GATE DA 2025 — Solved Walkthrough

Probability & Statistics

Bayes with three unequal-prior boxes. Three boxes hold white and black balls with priors (1/2, 1/6, 1/3). A white ball is drawn. Find P(Box 2 | white).

Apply Bayes with total probability in the denominator — three terms, one per box:

P(Box 2 | white) =            P(white | Box 2)·P(Box 2)
                    ───────────────────────────────────────────────
                    Σ over all three boxes of P(white | Boxᵢ)·P(Boxᵢ)

Plugging the box compositions and the priors (1/2, 1/6, 1/3) and simplifying gives 0.25. The recipe never changes: write priors and likelihoods, build the evidence denominator by total probability, divide.

Answer: P(Box 2 | white) = 0.25.

→ Taught in Bayes’ Theorem

Law of total expectation. A joint setup asks for E[ E[X | Y] ]. Evaluate it.

This is a one-line recognition problem. Averaging the conditional expectation E[X | Y] over Y recovers the plain expectation of X:

E[ E[X | Y] ] = E[X]      (law of total expectation)

Spot the nested expectation and you skip all the table algebra — the answer is just E[X], no computation needed.

Answer: E[ E[X | Y] ] = E[X].

→ Taught in Joint, Marginal & Conditional Distributions

CLT on a sum of 300 Bernoulli(0.25). Let Y be the sum of 300 independent Bernoulli(0.25) variables. Using the normal approximation, find P(60 ≤ Y ≤ 90).

First get the mean and variance of the sum, then standardize both endpoints:

mean     = 300 · 0.25        = 75
variance = 300 · 0.25 · 0.75 = 56.25,   std = 7.5

z_lower = (60 − 75)/7.5 = −2,   z_upper = (90 − 75)/7.5 = +2

P(60 ≤ Y ≤ 90) = Φ(2) − Φ(−2) = 0.9772 − 0.0228 = 0.9544

The trap to avoid: divide the variance by n and the standard deviation by √n — never the SD by n.

Answer: P(60 ≤ Y ≤ 90) ≈ 0.9544 (the expression Φ(2) − Φ(−2)).

→ Taught in Central Limit Theorem & Confidence Intervals

Linear Algebra

Rank of A versus A² when A³ = A. A real matrix satisfies A³ = A. Is it always true that A and A² have the same rank?

Work through the eigenvalues. From Av = λv, A³ = A forces λ³ = λ:

λ³ = λ  →  λ(λ² − 1) = 0  →  λ ∈ {0, 1, −1}

then squaring (eigenvalues of A²):  λ² ∈ {0, 1, 1}

A nonzero eigenvalue of A (±1) stays nonzero in A² (becomes 1), and a zero eigenvalue stays zero. The count of nonzero eigenvalues — hence the rank — is unchanged.

Answer: yes, A and A² always have the same rank.

→ Taught in Eigen-properties & Transforms

A norm-preserving matrix is orthogonal. A real n × n matrix A satisfies ‖Ax‖ = ‖x‖ for every x. Which properties must hold?

Preserving every length preserves every dot product, so AᵀA = I:

AᵀA = I       ⇒ A is orthogonal              (TRUE)
A⁻¹ = Aᵀ      ⇒ A is invertible, full rank   (TRUE)
det(A)² = det(AᵀA) = 1  ⇒  det(A) = ±1       (TRUE)
"eigenvalues are ±1"                          (FALSE)

The eigenvalue claim is the distractor: a 90° rotation [[0, −1], [1, 0]] is orthogonal and full rank, yet its eigenvalues are ±i — on the unit circle, but neither +1 nor −1.

Answer: orthogonal, full rank, det = ±1 (eigenvalues need not be ±1).

→ Taught in Orthogonality & Orthogonal Matrices

Programming & DSA

Hashing with linear probing. Table size m = 10, hash h(x) = 3x mod 10, collisions resolved by linear probing. Insert 1, 4, 5, 6, 14, 15 in order. Where do 14 and 15 land?

Compute each home slot; on a collision, step forward +1 (mod 10) until empty:

h(1)=3, h(4)=2, h(5)=5, h(6)=8    (all land in their home slots)

h(14) = 3·14 mod 10 = 2 → slot 2 (4), probe 3 (1), probe 4 empty  ⇒ slot 4
h(15) = 3·15 mod 10 = 5 → slot 5 (5), probe 6 empty               ⇒ slot 6

Answer: 14 → slot 4, 15 → slot 6.

→ Taught in Hash Tables & Linear Probing

append versus extend in Python. Start with A = [1, 2, 3] and B = [4, 5, 6]. Which single operation makes A equal to [1, 2, 3, 4, 5, 6]?

Check each candidate:

A.append(B) → [1, 2, 3, [4, 5, 6]]   (B added as one nested element, length 4)
A.extend(B) → [1, 2, 3, 4, 5, 6]      (each of 4,5,6 appended in turn — the answer)
A + B       → [1, 2, 3, 4, 5, 6]      but A itself is unchanged (no mutation)

Only extend mutates A into the flat six-element list; append nests and + builds a new list without touching A.

Answer: A.extend(B).

→ Taught in Lists, Tuples, Dicts, Sets & Gotchas

Databases

Cost of a non-dependency-preserving decomposition. A relation R is split so the decomposition is not dependency-preserving. To enforce the lost FDs, which relational-algebra operator must run more often?

When an FD X → Y has X and Y straddling two fragments, the DBMS cannot see both columns in a single table. To check that FD on each update it must reconstruct the original relation — that is, take the join of the fragments.

Answer: JOIN (⋈) runs more often.

→ Taught in Lossless-Join vs Dependency-Preservation

Machine Learning

Least-squares fit through the origin. Fit y = wx to the points (−1, 1), (2, −5), (3, 5) by least squares. Find w.

For a through-origin fit, w = Σ(x·y) / Σ(x²). Build both sums column by column:

Σ x·y = (−1)(1) + (2)(−5) + (3)(5) = −1 − 10 + 15 = 4
Σ x²  = 1 + 4 + 9                   = 14

w = 4 / 14 ≈ 0.286

The fit passes through no single point — it balances all three so the squared vertical gaps are collectively smallest.

Answer: w ≈ 0.286.

→ Taught in Simple Linear Regression

Naive-Bayes misclassification probability. Two classes with priors P(y1) = 1/3, P(y2) = 2/3; likelihoods P(x | y1) = 3/4, P(x | y2) = 1/4. Predict the class for x and find the probability the prediction is wrong.

Score each class with prior × likelihood (unnormalized posteriors):

y1 : (3/4)(1/3) = 1/4  ≈ 0.250   ← larger, so predict y1
y2 : (1/4)(2/3) = 1/6  ≈ 0.167

The prediction is wrong exactly when the truth is y2, so the misclassification probability is the normalized posterior P(y2 | x):

P(y2 | x) = (1/6) / (1/4 + 1/6) = (1/6) / (5/12) = 0.40

Answer: predict y1; misclassification probability = 0.40.

→ Taught in Naive Bayes