GATE DA 2025 - W3Colleges

Question 31

NAT · 1 mark

There are three boxes containing white balls and black balls.

Box-1 contains 2 black and 1 white balls.
Box-2 contains 1 black and 2 white balls.
Box-3 contains 3 black and 3 white balls.

In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is \(\frac{1}{2}\), Box-2 is \(\frac{1}{6}\), and Box-3 is \(\frac{1}{3}\). A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is ________(Round off to two decimal places)

Your answer

Explanation

Answer: 0.25

Let \(W\) be the event that the ball is white. The chance of white from each box is \(\frac{1}{3}\), \(\frac{2}{3}\) and \(\frac{3}{6} = \frac{1}{2}\). By the law of total probability,

\[P(W) = \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{6}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{3}{18} + \frac{2}{18} + \frac{3}{18} = \frac{4}{9}.\]

By Bayes' theorem,

\[P(\text{Box-2} \mid W) = \frac{\frac{1}{6}\cdot\frac{2}{3}}{\frac{4}{9}} = \frac{1/9}{4/9} = \frac{1}{4} = 0.25.\]

Box-2 has the highest share of white balls, but it is the least likely box to be chosen, which keeps its posterior low.

Answer: 0.25

Question 32

NAT · 1 mark

\(\lim\limits_{t \to +\infty} \sqrt{t^{2} + t} - t =\) ________
(Round off to one decimal place)

Your answer

Explanation

Answer: 0.5

Substituting directly gives the indeterminate form \(\infty - \infty\), so multiply and divide by the conjugate, then divide the numerator and denominator by \(t > 0\):

\[\sqrt{t^{2} + t} - t = \frac{(t^{2} + t) - t^{2}}{\sqrt{t^{2} + t} + t} = \frac{t}{\sqrt{t^{2} + t} + t} = \frac{1}{\sqrt{1 + \frac{1}{t}} + 1}.\]

As \(t \to +\infty\), \(\frac{1}{t} \to 0\), so the limit is \(\frac{1}{1 + 1} = 0.5\). As a check, the expression is about 0.4988 at \(t = 100\).

Answer: 0.5

Question 33

NAT · 1 mark

On a relation named Loan of a bank:

Loan
loan_number	branch_name	amount
L11	Banjara Hills	90000
L14	Kondapur	50000
L15	SR Nagar	40000
L22	SR Nagar	25000
L23	Balanagar	80000
L25	Kondapur	70000
L19	SR Nagar	65000

the following SQL query is executed.

SELECT L1.loan_number
FROM Loan L1
WHERE L1.amount > (SELECT MAX (L2.amount)
                   FROM Loan L2
                   WHERE L2.branch_name = 'SR Nagar');

The number of rows returned by the query is __________ (Answer in integer)

Your answer

Explanation

Answer: 3

The inner query does not use L1, so it produces a single value: the largest SR Nagar amount, \(\max(40000, 25000, 65000) = 65000\).

The outer query keeps the loans whose amount is strictly greater than 65000: L11 (90000), L23 (80000) and L25 (70000). L19 is left out because 65000 is not greater than itself, and every other loan is smaller. So the query returns 3 rows.

Answer: 3

Question 34

NAT · 1 mark

Given data \(\{(-1, 1), (2, -5), (3, 5)\}\) of the form \((x, y)\), we fit a model \(y = wx\) using linear least-squares regression. The optimal value of \(w\) is __________
(Round off to three decimal places)

Your answer

Explanation

Answer: 0.29

The model has no intercept, so least squares picks \(w\) to minimise \(L(w) = \sum_i (y_i - w x_i)^{2}\). Setting \(L'(w) = -2\sum_i x_i (y_i - w x_i) = 0\) gives

\[w = \frac{\sum_i x_i y_i}{\sum_i x_i^{2}}.\]

Here \(\sum x_i y_i = (-1)(1) + (2)(-5) + (3)(5) = 4\) and \(\sum x_i^{2} = 1 + 4 + 9 = 14\), so \(w = \frac{4}{14} = \frac{2}{7} \approx 0.286\).

