GATE 2024 Data Science and Artificial Intelligence Question Paper (Available)- Download Solution PDF with Answer Key

Solution: The sequence of words follows an increasing intensity, meaning each word represents a stronger or more intense form of the previous word. - sick → infirm → moribund implies a progression from a mild condition (sick) to a more severe one (moribund). - Similarly, for the word "silly," the word that follows in increasing intensity (more severe or extreme) would logically be "vain," as it represents a more extreme form of foolishness, and "daft" represents the most extreme form.
Thus, the correct option is (D) vain.

Solution: Step 1: We are given that there are 15 parts, and the objective is to assign colors such that no two adjacent parts (with shared boundaries) have the same color. This is a classical graph coloring problem where each part can be treated as a vertex, and each adjacent part (with shared boundaries) represents an edge.

Step 2: From the given figure, it appears to be a planar graph. The four color theorem for planar graphs states that the minimum number of colors required to color any planar graph is at most 4. However, since we need to color the regions such that no adjacent parts share the same color, the problem reduces to determining the chromatic number of this graph.

Step 3: By observing the structure of the graph and considering its adjacency, the minimum number of colors required to ensure no two adjacent parts share the same color is 3. This is the chromatic number of the given graph, meaning that three colors are sufficient and necessary.

Step 4: Therefore, the correct answer is that 3 colors are required.

Solution: Step 1: First, recall that a number is divisible by 3 if the sum of its digits is divisible by 3. We are tasked with forming 4-digit numbers from the digits {1,3,4,6,7} with no repetition of digits. The sum of the digits we are working with is:
1 + 3 + 4 + 6 + 7 = 21.

The remainder when 21 is divided by 3 is 21 mod 3 = 0, which means that the sum of the digits is divisible by 3.

Step 2: The total number of ways to choose 4 digits from the set {1,3, 4, 6, 7} is: ⁵C₄ = 5.

For each selection of 4 digits, there are 4! = 24 ways to arrange them.

Step 3: Since the sum of the digits is always divisible by 3, every combination of 4 digits will form a number divisible by 3. Thus, the total number of such 4-digit numbers is:
5 x 24 = 120.

Step 4: Therefore, the number of 4-digit numbers divisible by 3 is 48, as only half of the combinations will satisfy the condition (the sum of the digits divisible by 3).

Solution: Step 1: The series appears to consist of terms involving fractions with different denominators in the pattern ¹⁄₂, ¹⁄₃, ¹⁄₄, ¹⁄₈, ¹⁄₁₆, ¹⁄₂₇....

Step 2: We can observe that: - The first term is ¹⁄₂, - The second term is ¹⁄₃. - The third term is ¹⁄₄, - Then, powers of 2: ¹⁄₈, ¹⁄₁₆. - Then powers of 3: ¹⁄₂₇, and so on.

This series is a combination of different series, and by recognizing that the series sums to a known value for the pattern, we find that the sum converges to ⁷⁄₂.

Step 3: Thus, the sum of the infinite series is ⁷⁄₂.

Solution: Step 1: The total number of votes cast is given as 1,15,000, out of which 5,000 are invalid. Therefore, the number of valid votes is:
Total valid votes = 1,15,000 - 5,000 = 1,10,000.

Step 2: From the pie chart: - The share of votes for candidate B is 25%. - The share of votes for candidate C is 20%.

The number of valid votes received by candidates B and C is the sum of these two percentages:
Votes for B and C = 25% + 20% = 45%.

Step 3: The total valid votes for candidates B and C is 45:
Votes for B and C = 0.45 × 1,10,000 = 49,500.

Thus, the total number of valid votes received by candidates B and C is 49,500.

Solution: The passage mentions that some people developed the ability to digest dairy milk due to a mutation in a particular gene. This suggests that the digestion of dairy milk is not inherent in all human beings, and it resulted from a genetic mutation in certain individuals. Hence, option (D) is the correct inference.

Solution: We are dealing with a situation where each child has an independent probability of being a boy or a girl, with each having a probability of ¹⁄₂. The family has three children, and we want to know the probability of having exactly two girls and one boy.

This is a binomial probability problem where the number of trials n = 3, the number of successes (girls) k = 2, and the probability of success (having a girl) p = ¹⁄₂.

The probability mass function for a binomial distribution is given by: P(X = k) = ⁿC_k p^k(1-p)^n-k

Substituting the values: P(2 girls) = ³C₂ (¹⁄₂)²(¹⁄₂)^3-2 = 3 x ¹⁄₄ x ¹⁄₂ = ³⁄₈

Thus, the probability of having exactly two girls and one boy is ³⁄₈.

Solution: Let the annual rate of return for mutual fund A be r.

- For Person 1, the total amount after one year will be:
Total for Person 1 = 10000 × (1 + ^r⁄₁₀₀) + 20000 × (1 + ¹⁵⁄₁₀₀) + 20000 × (1 + ¹⁵⁄₁₀₀)
= 10000 × (1 + ^r⁄₁₀₀) + 20000 × 1.15 + 20000 × 1.15

- For Person 2, the total amount after one year will be:
Total for Person 2 = 20000 × (1 + ^r⁄₁₀₀) + 15000 × (1 + ¹⁵⁄₁₀₀) + 15000 × (1 + ¹⁵⁄₁₀₀)
= 20000 × 1.15 + 15000 × 1.15 + 15000 × 1.15

We are given that Person 1 gets Rs 500 more than Person 2.
Total for Person 1 – Total for Person 2 = 500

Substitute the expressions for the totals and simplify to find r.

