4×0.6=2.88.
Q49 medium

Discrete uniform on 1,…,n has E(X) equal to
1. A n/2
2. B (n+1)/2
3. C n
4. D 1
💡 Explanation:
Mean of 1..n is (n+1)/2.
Q50 hard

For X ~ Poisson(10), P(X ≤ 10) is roughly 0.5 because
1. A Poisson is always exactly symmetric
2. B λ=0
3. C Poisson has no mean
4. D Poisson is skewed but median near λ for moderate λ
💡 Explanation:
For larger λ, Poisson approaches symmetric around λ.
Q51 Past Paper · PPSC/FPSC/CSS easy

A Bernoulli trial with p=0.7 repeated 8 times: total successes ~
1. A Bin(8, 0.7)
2. B Poisson(5.6) primarily
3. C Hypergeometric(8,7,8)
4. D Geometric(0.7)
💡 Explanation:
Fixed 8 independent trials → binomial.
Q52 Past Paper · PPSC/FPSC/CSS easy

The factorial 0! is defined as
1. A 1
2. B 0
3. C undefined
4. D −1
💡 Explanation:
0! = 1 by convention for combinatorics.
Q53 hard

Stirling's approximation helps estimate
1. A small n exactly
2. B only Poisson λ
3. C n! for large n
4. D hypergeometric N only
💡 Explanation:
Stirling approximates factorials for large n in binomial coefficients.
Q54 Past Paper · PPSC/FPSC/CSS medium

For rare disease screening with n=10000, p=0.001, expected cases E(X) equals
1. A 10
2. B 1
3. C 100
4. D 0.1
💡 Explanation:
E(X)=np=10; Poisson(10) may approximate.
Q55 hard

Multinomial with n trials and probabilities p1,…,pk requires
1. A all pi equal 0.5
2. B p1 + … + pk = 1
3. C k = 2 only
4. D n = k
💡 Explanation:
Category probabilities must sum to 1.
Q56 Past Paper · PPSC/FPSC/CSS easy

A store gets avg 6 customers per 10 minutes. In 5 minutes, λ for Poisson equals
1. A 6
2. B 1
3. C 3
4. D 0.5
💡 Explanation:
Rate scales: 6×(5/10)=3.
Q57 Past Paper · PPSC/FPSC/CSS medium

For X ~ Bin(n,0.5), Var(X) equals
1. A n/2
2. B np
3. C n/4
4. D 0.25
💡 Explanation:
Var = np(1−p) = n×0.25 = n/4.
Q58 hard

Hypergeometric variance is smaller than binomial with same n and p when
1. A N is infinite
2. B p = 1
3. C sampling is without replacement from finite N
4. D n = 0
💡 Explanation:
Finite population correction reduces variance.
Q59 hard

The PMF of Poisson(0) is
1. A uniform on 0,1
2. B undefined
3. C e^{−1}
4. D degenerate at 0 (P(X=0)=1)
💡 Explanation:
λ=0 implies no events: P(X=0)=1.

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