Fitting a line with an intercept would give a different slope (about 0.308), but this model is forced through the origin.

Answer: 0.286

Question 35

NAT · 1 mark

The naive Bayes classifier is used to solve a two-class classification problem with class-labels \(y_1, y_2\). Suppose the prior probabilities are \(P(y_1) = \frac{1}{3}\) and \(P(y_2) = \frac{2}{3}\). Assuming a discrete feature space with

\[P(x|y_1) = \frac{3}{4} \quad \text{and} \quad P(x|y_2) = \frac{1}{4}\]

for a specific feature vector \(x\). The probability of misclassifying \(x\) is __________
(Round off to two decimal places)

Your answer

Explanation

Answer: 0.39 to 0.41

Naive Bayes compares prior times likelihood for the two classes:

\(P(y_1)P(x|y_1) = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\)
\(P(y_2)P(x|y_2) = \frac{2}{3} \cdot \frac{1}{4} = \frac{1}{6}\)

So \(P(x) = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}\), giving posteriors \(P(y_1|x) = \frac{3}{5}\) and \(P(y_2|x) = \frac{2}{5}\). The classifier picks the larger posterior, \(y_1\), and it is wrong exactly when \(x\) really belongs to \(y_2\), which has probability \(P(y_2|x) = 0.40\). The larger prior of \(y_2\) is outweighed by the much higher likelihood under \(y_1\).

Answer: 0.40

Question 36

MCQ · 2 marks · −0.66

Let \(Y = Z^{2}\), \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\). The variance of \(Y\) is

1 2 3 4

Explanation

Answer: (B)

Standardising a normal variable gives \(Z \sim N(0, 1)\), so \(E[Z^{2}] = \operatorname{Var}(Z) = 1\). The fourth moment of a standard normal is \(E[Z^{4}] = 3\): its moment generating function \(e^{t^{2}/2}\) has \(t^{4}\) coefficient \(\frac{1}{8}\), and \(4! \cdot \frac{1}{8} = 3\). Therefore

\[\operatorname{Var}(Y) = E[Z^{4}] - \left(E[Z^{2}]\right)^{2} = 3 - 1 = 2.\]

Equivalently, \(Y = Z^{2}\) is chi-squared with 1 degree of freedom, whose variance is twice the degrees of freedom. Option C is \(E[Z^{4}]\) alone, not the variance.

Answer: B

Question 37

MCQ · 2 marks · −0.66

Let \(A \in \mathbb{R}^{n \times n}\) be such that \(A^{3} = A\). Which one of the following statements is ALWAYS correct?

\(A\) is invertible Determinant of \(A\) is \(0\) The sum of the diagonal elements of \(A\) is \(1\) \(A\) and \(A^{2}\) have the same rank

Explanation

Answer: (D)

The zero matrix and the identity both satisfy \(A^{3} = A\), and they rule out the first three options: \(A = 0\) is not invertible (A), \(A = I\) has determinant 1 (B), and \(A = 0\) has trace 0 (C).

For (D), multiplying by a matrix never increases rank, so \(\operatorname{rank}(A^{2}) \le \operatorname{rank}(A)\). Also \(A = A^{3} = A \cdot A^{2}\), so \(\operatorname{rank}(A) \le \operatorname{rank}(A^{2})\). The two ranks are therefore always equal.

Another view: \(A\) is diagonalisable with eigenvalues in \(\{-1, 0, 1\}\), and squaring keeps exactly the same eigenvalues non-zero.

Answer: D

Question 38

MCQ · 2 marks · −0.66

Let \(\{x_1, x_2, \ldots, x_n\}\) be a set of linearly independent vectors in \(\mathbb{R}^{n}\). Let the \((i, j)\)-th element of matrix \(A \in \mathbb{R}^{n \times n}\) be given by \(A_{ij} = x_i^{\top} x_j\), \(1 \le i, j \le n\). Which one of the following statements is correct?