Hence, the correct answer is (B) 10%

Solution: To solve this question, we need to determine which arrangement of the numbers on the die satisfies the given views. In a standard die, opposite faces sum to 7, so: 5 and 2 are opposite faces. - 4 and 3 are opposite faces. - 6 and 1 are opposite faces.

By checking the options, we find that option (A) satisfies the conditions:

Thus, the correct answer is (A).

Solution: In this question, we have two identical right circular cones. One cone is inverted over the other, sharing a common circular base. When a cutting plane passes through the vertices of the assembled cones, the outer boundary of the resulting cross-section will be shaped by the slant heights of the cones.

Since the cones are identical and the cutting plane intersects through the vertices of both cones, the resulting cross-section will form a rhombus. The slant heights of the cones create the boundary of the rhombus, as the cutting plane passes through both cone vertices symmetrically.

Thus, the shape of the outer boundary of the cross-section is a rhombus.

Solution: Statement (i) is true for a Poisson random variable, as for a Poisson distribution with parameter λ, both the mean and variance are equal to λ.

- Statement (ii) is also true for a standard normal random variable, where the mean is zero and the variance is one.

Therefore, the correct answer is (A) Both (i) and (ii) are true.

Solution: The possible outcomes of tossing three fair coins are:
{HHH, HHT, HTH, HTT,THH,THT,TTH,TTT}

- T (two or more heads) occurs for: {HHH, HHT, HTH,THH}. - S (two or more tails) occurs for: {HTT,THT,TTH,TTT}.

The intersection T∩S represents the event where there are exactly two heads and two tails. From the list of outcomes, the only outcome satisfying both events is {HTT,THT,TTH}, which occurs with probability ³⁄₈.

Thus, the probability of the event T ∩ S is:
P(T∩S) = ³⁄₈ = 0.375 -> 0

Solution: The eigenvalues of a matrix M are the solutions to the characteristic equation:
det(M – λI) = 0

For the matrix M = ^{2 3}⁄_{-1 -1}, the characteristic equation is:
det(^{2 - λ 3}⁄_{-1 -1-λ}) = 0
(2 – λ)(−1 – λ) – (−3) = 0
λ² − λ + 2 + 3 = 0
λ² + λ + 5 = 0

The discriminant of this quadratic equation is:
Δ = 1² − 4 × 1 × (−5) = 1 + 20 = 21

Since the discriminant is positive, the eigenvalues are real and distinct. Thus, the eigenvalues of M are real and non-negative.

Thus, the correct answer is (D) One eigenvalue of M is non-negative and real, and another eigenvalue of M is negative and real.

Solution: In depth-first search (DFS), edges are classified into the following types: - Tree edges: Edges that are part of the DFS tree. - Back edges: Edges that connect a vertex to one of its ancestors in the DFS tree. - Cross edges: Edges that connect vertices in different DFS trees. - Forward edges: Edges that connect a vertex to a descendant in the DFS tree.

Since (u, v) is explored first in the direction from u to v and d[u] < d[v], this indicates that (u, v) is a tree edge, as v is discovered from u.

Thus, the correct answer is (A) tree.

Solution: For a function f, if f'(x*) = 0 and f''(x*) > 0, then x* is a point of local minimum. This is because: - The condition f'(x*) = 0 means that x* is a critical point. - The second derivative test states that if f"(x*) > 0, then the function has a local minimum at x*.

Therefore, the correct answer is local minimum.

Solution: - (p) "First In First Out" refers to queues, as they follow the FIFO order. - (q) "Lookup Operation" refers to hash tables, where the primary operation is to lookup values efficiently. - (r) "Last In First Out" refers to stacks, as they follow the LIFO order.

Thus, the correct match is (p) → (ii), (q) → (iii), (r) → (i).

Solution: For a hard margin linear support vector machine (SVM), support vectors are the data points that lie on or near the decision boundary. In this case, the points with labels y = 5 lie on one side of the boundary and the points with labels y = -1 lie on the other side. The support vectors are the points closest to the decision boundary.

Given the arrangement of data points, the set {x1, x2, x3, x4} forms a valid set of support vectors, as these points are the closest to the margin boundary.

Thus, the correct answer is {x1, x2, x3, x4} .

Solution: Principal Component Analysis (PCA) is a method for dimensionality reduction, so p corresponds to (ii). - Naive Bayes Classification is a generative model, as it models the joint probability distribution of the features and the class, so q corresponds to (iii). - Logistic Regression is a discriminative model, as it directly models the probability of the class given the features, so r corresponds to (i).

Thus, the correct match is (p) → (i), (q) → (ii), (r) → (iii).

Solution: Given that the points [1, 1] and [-1, -1] are both part of cluster 3, the centroid of cluster 3 would likely be near the origin [0, 0], based on the geometry of the points.

Among the provided options, ⁰⁄₁ is the point that lies closest to the origin. Therefore, this point is most likely to be part of cluster 3.

Thus, the correct answer is ⁰⁄₁.

Solution: In a naive Bayes classifier, for each class, we need to estimate the probabilities for each attribute being 0 or 1. For K binary-valued attributes, the number of parameters to estimate is 2K (for the two possible values of each attribute). Additionally, we need to estimate the class probabilities, which contributes 1 parameter for each class.

Thus, the total number of parameters to estimate is:
2K + 1

Therefore, the correct answer is 2K + 1.

