Probability Theory & Statistics
Level: 1 — Recognition (MCQ + Matching + True/False with justification) Time limit: 20 minutes Total marks: 30
Section A — Multiple Choice (1 mark each)
Choose the single best answer.
Q1. Which is NOT one of Kolmogorov's axioms of probability? (a) (b) (c) Countable additivity for disjoint events (d) for all events
Q2. For events , the inclusion–exclusion principle gives (a) (b) (c) (d)
Q3. If , then and are (a) (b) (c) (d)
Q4. The MGF of a random variable is defined as (a) (b) (c) (d)
Q5. For , equals (a) (b) (c) (d)
Q6. The PDF of (rate form), for , is (a) (b) (c) (d)
Q7. A -distribution with degrees of freedom arises as (a) sum of squared normals (b) with independent (c) ratio of two chi-squares (d) product of normals
Q8. Two random variables are independent if and only if (a) (b) (c) for all (d) they have equal variance
Q9. The correlation coefficient always satisfies (a) (b) (c) (d) unbounded
Q10. The Central Limit Theorem states that for i.i.d. with mean , variance , the quantity converges in distribution to (a) (b) (c) (d)
Q11. An estimator is unbiased if (a) (b) (c) a.s. (d) is the MLE
Q12. In Bayesian statistics, posterior (a) prior likelihood (b) prior likelihood (c) likelihood only (d) prior only
Section B — Matching (1 mark each)
Q13–Q17. Match each distribution/quantity (left) to its correct property (right). Write pairs like 13→(x).
| # | Item |
|---|---|
| 13 | for (support ) |
| 14 | Mean of |
| 15 | MGF of |
| 16 | Degrees of freedom of chi-squared sum of standard normals squared |
| 17 | Method of moments principle |
| Option | Property |
|---|---|
| (p) | |
| (q) | |
| (r) | |
| (s) | |
| (t) | equate sample moments to population moments |
Section C — True/False WITH justification (2 marks each: 1 mark T/F, 1 mark justification)
Q18. "If then and are independent." True or False? Justify.
Q19. "For any random variable, ." True or False? Justify.
Q20. "A 95% confidence interval means the true parameter lies in the computed interval with probability 0.95." True or False? Justify.
Q21. "The p-value is the probability that the null hypothesis is true." True or False? Justify.
Q22. "By the Weak Law of Large Numbers, the sample mean converges in probability to the population mean." True or False? Justify.
Answer keyMark scheme & solutions
Section A (1 mark each)
Q1. (d). Additivity only holds for disjoint (mutually exclusive) events; the stated equality without is false in general. 1 mark for (d).
Q2. (b). Inclusion–exclusion subtracts the double-counted intersection. 1 mark.
Q3. (b). Binomial mean , variance . 1 mark.
Q4. (b). MGF . (Option (d) is the cumulant generating function.) 1 mark.
Q5. (c). Poisson mean = variance = . 1 mark.
Q6. (a). Rate-form exponential density . 1 mark.
Q7. (b). with independence. 1 mark.
Q8. (c). Full independence requires factorization of the joint density/PMF for all points; (a),(b) are necessary not sufficient. 1 mark.
Q9. (b). By Cauchy–Schwarz . 1 mark.
Q10. (b). Standardized sample mean → . 1 mark.
Q11. (b). Unbiasedness means expected value equals the parameter. 1 mark.
Q12. (a). Bayes: posterior prior likelihood. 1 mark.
Section B (1 mark each)
Q13 → (r): Geometric variance . Q14 → (p): Uniform mean . Q15 → (q): Normal MGF . Q16 → (s): Sum of squared standard normals is , df . Q17 → (t): Method of moments equates sample and population moments.
Section C (2 marks each: 1 T/F + 1 justification)
Q18. False. Zero covariance does not imply independence in general (only the converse holds). Counterexample: , has but they are dependent. 1 (False) + 1 (justification).
Q19. True. Adding constant shifts but doesn't change spread; scaling by multiplies variance by . 1 + 1.
Q20. False. The parameter is fixed; randomness is in the interval. The 95% refers to the long-run coverage frequency of the procedure, not the probability for one computed interval. 1 + 1.
Q21. False. The p-value is , not . 1 + 1.
Q22. True. WLLN: for i.i.d. with finite mean. 1 + 1.
[
{"claim":"Binomial(n,p) variance = np(1-p), check n=10,p=0.3 gives 2.1","code":"n,p=10,Rational(3,10); result = (n*p*(1-p))==Rational(21,10)"},
{"claim":"Uniform(a,b) mean = (a+b)/2 for a=2,b=8 is 5","code":"a,b=2,8; result = Rational(a+b,2)==5"},
{"claim":"Geometric(p) variance (1-p)/p^2 for p=1/4 equals 12","code":"p=Rational(1,4); result = ((1-p)/p**2)==12"},
{"claim":"Normal MGF exp(mu t + sigma^2 t^2/2) has second derivative at 0 = mu^2+sigma^2","code":"t,mu,s=symbols('t mu s'); M=exp(mu*t+s**2*t**2/2); result = simplify(diff(M,t,2).subs(t,0)-(mu**2+s**2))==0"}
]