Probability & Statistics

Subject: AI-ML | Chapter: Probability & Statistics Time Limit: 20 minutes | Total Marks: 30

Instructions: Answer all questions. For True/False items, a correct justification is required for full marks. Use $...$ for any math you write.

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Section A — Multiple Choice (1 mark each) [10 marks]

Q1. The three Kolmogorov axioms require that for any event $A$, $P(A)\geq 0$; that $P(\Omega)=1$; and that for mutually exclusive events, probability is: (a) multiplicative (b) additive (c) subtractive (d) constant

Q2. Conditional probability $P(A\mid B)$ is defined (for $P(B)>0$) as: (a) $P(A)P(B)$ (b) $\dfrac{P(A\cap B)}{P(B)}$ (c) $\dfrac{P(B)}{P(A\cap B)}$ (d) $P(A)+P(B)$

Q3. Bayes' theorem expresses $P(H\mid E)$ as: (a) $\dfrac{P(E\mid H)P(H)}{P(E)}$ (b) $\dfrac{P(H)P(E)}{P(E\mid H)}$ (c) $P(E\mid H)+P(H)$ (d) $\dfrac{P(E)}{P(H)}$

Q4. Two events $A$ and $B$ are independent if and only if: (a) $P(A\cap B)=0$ (b) $P(A\cup B)=P(A)+P(B)$ (c) $P(A\cap B)=P(A)P(B)$ (d) $A=B$

Q5. For a continuous random variable, the probability density function $f(x)$ satisfies: (a) $f(x)$ is a probability (b) $\sum f(x)=1$ (c) $\int_{-\infty}^{\infty} f(x)\,dx = 1$ (d) $f(x)\leq 1$ always

Q6. The variance of a random variable $X$ equals: (a) $E[X]^2$ (b) $E[X^2]-(E[X])^2$ (c) $E[X^2]+E[X]$ (d) $(E[X])^2-E[X^2]$

Q7. For a Binomial$(n,p)$ distribution, the mean is: (a) $np$ (b) $np(1-p)$ (c) $p$ (d) $n/p$

Q8. The Poisson distribution with parameter $\lambda$ has mean and variance: (a) mean $\lambda$, variance $\lambda^2$ (b) mean $\lambda$, variance $\lambda$ (c) mean $\lambda^2$, variance $\lambda$ (d) both equal to $1$

Q9. The Central Limit Theorem states that the distribution of the sample mean of i.i.d. variables approaches: (a) uniform (b) exponential (c) Gaussian/normal (d) Poisson

Q10. In hypothesis testing, the p-value is: (a) the probability the null hypothesis is true (b) the probability of observing data at least as extreme as observed, assuming $H_0$ is true (c) the significance level $\alpha$ (d) the power of the test

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Section B — Matching (1 mark each) [6 marks]

Q11. Match each distribution/quantity (i–vi) to its property (A–F).

#	Term		Property
i	Bernoulli$(p)$	A	Memoryless continuous distribution
ii	Exponential$(\lambda)$	B	Mean $= \tfrac{a+b}{2}$
iii	Uniform$(a,b)$	C	Single trial, values $\{0,1\}$
iv	KL divergence	D	Always symmetric, bell-shaped
v	Normal	E	$\sum p(x)\log\frac{p(x)}{q(x)}$, $\geq 0$
vi	Correlation	F	Normalized covariance, in $[-1,1]$

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Section C — True/False with Justification (2 marks each: 1 verdict + 1 justification) [14 marks]

Q12. Mutually exclusive events with nonzero probability are always independent. (T/F + justify)

Q13. The CDF $F(x)$ of any random variable is non-decreasing and satisfies $F(-\infty)=0$, $F(+\infty)=1$. (T/F + justify)

Q14. For independent random variables $X$ and $Y$, the covariance $\mathrm{Cov}(X,Y)=0$. (T/F + justify)

Q15. The Law of Large Numbers guarantees the sample mean equals the true mean for any finite sample size. (T/F + justify)

Q16. MAP estimation reduces to MLE when the prior is uniform (flat). (T/F + justify)

Q17. Cross-entropy between distributions $p$ and $q$ equals the entropy of $p$ plus the KL divergence $D_{KL}(p\|q)$. (T/F + justify)

Q18. A 95% confidence interval means there is a 95% probability that the true parameter lies in that specific computed interval. (T/F + justify)

[
  {"claim":"Binomial(n,p) mean equals n*p (Q7)","code":"n,p=symbols('n p',positive=True); mean=n*p; result = simplify(mean - n*p)==0"},
  {"claim":"Variance identity Var=E[X^2]-(E[X])^2 for X uniform on {1,2,3} (Q6)","code":"vals=[1,2,3]; EX=Rational(sum(vals),3); EX2=Rational(sum(v*v for v in vals),3); var=EX2-EX**2; direct=Rational(sum((v-EX)**2 for v in vals),3); result = simplify(var-direct)==0"},
  {"claim":"Mutually exclusive nonzero events cannot be independent: P(A)P(B)!=0 while P(A cap B)=0 (Q12)","code":"PA=Rational(1,2); PB=Rational(1,3); inter=0; result = (PA*PB != inter)"},
  {"claim":"Cross-entropy = H(p)+KL(p||q) for sample p,q (Q17)","code":"p=[Rational(1,2),Rational(1,2)]; q=[Rational(1,4),Rational(3,4)]; H=-sum(pi*log(pi) for pi in p); KL=sum(pi*log(pi/qi) for pi,qi in zip(p,q)); CE=-sum(pi*log(qi) for pi,qi in zip(p,q)); result = simplify(CE-(H+KL))==0"}
]