RBI Grade B (DSIM) - Department of Statistics and Information Management · India · Reserve Bank of India Services Board
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)
How our questions for RBI Grade B spread across the units of Probability Theory. Difficulty: 7 easy, 9 medium, 3 hard.
| Sample spaces, sigma-fields and the axioms of probability | 3 | |
| Conditional probability, independence and Bayes' theorem | 3 | |
| Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz | 3 | |
| Joint, marginal and conditional distributions; conditional expectation | 2 | |
| Functions of random variables and transformations (Jacobian method) | 2 | |
| Weak and strong laws of large numbers | 2 | |
| Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method | 2 | |
| Expectation, moments and moment-generating, probability-generating and characteristic functions | 1 | |
| Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Which of these is NOT one of Kolmogorov's axioms but a theorem derived from them?
D. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) for any events
The axioms are non-negativity, P(Ω) = 1 and countable additivity for disjoint events. The general addition rule for overlapping events is derived by splitting A ∪ B into the disjoint pieces A and B − A.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
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
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
√n(Tₙ − θ) converges in distribution to N(0, σ²), and g′(θ) = 0 while g″(θ) ≠ 0. Which statement is correct?
A. √n(g(Tₙ) − g(θ)) → 0 in probability, and n(g(Tₙ) − g(θ)) has a scaled chi-square(1) limit
First-order delta method gives variance g′(θ)²σ² = 0, a degenerate limit. The second-order expansion g(Tₙ) − g(θ) ≈ g″(θ)(Tₙ − θ)²/2 gives n(g(Tₙ) − g(θ)) → (g″(θ)σ²/2)χ²₁.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
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