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Indian Statistical Service Examination Probability Theory: practice questions

Indian Statistical Service Examination (with Indian Economic Service) · India · Union Public Service Commission

Original practice questions written to the published syllabus · not past papers
8worked here, free
27written to this exam in this subject
10syllabus units

What this subject covers

Probability 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)

Where the 27 questions fall

How our questions for Indian Statistical Service Examination spread across the units of Probability Theory. Difficulty: 7 easy, 14 medium, 6 hard.

Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz4
Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method4
Sample spaces, sigma-fields and the axioms of probability3
Conditional probability, independence and Bayes' theorem3
Joint, marginal and conditional distributions; conditional expectation3
Weak and strong laws of large numbers3
Expectation, moments and moment-generating, probability-generating and characteristic functions2
Functions of random variables and transformations (Jacobian method)2
Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma2
Elements of measure theory: measurable functions and Lebesgue integral1

8 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Sample spaces, sigma-fields and the axioms of probability · easy

Which of these is NOT one of Kolmogorov's axioms but a theorem derived from them?

  1. P(∪Aᵢ) = ΣP(Aᵢ) for pairwise disjoint events
  2. P(Ω) = 1 for the whole sample space
  3. P(A) ≥ 0 for every event A in the σ-field
  4. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) for any events
Answer and working

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

2 · Sample spaces, sigma-fields and the axioms of probability · medium

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?

  1. 0.7
  2. 0.8
  3. 0.784
  4. 0.75
Answer and working

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

3 · Conditional probability, independence and Bayes' theorem · medium

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?

  1. 0.162
  2. 0.950
  3. 0.100
  4. 0.019
Answer and working

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

4 · Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz · easy

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?

  1. 0.75
  2. 0.25
  3. 0.05
  4. 0.0625
Answer and working

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

5 · Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz · medium

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)?

  1. 0.40
  2. 0.84
  3. 0.0124
  4. 0.16
Answer and working

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

6 · Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz · hard

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?

  1. 2000
  2. 385
  3. 100
  4. 8000
Answer and working

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

7 · Weak and strong laws of large numbers · medium

Chebyshev's form of the weak law of large numbers requires the variables X₁, X₂, ... to be:

  1. Identically distributed, with any form of dependence
  2. Uncorrelated, with variances bounded by a common constant
  3. Normally distributed with a common variance
  4. Independent and identically distributed with finite mean
Answer and working

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

8 · Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method · medium

√n(Tₙ − θ) converges in distribution to N(0, σ²), and g′(θ) = 0 while g″(θ) ≠ 0. Which statement is correct?

  1. √n(g(Tₙ) − g(θ)) → 0 in probability, and n(g(Tₙ) − g(θ)) has a scaled chi-square(1) limit
  2. √n(g(Tₙ) − g(θ)) → N(0, σ²), the same limit as for Tₙ itself
  3. No limit exists, and the delta method cannot be adapted to this case
  4. √n(g(Tₙ) − g(θ)) → N(0, g″(θ)²σ²), with g″ replacing g′
Answer and working

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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