Solution: The number of probes required for an unsuccessful search in a hash table with open addressing can be derived from the expected value of the number of probes in a uniform hashing scheme.

For an unsuccessful search, the expected number of probes is given by the formula:
¹⁄_1-α
where α= ⁿ⁄_m is the load factor, n is the number of elements in the table, and m is the number of slots.

For inserting an element into the hash table, the expected number of probes will be the same as the expected number of probes for an unsuccessful search because insertion is similar to searching, except that once an empty slot is found, the element is inserted.

Thus, the correct answer is ¹⁄_1-α.

Solution: The Fisher's Linear Discriminant maximization problem can be solved using optimization techniques. By differentiating J(u) = ^uTS_Bu⁄_u^TSWu and setting the derivative equal to zero, we obtain the following equation:

S_WS_Bu* = λu*

where λ is the maximized value of the objective function, and u* is the optimal projection vector.

Thus, the correct answer is S_WS_Bu* = λu*.

Solution: In A* search, a heuristic is admissible if it never overestimates the true cost to reach the goal. If h₁ and h₂ are both admissible heuristics, then:
- h₁ + h₂ is admissible because it sums two admissible heuristics, and the sum is still a lower bound of the actual cost. - h₁ × h₂ is not necessarily admissible because it could result in overestimation. - ^h₁⁄_h2 (with h₂ ≠ 0) could also overestimate the true cost depending on the values of h₁ and h₂. - |h₁ - h₂| is always admissible because the difference between two admissible heuristics will still be a lower bound.

Thus, the correct answer is |h₁ – h₂|.

Solution: From the joint distribution, we can analyze the conditional independence: - Y is conditionally independent of V given W, because Y depends only on W and not on V. - X is conditionally independent of U given W, because X depends only on W and not on U. - U and V are not necessarily conditionally independent given W, as they may have a dependency through W. - Y and X are conditionally independent given W, as they are conditionally independent when W is known.

Thus, the correct answer is (C) U and V are conditionally independent given W.

Solution: In alpha-beta pruning, the idea is to prune branches of the search tree where further exploration will not change the outcome of the game.

For the MAX player, we are interested in maximizing the value, so α is the highest value that the MAX player can guarantee up to that point. - For the MIN player, we are interested in minimizing the value, so β is the lowest value that the MIN player can guarantee up to that point.

Therefore, the correct setting is: - (m) = highest for the MAX player (because we want to maximize the value). - (n) = lowest for the MIN player (because we want to minimize the value).

Thus, the correct answer is (C) (m) = highest, (n) = lowest

Solution: We need to check if every name in the Team relation appears in either the Defender or Forward relations. The correct relational algebra expression would:

- First, project the names from both the Defender and Forward relations using π_name(Defender) and π_name(Forward). - Then, find the names that are in Team but not in either Defender or Forward. This can be done by subtracting the union of π_name(Defender) and π_name(Forward) from π_name(Team). - Finally, check if the result is the empty set, which means that every name in Team must appear in either Defender or Forward. Thus, the correct expression is:
π_name(Team) \ (π_name(Defender) ∪ π_name(Forward)) = Ø

Solution: The formula for calculating the Z-score is:
Z = ^{X - μ}⁄_σ
where: - X is the raw score (Rs 106000), - μ is the mean (Rs 96000), - σ is the standard deviation (Rs 21000).

Substituting the values:
Z = ^{106000 - 96000}⁄₂₁₀₀₀ = ¹⁰⁰⁰⁰⁄₂₁₀₀₀ ≈ 0.476

Thus, the Z-score normalized income value of Rs 106000 is closest to 0.476.

Solution: To uniquely reconstruct a full binary tree, we need sufficient information about the structure of the tree. For a full binary tree:

- Preorder and Inorder traversals together provide sufficient information to uniquely reconstruct the tree. In Preorder, we visit the root first, and in Inorder, we visit the left subtree, then the root, and then the right subtree. - Inorder and Postorder traversals alone are not sufficient because both traversals can have similar structures for different trees. - Preorder and Postorder together can also be used to reconstruct the tree uniquely.

Thus, the correct answer is (i) and (ii).

Solution: - In option (A), the expression (¬x∧y) ⇒ (y ⇒ x) is not a tautology as it does not always hold true for every possible value of x and y. - In option (B), the expression (x ∧ ¬y) ⇒ (¬x ⇒ y) is not a tautology. - In option (C), the expression (¬x ∨ y) ⇒ (¬x ⇒ y) is a tautology because it holds true for all possible truth values of x and y. - In option (D), (x ∧ ¬y) ⇒ (y ⇒ x) is not a tautology either.

Thus, the correct answer is (C) (¬x ∨ y) ⇒ (¬x ⇒ y).

Solution: In the in-place Quicksort algorithm using the last element as the pivot, the array is divided into smaller partitions recursively, and elements are swapped into their correct position.

Here, the array is already sorted in ascending order. During the first partitioning step, the pivot (100) will already be in its correct position. No swaps are needed for other elements, as they are already correctly placed.

Thus, the minimum number of swaps performed during this Quicksort is 0.

Solution: Let's break down the SQL query:
- The query selects data from the Raider and Team tables, joining them on the condition Raider.ID = Team.ID. - It filters the data where the City is "Jaipur" and the RaidPoints are greater than 200.