\(A\) is invertible \(0\) is a singular value of \(A\) Determinant of \(A\) is \(0\) \(z^{\top} A z = 0\) for some non-zero \(z \in \mathbb{R}^{n}\)

Explanation

Answer: (A)

Put the vectors in the columns of \(X = [x_1\ x_2\ \cdots\ x_n]\). Then \(A_{ij} = x_i^{\top} x_j\) says \(A = X^{\top} X\), a Gram matrix.

For any non-zero \(z\), \(z^{\top} A z = (Xz)^{\top}(Xz) = \|Xz\|^{2}\). The columns of \(X\) are linearly independent, so \(Xz \ne 0\) and \(z^{\top} A z \gt 0\). Hence \(A\) is positive definite, which rules out (D).

A positive definite matrix has only positive eigenvalues, so \(\det A = (\det X)^{2} \gt 0\) and \(A\) is invertible. Being symmetric, its singular values are these same positive eigenvalues, so 0 is not one of them. Options B and C fail.

Answer: A

Question 39

MCQ · 2 marks · −0.66

Consider the cumulative distribution function (CDF) of a random variable \(X\):

\[F_X(x) = \begin{cases} 0 & x \le -1 \\ \frac{1}{4}(x+1)^{2} & -1 \le x \le 1 \\ 1 & x \ge 1 \end{cases}\]

The value of \(P(X^{2} \le 0.25)\) is

0.625 0.25 0.5 0.5625

Explanation

Answer: (C)

\(X^{2} \le 0.25\) means \(-0.5 \le X \le 0.5\). The CDF has no jumps (it rises continuously from 0 at \(x = -1\) to 1 at \(x = 1\)), so

\[P(-0.5 \le X \le 0.5) = F_X(0.5) - F_X(-0.5) = \frac{(1.5)^{2} - (0.5)^{2}}{4} = \frac{2}{4} = 0.5.\]

Two slips lead to the distractors: \(F_X(0.5) = 0.5625\) alone forgets the lower limit (D), and reading the event as \(|X| \le 0.25\) gives \(F_X(0.25) - F_X(-0.25) = 0.25\) (B).

Answer: C

Question 40

MCQ · 2 marks · −0.66

A random variable \(X\) is said to be distributed as \(Bernoulli(\theta)\), denoted by \(X \sim Bernoulli(\theta)\), if

\[P(X = 1) = \theta, \quad P(X = 0) = 1 - \theta\]

for \(0 \lt \theta \lt 1\). Let \(Y = \sum_{i=1}^{300} X_i\), where \(X_i \sim Bernoulli(\theta)\), \(i = 1, 2, \ldots, 300\) be independent and identically distributed random variables with \(\theta = 0.25\). The value of \(P(60 \le Y \le 90)\), after approximation through Central Limit Theorem, is given by
(Recall that \(\phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-\frac{t^{2}}{2}}\, dt\))

\(\phi(2) - \phi(-2)\) \(\phi(1) - \phi(-1)\) \(\phi(3) - \phi(-3)\) \(\phi(90) - \phi(60)\)

Explanation

Answer: (A)

\(Y\) is a sum of 300 independent Bernoulli variables with \(\theta = 0.25\), so

\(E[Y] = 300 \times 0.25 = 75\);
\(\operatorname{Var}(Y) = 300 \times 0.25 \times 0.75 = 56.25\), a standard deviation of \(7.5\).

By the Central Limit Theorem, \(\frac{Y - 75}{7.5}\) is approximately standard normal. The limits standardise to \(\frac{60 - 75}{7.5} = -2\) and \(\frac{90 - 75}{7.5} = 2\), so

\[P(60 \le Y \le 90) \approx \phi(2) - \phi(-2).\]

Option D puts the raw values of \(Y\) into \(\phi\) without standardising.

Answer: A

Data Science and Artificial Intelligence