Level 1 — RecognitionProbability Theory & Statistics

Probability Theory & Statistics

20 minutes30 marksprintable — key stays hidden on paper

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) P(Ω)=1P(\Omega)=1 (b) P(A)0P(A)\ge 0 (c) Countable additivity for disjoint events (d) P(AB)=P(A)+P(B)P(A\cup B)=P(A)+P(B) for all events

Q2. For events A,BA,B, the inclusion–exclusion principle gives P(AB)=P(A\cup B)= (a) P(A)+P(B)P(A)+P(B) (b) P(A)+P(B)P(AB)P(A)+P(B)-P(A\cap B) (c) P(A)P(B)P(A)P(B) (d) P(A)+P(B)+P(AB)P(A)+P(B)+P(A\cap B)

Q3. If XBinomial(n,p)X\sim\text{Binomial}(n,p), then E[X]E[X] and Var(X)\text{Var}(X) are (a) np, npnp,\ np (b) np, np(1p)np,\ np(1-p) (c) p, p(1p)p,\ p(1-p) (d) np(1p), npnp(1-p),\ np

Q4. The MGF of a random variable XX is defined as (a) E[Xt]E[X^t] (b) E[etX]E[e^{tX}] (c) E[tX]E[t^X] (d) lnE[etX]\ln E[e^{tX}]

Q5. For XPoisson(λ)X\sim\text{Poisson}(\lambda), E[X]E[X] equals (a) λ2\lambda^2 (b) λ\sqrt{\lambda} (c) λ\lambda (d) 1/λ1/\lambda

Q6. The PDF of XExponential(λ)X\sim\text{Exponential}(\lambda) (rate form), for x0x\ge 0, is (a) λeλx\lambda e^{-\lambda x} (b) ex/λe^{-x/\lambda} (c) 1λeλx\frac{1}{\lambda}e^{-\lambda x} (d) λxeλx\lambda x e^{-\lambda x}

Q7. A tt-distribution with kk degrees of freedom arises as (a) sum of kk squared normals (b) Z/χk2/kZ/\sqrt{\chi^2_k/k} with Z,χk2Z,\chi^2_k independent (c) ratio of two chi-squares (d) product of kk normals

Q8. Two random variables X,YX,Y are independent if and only if (a) Cov(X,Y)=0\text{Cov}(X,Y)=0 (b) E[XY]=E[X]E[Y]E[XY]=E[X]E[Y] (c) fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x,y)=f_X(x)f_Y(y) for all x,yx,y (d) they have equal variance

Q9. The correlation coefficient ρ\rho always satisfies (a) 0ρ10\le\rho\le1 (b) 1ρ1-1\le\rho\le1 (c) ρ0\rho\ge0 (d) ρ\rho unbounded

Q10. The Central Limit Theorem states that for i.i.d. XiX_i with mean μ\mu, variance σ2\sigma^2, the quantity Xˉnμσ/n\frac{\bar X_n-\mu}{\sigma/\sqrt n} converges in distribution to (a) Exponential(1)\text{Exponential}(1) (b) N(0,1)N(0,1) (c) χ12\chi^2_1 (d) Uniform(0,1)\text{Uniform}(0,1)

Q11. An estimator θ^\hat\theta is unbiased if (a) Var(θ^)0\text{Var}(\hat\theta)\to0 (b) E[θ^]=θE[\hat\theta]=\theta (c) θ^θ\hat\theta\to\theta a.s. (d) θ^\hat\theta is the MLE

Q12. In Bayesian statistics, posterior \propto (a) prior ×\times 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 Var(X)\text{Var}(X) for XGeometric(p)X\sim\text{Geometric}(p) (support 1,2,1,2,\dots)
14 Mean of Uniform(a,b)\text{Uniform}(a,b)
15 MGF of N(μ,σ2)N(\mu,\sigma^2)
16 Degrees of freedom of chi-squared sum of nn standard normals squared
17 Method of moments principle
Option Property
(p) a+b2\frac{a+b}{2}
(q) exp(μt+12σ2t2)\exp(\mu t+\tfrac12\sigma^2 t^2)
(r) 1pp2\frac{1-p}{p^2}
(s) nn
(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 Cov(X,Y)=0\text{Cov}(X,Y)=0 then XX and YY are independent." True or False? Justify.

Q19. "For any random variable, Var(aX+b)=a2Var(X)\text{Var}(aX+b)=a^2\text{Var}(X)." 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 P(AB)-P(A\cap B) is false in general. 1 mark for (d).

Q2. (b). Inclusion–exclusion subtracts the double-counted intersection. 1 mark.

Q3. (b). Binomial mean npnp, variance np(1p)np(1-p). 1 mark.

Q4. (b). MGF MX(t)=E[etX]M_X(t)=E[e^{tX}]. (Option (d) is the cumulant generating function.) 1 mark.

Q5. (c). Poisson mean = variance = λ\lambda. 1 mark.

Q6. (a). Rate-form exponential density λeλx\lambda e^{-\lambda x}. 1 mark.

Q7. (b). tk=Z/χk2/kt_k = Z/\sqrt{\chi^2_k/k} 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 ρ1|\rho|\le1. 1 mark.

Q10. (b). Standardized sample mean → N(0,1)N(0,1). 1 mark.

Q11. (b). Unbiasedness means expected value equals the parameter. 1 mark.

Q12. (a). Bayes: posterior \propto prior ×\times likelihood. 1 mark.

Section B (1 mark each)

Q13 → (r): Geometric variance 1pp2\frac{1-p}{p^2}. Q14 → (p): Uniform mean a+b2\frac{a+b}{2}. Q15 → (q): Normal MGF exp(μt+12σ2t2)\exp(\mu t+\tfrac12\sigma^2t^2). Q16 → (s): Sum of nn squared standard normals is χn2\chi^2_n, df =n=n. 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: XN(0,1)X\sim N(0,1), Y=X2Y=X^2 has Cov=0\text{Cov}=0 but they are dependent. 1 (False) + 1 (justification).

Q19. True. Adding constant bb shifts but doesn't change spread; scaling by aa multiplies variance by a2a^2. 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 P(test statistic as or more extremeH0 true)P(\text{test statistic as or more extreme}\mid H_0\text{ true}), not P(H0 true)P(H_0\text{ true}). 1 + 1.

Q22. True. WLLN: XˉnPμ\bar X_n \xrightarrow{P}\mu 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"}
]