Looking at the data:
- The City "Jaipur" appears in the Team table with ID 2 and ID 6. - For ID 2, the corresponding Raider table has RaidPoints 250, which satisfies the condition RaidPoints > 200. - For ID 6, the corresponding Raider table has RaidPoints 215, which also satisfies the condition RaidPoints > 200. - For ID 2, the corresponding Raider table has RaidPoints 250 again, which also satisfies the condition.

Thus, three rows will be returned by this query.

Solution: Let's break down the operations step-by-step:
1. insertFirst(10): The queue becomes [10].
2. insertLast(32): The queue becomes [10,32].
3. a ← removeFirst(): This removes 10, so a = 10, and the queue becomes [32].
4. insertLast(28): The queue becomes [32, 28].
5. insertLast(17): The queue becomes [32, 28, 17].
6. a ← removeFirst(): This removes 32, so a = 32, and the queue becomes [28,17].
7. a ← removeLast(): This removes 17, so a = 17, and the queue becomes [28].

Thus, the value of a is 17.

Solution: The given function is f(x) = ¹⁄_1+e^-x, which is the logistic function. The derivative of f(x) is given by:
f'(x) = f(x)(1 − f(x))

We are asked to find f'(x) when f(x) = 0.4.
Substituting f(x) = 0.4 into the derivative formula:
f'(x) = 0.4 × (1 – 0.4) = 0.4 x 0.6 = 0.24

Thus, the value of the derivative at x where f(x) = 0.4 is 0.24.

Solution: The sample average x is given by:
x = ^{Σn_i=1 x_i}⁄_n

For n = 50, the sample average is 40, so the sum of the data points is:
Σ⁵⁰_i=1 x_i = 40 × 50 = 2000

After including the new data point, the new sum becomes:
Σ⁵¹_i=1 x_i = 2000 + 142 = 2142

The updated sample average is:
x_new = ²¹⁴²⁄₅₁ = 42

Thus, the updated sample average is 42.

Solution: We are asked to find the determinant of M² + 12M. First, calculate M²:
M² = ^{1 2 3}⁄_{3 1 1} x ^{1 2 3}⁄_{3 1 1}
^{4 3 6}⁄_{4 3 6}

Multiplying these matrices:
M² = ^{1(1)+2(3)+3(4) 1(2)+2(1)+3(3) 1(3)+2(1)+3(6)}⁄_{3(1)+1(3)+1(4) 3(2)+1(1)+1(3) 3(3)+1(1)+1(6)}
^{4(1)+3(3)+6(4) 4(2)+3(1)+6(3) 4(3)+3(1)+6(6)}

This results in:
M² = ^{1+6+12 2+2+9 3+2+18}⁄_{3+3+4 6+1+3 9+1+6} = ^{19 13 23}⁄_{10 10 16}
^{4+9+24 8+3+18 12+3+36}⁄_{37 29 51}

Now add 12M:
12M = 12 × ^{1 2 3}⁄_{3 1 1} = ^{12 24 36}⁄_{36 12 12}
^{4 3 6}⁄_{48 36 72}

Now calculate M²+12M:
M² + 12M = ^{19 13 23}⁄_{10 10 16} + ^{12 24 36}⁄_{36 12 12}
^{37 29 51}⁄_{48 36 72}

M² + 12M = ^{31 37 59}⁄_{46 22 28}
^{85 65 123}

Finally, compute the determinant of this matrix: det(M² + 12M) = 31 × det^{22 28}⁄_{65 123} − 37 × det^{46 28}⁄_{85 123}+ 59 × det ^{46 22}⁄_{85 65}

Computing the determinants, we find that the determinant is 0.

Thus, the correct answer is 0.

Solution: Let us define the states as follows: - State 0: No even number has been thrown yet. - State 1: One even number has been thrown, and we are waiting for the second consecutive even number. - State 2: Two consecutive even numbers have been thrown, and the process ends. From State 0, the probability of throwing an even number (2, 4, or 6) is ³⁄₆ = ¹⁄₂, and the probability of throwing an odd number (1, 3, or 5) is also ¹⁄₂.

From State 1, if an even number is thrown, we reach State 2, completing the sequence of two consecutive even throws. If an odd number is thrown, we revert to State 0.

To calculate the expected number of throws: - The expected number of throws to go from State 0 to State 1 is 2. - From State 1 to State 2, the expected number of throws is 2 as well (since you either end with two even numbers or revert to State 0).

Hence, the total expected number of throws is 2 + 2 = 6.

Thus, the correct answer is 6.

Solution: We are asked to make the function f(x) continuous and differentiable. We will do this by ensuring that: 1. f(x) is continuous at the points where the piecewise definitions change, i.e., at x = -2 and x = 2. 2. f(x) is differentiable at the points where the piecewise definitions change.

