State statistical services: Statistical Officer / Assistant Statistical Officer (State PSCs and selection boards) · India · State Public Service Commissions and Subordinate Service Selection Boards
Original practice questions written to the published syllabus · not past papersProbability from its axioms to the limit theorems: random variables, expectation, generating functions, inequalities and modes of convergence, ending with the laws of large numbers and the central limit theorem that justify most of applied statistics.
In the exams: AICE-JRF Ag Stat Unit 1 (heavy); GATE ST 'Probability' and limit theorems (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I (heavy); ISI admission (heavy); CSIR-NET Mathematical Sciences Unit 4 (present)
ICAR AIEEA-PG Group 07, CUET-PG Statistics, SSC JSO and state statistical services
Classical and conditional probability, addition and multiplication rules, Bayes' theorem, expectation and variance, Chebyshev's inequality and the normal approximation.
How our questions for State statistical services spread across the units of Probability Theory. Difficulty: 25 easy, 12 medium, 3 hard.
| Conditional probability, independence and Bayes' theorem | 7 | |
| Random variables, distribution functions, mass and density functions | 5 | |
| Expectation, moments and moment-generating, probability-generating and characteristic functions | 5 | |
| Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz | 5 | |
| Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method | 5 | |
| Sample spaces, sigma-fields and the axioms of probability | 4 | |
| Joint, marginal and conditional distributions; conditional expectation | 4 | |
| Weak and strong laws of large numbers | 2 | |
| Functions of random variables and transformations (Jacobian method) | 1 | |
| Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma | 1 | |
| Elements of measure theory: measurable functions and Lebesgue integral | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Three pests attack a crop. Each pest is present on a plant with probability 0.4, each pair is present together with probability 0.15, and all three with probability 0.05. What is the probability that at least one pest is present?
B. 0.8
Inclusion-exclusion: 3(0.4) − 3(0.15) + 0.05 = 1.2 − 0.45 + 0.05 = 0.8. Dropping the triple term gives 0.75; subtracting it gives 0.7; 0.784 assumes independence, which the given pair probability 0.15 ≠ 0.16 contradicts.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
A test for a seed-borne virus has sensitivity 0.95 and specificity 0.9. The virus is present in 2% of lots. A lot tests positive. What is the probability that it is infected?
A. 0.162
Bayes: P(infected | +) = (0.02 × 0.95)/(0.02 × 0.95 + 0.98 × 0.10) = 0.019/0.117 = 0.162. 0.950 is the sensitivity P(+ | infected), the commonest confusion; 0.019 is the joint probability P(infected and +), not the conditional.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
A seed company buys 50%, 30% and 20% of its seed from growers G1, G2 and G3, whose seed fails to germinate with probabilities 0.02, 0.03 and 0.05. A seed fails. What is the probability that it came from G3?
A. 0.345
P(failure) = 0.010 + 0.009 + 0.010 = 0.029. Bayes: P(G3 | fail) = 0.2 × 0.05 / 0.029 = 0.345. 0.200 is the prior share, 0.050 is P(fail | G3) and 0.010 is the joint probability.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
X is uniform on (0, 0.5), so its density equals 2 on that interval. Which statement is correct?
B. This is valid: a density must integrate to 1, and its values are allowed to exceed 1
A pdf must be non-negative and integrate to 1; 2 × 0.5 = 1, so it is a valid density. Density values are not probabilities: P(X = x) = 0 for a continuous variable, and probabilities come from areas.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
X has density f_X, and Y = 3X + 2. What is the density of Y?
C. f_Y(y) = (1/3) f_X((y − 2)/3)
Inverse x = (y − 2)/3, |dx/dy| = 1/3, so f_Y(y) = f_X((y − 2)/3)/3. Multiplying by 3 instead of dividing gives a function that integrates to 9, not 1.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Yield loss X per plot is non-negative with mean 5 kg. Nothing else is known. What is the best upper bound on P(X ≥ 20) that holds for every such distribution?
B. 0.25
Markov's inequality: P(X ≥ 20) ≤ E(X)/20 = 5/20 = 0.25, and it is attained by a two-point distribution on 0 and 20. 0.0625 would need a variance (Chebyshev); 0.75 is the complementary lower bound P(X < 20) ≥ 0.75.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Plant height has mean 50 cm and standard deviation 4 cm; the distribution is unknown. What does Chebyshev's inequality give as an upper bound for P(|X − 50| ≥ 10)?
D. 0.16
P(|X − μ| ≥ c) ≤ σ²/c² = 16/100 = 0.16. 0.40 uses σ/c without squaring; 0.84 is the lower bound for the complement; 0.0124 is the exact value only if X were normal.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
A germination rate p is to be estimated by the sample proportion, within ±0.05 with probability at least 0.95. Using Chebyshev's inequality and the bound p(1 − p) ≤ 1/4, how many seeds are needed?
A. 2000
P(|p̂ − p| ≥ ε) ≤ p(1 − p)/(nε²) ≤ 1/(4nε²). Setting 1/(4n × 0.0025) ≤ 0.05 gives n ≥ 1/(4 × 0.0025 × 0.05) = 2000. The normal approximation gives 1.96² × 0.25/0.0025 ≈ 385; 8000 forgets the bound p(1 − p) ≤ 1/4 and uses p(1 − p) = 1.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Chebyshev's form of the weak law of large numbers requires the variables X₁, X₂, ... to be:
B. Uncorrelated, with variances bounded by a common constant
Var(X̄ₙ) = (1/n²)ΣVar(Xᵢ) ≤ C/n → 0 when the Xᵢ are uncorrelated with bounded variances, and Chebyshev's inequality finishes the proof. Identical distributions are not needed; dependence of any form can destroy the result.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Plot yields have mean 40 kg and standard deviation 8 kg. For the mean of 64 independent plots, what does the central limit theorem give for P(X̄ > 42)?
A. 0.0228
SE = 8/√64 = 1, so z = (42 − 40)/1 = 2 and P(Z > 2) = 0.0228. Using σ = 8 instead of the standard error gives z = 0.25 and 0.4013; 0.0455 is a two-sided probability.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
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