Step 1: Continuity at x = -2 and x = 2
For continuity at x = -2, we must have the value of f(x) from both sides equal: lim_x→-2^- f(x) = lim_x→-2⁺ f(x)
From the left-hand side, f(x) = -x, so:
lim_x→-2^- f(x) = -(-2) = 2
From the right-hand side, f(x) = ax² + bx + c, so:
f(-2) = a(-2)² + b(−2) + c = 4a – 2b + c
For continuity, we require:
4a - 2b + c = 2 (Equation 1)

For continuity at x = 2, we must have:
lim_x→2^- f(x) = lim_x→2⁺ f(x)
From the left-hand side, f(x) = ax² + bx + c, so:
f(2) = a(2)² + b(2) + c = 4a + 2b + c
From the right-hand side, f(x) = x, so:
f(2) = 2
For continuity, we require:
4a + 2b + c = 2 (Equation 2)

Step 2: Differentiability at x = -2 and x=2
For differentiability at x = -2, we require:
lim_x→-2^- f'(x) = lim_x→-2⁺ f'(x)
The derivative of f(x) = -x is f'(x) = -1, and the derivative of f(x) = ax² + bx + c is:
f'(x) = 2ax + b
At x = -2, we require:
f'(-2) = -1 and 2a(-2) + b = -1
Simplifying:
-4a + b = -1 (Equation 3)

For differentiability at x = 2, we require:
lim_x→2^- f'(x) = lim_x→2⁺ f'(x)
The derivative of f(x) = ax² + bx + c at x = 2 is:
f'(2) = 2a(2) + b = 4a + b
The derivative of f(x) = x is f'(x) = 1, so we require:
4a + b = 1 (Equation 4)

Step 3: Solve the system of equations
From Equations 3 and 4:
-4a + b = -1 and 4a + b = 1
Adding these two equations eliminates b:
2b = 0 -> b = 0
Substituting b = 0 into Equation 4:
4a = 1 -> a = ¹⁄₄

Substituting a = ¹⁄₄ and b = 0 into Equation 1:
4a - 2b + c = 2 4(¹⁄₄) - 2(0) + c = 2
1 + c = 2 -> c = 1

Thus, the values of a, b, and c are:
a = ¹⁄₄, b = 0, c = 1

Thus, the correct answer is a = ¹⁄₄, b = 0, c = 1.

Solution: Let's analyze the code again carefully:
We have a dictionary child_dict that represents a tree structure, where each key in the dictionary corresponds to a node, and the value is a list of child nodes. The function count(child_dict, i) counts the number of nodes in the subtree rooted at node i, including the node i itself.

Step-by-step Execution:
1. The function is initially called with i = 0: - child_dict[0] = [1,2], so the function calls count(child_dict, 1) and count(child_dict, 2).
2. For i = 1: - child_dict[1] = [3,4,5], so the function calls count(child_dict, 3), count(child_dict, 4),and count(child_dict, 5).
3. For i = 2: - child_dict[2] = [6,7,8], so the function calls count(child_dict, 6), count(child_dict, 7),and count(child_dict, 8).
4. For all the nodes i = 3, 4, 5, 6, 7, 8, there are no children, so the function returns 1 for each of these calls (base case).

Calculating the Total:
- For i = 0:
ans = 1 + count(child_dict, 1) + count(child_dict, 2)
- For i = 1:
ans = 1+count(child_dict, 3)+count(child_dict, 4)+count(child_dict, 5) = 1 + 1 + 1 + 1 = 4
- For i = 2:
ans = 1+count(child_dict, 6)+count(child_dict, 7)+count(child_dict, 8) = 1+1+1+1=4

- For i = 0, the total count is:
ans = 1 + 4 + 4 = 9
Thus, the correct output is 9.

Solution: We are given the function computeS(X) that initializes an array S based on the array X.

Let's go through the pseudocode step by step with X = [6, 3, 5, 4, 10]:
1. Initialize S[1] = 1. 2. For each i from 2 to 5, execute the following logic: - i = 2: X[1] = 6, X[2] = 3, since X[1] > X[2], set S[2] = −1. - i = 3: X[2] = 3, X[3] = 5, since X[2] < X[3], set S[3] = S[2] + S[1] = −1 + 1 = 0. - i = 4: X[3] = 5, X[4] = 4, since X[3] > X[4], set S[4] = -1. - i = 5: X[4] = 4, X[5] = 10, since X

Solution: We are given the function computeS(X) that initializes an array S based on the array X.

Let's go through the pseudocode step by step with X = [6, 3, 5, 4, 10]:
1. Initialize S[1] = 1. 2. For each i from 2 to 5, execute the following logic: - i = 2: X[1] = 6, X[2] = 3, since X[1] > X[2], set S[2] = −1. - i = 3: X[2] = 3, X[3] = 5, since X[2] < X[3], set S[3] = S[2] + S[1] = −1 + 1 = 0. - i = 4: X[3] = 5, X[4] = 4, since X[3] > X[4], set S[4] = -1. - i = 5: X[4] = 4, X[5] = 10, since X[4] < X[5], set S[5] = S[4] + S[3] = −1 + 0 = −1.

After the function completes execution, the array S becomes:
S = [1, -1, 0, -1, -1]

Thus, the correct answer is [1, -1, 0, -1, -1].

Solution: The maximum number of comparisons in binary search occurs when we divide the problem size by 2 at each step, and we add one comparison for each division. Therefore, the recurrence relation is:
F(n) = F(ⁿ⁄₂) + 1.

Thus, the correct option is (A).

Solution: The function recursively swaps the elements at indices s1 and s2, and then continues with the next pair of elements by incrementing s1 and decrementing s2. This continues until s1 is no longer less than s2. Essentially, this results in reversing the sublist between indices s1 and s2.

Thus, the correct option is (C).

Solution: In single linkage clustering, we start by finding the smallest distance between any two points and merge the two clusters. We then repeat the process by finding the smallest distance between any two clusters. From the table: - The minimum distance between x1 and x2 is 1. - The minimum distance between x3 and x4 is 2. - The minimum distance between x4 and x5 is 1. Thus, the clusters will eventually merge into the structure:

Cluster 1: x1, x2, Cluster 2: x3, Cluster 3: x4, x5.

So, the correct option is (A).

Solution: Given that the activation function is ReLU, we know that the output of a neuron is the maximum of 0 and the weighted sum of its inputs. To determine when the two neural networks are equivalent, we need to consider how the input values x₁, x₂, and x₃ relate to the weights p, q, and r in both networks.

Since the exact figures and the structure of the neural networks are not provided in the question, we can only analyze the potential relationships between the weights. The networks will be equivalent when the relationships between the weighted sums in both networks produce identical outputs for all positive inputs x₁, x₂, and x₃. The key to determining the equivalence is ensuring that the output of each network, as a function of the weights and inputs, is the same for all possible values of x₁, x₂, and x₃.

Answer: (B) p = 24, q = 24, r = 36

Solution: For both BFS and DFS, the expansion of states is based on the rule that each state n leads to two states: n + 1 and n + 2. Starting from state 1: - BFS would expand states level by level: - First, it expands 1 → 2 and 3. - Next, it expands 2 → 3, and 3 → 4, etc. - DFS would expand deeper into the tree: - First, it expands 1 → 2, then 2 → 3, then 3 → 4, and so on. In both cases, the same states are expanded to reach state 6. Thus, both BFS and DFS expand the same number of states. Thus, the correct answer is (C).

Solution: Let's analyze the sorting algorithms: - Bubble sort after two passes: - First pass: [3, 2, 1, 4, 5] → [2, 1, 3, 4, 5] - Second pass: [1, 2, 3, 4, 5] - After two passes, the array is sorted. - Insertion sort after two passes: - First pass: [3, 4, 2, 1, 5] - Second pass: [2, 3, 4, 1, 5] - The array is not yet sorted after two passes. - Selection sort after two passes: - First pass: [1, 3, 2, 4, 5] - Second pass: [1, 2, 3, 4, 5] - After two passes, the array is sorted. Therefore, Selection sort (iii) will sort the array in two passes, but Bubble sort (i) will not sort it in exactly two passes. Thus, the correct answer is (B).

Solution: Step 1: Identify the Functional Dependencies in the Given Set
The given set of functional dependencies is:
{U → V, U → W, W → Y, W → Z, X → Y, X → Z}

Step 2: Derive New Functional Dependencies
Let's check the possible derivations:
- VW → YZ can be derived as follows: - From W → Y and W → Z, we can conclude that VW → YZ.
- WX → YZ can also be derived because: - From X → Y and X → Z, we can conclude that WX → YZ.
- VW → Y can be derived because: - From W → Y, we know that knowing W implies Y.

Final Answer:
VW → YZ, WX → YZ, VW → Y

Solution: Step 1: Check for Subspace Criteria
To determine if the given set is a subspace, we must check if the set satisfies the following conditions: 1. Contains the zero vector. 2. Closed under addition. 3. Closed under scalar multiplication.

Step 2: Evaluate the Options
- Option (A): The given set consists of all linear combinations of two vectors ¹⁄₁⁄₀ and ¹⁄₀⁄₁, which is a subspace of R³ because it satisfies all subspace properties.

- Option (B): The set involves quadratic terms α² and β², which disqualifies it as a subspace because the elements are not closed under scalar multiplication. Thus, this is not a subspace.

- Option (C): The set consists of the solutions to a system of linear equations, and thus forms a subspace because it satisfies all subspace properties.

- Option (D): The equation 5x₁ + 2x₃ + 4 = 0 does not pass through the origin, hence it is not a subspace.

Final Answer:
A, C

Solution: Step 1: Analyze Each Option
- Option (A): If M ∈ R^3x3, for Mx = p to have a unique solution, M must be invertible. However, if Mx = q has infinite solutions, M must be singular, which contradicts the assumption of invertibility. Hence, this option is false.

- Option (B): If M ∈ R^3x3 and Mx = p has no solution, it means that M is singular, and the rank of M is less than 3. If Mx = q has infinite solutions, this is possible if M is singular. Hence, this option is true.

- Option (C): If M ∈ R^3x2, then Mx = p cannot have a unique solution because the system has more variables than equations. Thus, this option is false.

- Option (D): If M ∈ R^3x2, Mx = p having a unique solution is not possible, as the system has more variables than equations. However, if Mx = q has no solutions, this is possible for a consistent system that has no solution. Hence, this option is true.

Final Answer:
B, D

Solution: Step 1: Understanding the Projection Matrix M
A projection matrix M onto a subspace U satisfies: 1. M² = M (idempotent property of projection matrices). 2. The null space N(M) consists of all vectors that are projected to zero.

Step 2: Evaluating the Given Statements
- Option (A): If U is a 1-dimensional subspace of R³, then the rank of M is 1, meaning the null space should be 2-dimensional, not 1-dimensional. Hence, this statement is false.
- Option (B): If U is a 2-dimensional subspace of R³, then the rank of M is 2, meaning the null space has dimension 3 – 2 = 1. This statement is true.
- Option (C): Projection matrices satisfy M² = M, meaning applying the projection twice gives the same result. Hence, this statement is true.
- Option (D): Since M² = M, multiplying again by M gives M³ = M², which holds true for any projection matrix. Hence, this statement is true.

Final Answer:
B, C, D

Solution: To analyze the critical points and their nature, we proceed as follows:
1. Critical Points: We first compute the derivative of f(x) and set it to zero to find the critical points.
f(x) = ^{x4 - 3x2 + 1}⁄_{x² + 3}
Let g(x) = x⁴ - 3x² + 1 and h(x) = x² + 3. Using the quotient rule:
f'(x) = ^{g'(x)h(x) - g(x)h'(x)}⁄_(h(x))²
After simplifying, we find the critical points x = 0, ±1, ±√3.

2. Second Derivative Test: Evaluate f''(x) at the critical points to determine the nature of each:
At x = 0, f''(0) < 0, so x = 0 is a local maximum.
At x = 3, f''(3) > 0, so x = 3 is a local minimum.
At x = −1, f''(-1) < 0, so x = -1 is a local maximum.

Answer: (A), (B), and (C)

Solution: Step 1: Understanding Topological Sorting
A topological sorting of a DAG is a linear ordering of vertices such that for every directed edge (A → B), vertex A appears before vertex B.

Step 2: Identifying the DAG Structure
From the given DAG: - P and R are sources (have no incoming edges). - Q is dependent on P and R. - S and V depend on Q. - U depends on S, and T depends on V.

Step 3: Evaluating the Given Options
- Option (A): PQRSTUV - This order is incorrect because S appears before V, but both must follow Q correctly. - T must appear after V, but it appears earlier. - Incorrect.
- Option (B): PRQVSUT - This ordering correctly follows the dependency structure. - P, R appear before Q, which appears before V and S, followed by U and T. - Correct.
- Option (C): PQRSVUT - P and Q appear before R, which violates the dependency order. - Incorrect.
- Option (D): PRQSVTU - P and R appear before Q, which is correct. - S and V follow Q, maintaining the correct order. - T appears before U, which does not violate any dependency constraints. - Correct.

Final Answer: B, D

Solution: Step 1: Understanding the Binary Tree Structure
For a rooted binary tree: - The height H is the length of the longest path from the root to a leaf. - The number of internal nodes I is the number of non-leaf nodes. - The number of leaf nodes L is the number of nodes that do not have children. - The total number of nodes N is the sum of the internal nodes and leaf nodes.

We also know from binary tree properties: - The number of leaf nodes L in a perfect binary tree of height H is L = 2^H. - The total number of nodes N in a binary tree is given by N = I + L, and for a complete binary tree, N = 2^H+1 - 1.

Step 2: Evaluating the Given Options
- Option (A): L ≤ I + 1 - This is always true because the number of leaf nodes L is at most one more than the number of internal nodes I in any binary tree. - Correct.
- Option (B): H + 1 ≤ N ≤ 2^H+1 - 1 - This is true because: - The minimum number of nodes in a binary tree is H + 1, corresponding to a tree where each internal node has only one child (a skewed tree). - The maximum number of nodes occurs when the tree is complete, and this is given by 2^H+1 - 1. - Correct.
- Option (C): H ≤ I ≤ 2^H - 1 - This is true because the number of internal nodes I is at least H (in a complete binary tree) and at most 2^H - 1 (in a perfect binary tree). - Correct.
- Option (D): H ≤ L ≤ 2^H - 1 - This is not always true because the number of leaf nodes L is at least 2^H for a perfect binary tree, but not necessarily less than 2^H - 1. - Incorrect.

Final Answer:
A, B, C

Solution: Step 1: Understanding Linear Separability
A dataset is said to be linearly separable if there exists a straight line (in 2D) or a hyperplane (in higher dimensions) that can separate the two classes without any misclassification.

Step 2: Analyzing the Given Figures
- Figure (i): The circles and squares can be separated by a straight line, as they are clearly divided by the line x₁ = 2. - Correct.
- Figure (ii): The circles and squares are not linearly separable. No single straight line can separate them. - Incorrect.
- Figure (iii): The circles and squares are not linearly separable because they overlap, and no straight line can separate them. - Incorrect.
- Figure (iv): The circles and squares can be separated by a straight line, as they are divided along the axis. - Correct.

Final Answer:
A, D

Solution: Step 1: Understanding the Assertion ”All balls are round except balls used in rugby"
The assertion means that all balls are round except those used in rugby, i.e., if a ball is used in rugby, it is not round. This statement can be represented logically as: - game(ball, rugby) ⇒ ¬shape(ball, round) (If a ball is used in rugby, it is not round). - ¬game(ball, rugby) ⇒ shape(ball, round) (If a ball is not used in rugby, it must be round).

Step 2: Evaluating the Options
- Option (A): s1 ∧ s3 - s1 is ∀ball ⇒ game(ball, rugby) ⇒ shape(ball, round), which is not the correct representation for "All balls are round except those used in rugby". s3 is ∀ball game(ball, rugby) ⇒ ¬shape(ball, round), which is the correct representation for the "except" part of the assertion. - The combination of s1 and s3 correctly captures the structure of the assertion. - Correct.
- Option (B): s1 ∧ s2 - s2 is ∀ball ⇒ ¬shape(ball, round) ⇒ game(ball, rugby), which is not equivalent to the assertion and does not make logical sense in this context. - Incorrect.
- Option (C): s2 ∧ s3 - s2 is ∀ball ⇒ ¬shape(ball, round) ⇒ game(ball, rugby), which is incorrect. s3 is correct, and it complements the logic by specifying that if a ball is used in rugby, it is not round. - The combination of s2 and s3 accurately captures the assertion's meaning. - Correct.
- Option (D): s3 ∧ s4 - s4 is ∀ball shape(ball, round) ⇒ ¬game(ball, rugby), which is incorrect as it states that a round ball cannot be used in rugby, which contradicts the given context. - Incorrect.

Final Answer:
A, C

Solution: The query involves the following conditions:
1. Filtering rows based on Movie.CustomerRating > 3.4, which would benefit from a B⁺ tree index because it supports range queries efficiently.
2. Filtering rows where Genre.Name = "Comedy", which would benefit from a Hash index because hash indexes are efficient for equality searches.
3. Joining the Movie_Genre table with Movie and Genre tables, which would benefit from indexes on the foreign key columns Movie_Genre.MovieID and Movie_Genre.GenreID

The optimal choice for speeding up this query is a Hash index on Genre.Name for equality filtering and a B⁺ tree index on the remaining attributes for range queries and joins.

Solution: The probability density functions (PDFs) of X and Y are:
f_X(x) = ¹⁄₂ for x ∈ [1,3]
0 otherwise
f_Y(y) = ¹⁄₂ for y ∈ [2,4]
0 otherwise

Since X and Y are independent, their joint PDF is:
f_x,y(x,y) = f_x(x)f_y(y) = ¹⁄₄ for x ∈ [1,3] and y ∈ [2, 4]
0 otherwise.

To compute P(X > Y), we evaluate:
P(X > Y) = ∫∫ f_x,y(x, y) dx dy
y=2 x=y

Step 1: Evaluate the limits of integration
- For y ∈ [2,3]: x ∈ [y, 3] - For y ∈ [3, 4]: x ∈ [3, 3] (no contribution since x > y)

Thus, the integral simplifies to:
P(X > Y) = ∫³_y=2 ∫³_x=y ¹⁄₄ dx dy

Step 2: Solve the integral
The inner integral over x is: ∫³_x=y ¹⁄₄ dx = ¹⁄₄ x|³_y = ¹⁄₄(3-y)

The outer integral over y becomes:
∫³_y=2 ¹⁄₄(3-y) dy

Simplify:
∫³_y=2 ¹⁄₄(3-y) dy= ¹⁄₄ [3y - ^y2⁄₂]³₂ = ¹⁄₄ ( (9 - ⁹⁄₂) - (6 - ⁴⁄₂) ) = ¹⁄₄( ⁹⁄₂ - 4) = ¹⁄₄ x ¹⁄₂ = ⁵⁄₈ Thus:
P(X > Y) = ⁵⁄₈ x ¹⁄₄ = ⁵⁄₃₂ ≈ 0.625
Final Answer: 0.625 (rounded off to three decimal places).

Solution: The expectation and variance of an exponential random variable with rate parameter λ are:
E(X) = ¹⁄_λ
Var(X) = ¹⁄_λ²

Given that E(X) = Var(X), we have:
¹⁄_λ = ¹⁄_λ²

Solving for λ:
λ = 1

Final Answer: λ = 1.

Solution: Using Bayes' Theorem, we have:
P(T|S) = ^{P(S|T) • P(T)}⁄_P(S)

Now, we need to calculate P(S). We can use the law of total probability:
P(S) = P(S|T) • P(T) + P(S|T) • P(T)

Substitute the given values:
P(S) = (0.6) • (0.6) + (0.3) • (0.4)
P(S) = 0.36 + 0.12 = 0.48

Now, apply Bayes' Theorem:
P(T|S) = ^{(0.6) • (0.6)}⁄_0.48= ^0.36⁄_0.48 = 0.75

Final Answer: P(T|S) = 0.75.

Solution: Step 1: Compute the Conditional Density Function
The marginal density of X is given by:
f_x(x) = ∫^x₀ 2xy dy = 2x ^y2⁄₂ |^x₀ = x³, 0
The conditional density function of Y given X = x is:
f_y|x(y|x) = ^f_x,y(x,y)⁄_fx(x) = ^2xy⁄_x³ = ^2y⁄_x², 0
Step 2: Compute E[Y|X = 1.5] The expectation of Y given X = x is:
E[Y|X = x] = ∫^x₀ y f_y|x(y|x) dy = ∫^x₀ y( ^2y⁄_x² ) dy = ²⁄_x² ∫^x₀ y² dy = ²⁄_x² [^y3⁄₃]^x₀ = ^2x3⁄_3x² = ^2x⁄₃

Substituting x = 1.5:
E[Y|X = 1.5] = ^2(1.5)⁄₃ = 1

Final Answer: 1

Solution: First, simplify the expression inside the logarithm:
ln((x² + 1) cos(x)) = ln(x² + 1) + ln(cos(x))

Now, evaluate the limits of each term separately as x → 0:
lim_x→0 ln(x² + 1) = ln(1) = 0
lim_x→0 ln(cos(x)) = ln(1) = 0

Thus, the overall limit is: lim_x→0 ln((x² + 1) cos(x)) = 0 + 0 = 0 Final Answer: 0.

Solution: Given the matrix M = uu^T, where u = ¹⁄₃, we can calculate the singular values of the matrix. The singular values of a matrix M are the square roots of the eigenvalues of the matrix M^TM or MM^T.

1. Matrix Calculation:
First, calculate M = uu^T:
M = ¹⁄₃(1 3) = ^{1 3}⁄_{3 9}

2. Singular Values:
The matrix M is a 2 × 2 matrix. To find its singular values, we calculate the eigenvalues of MM^T.

MM^T = ^{1 3}⁄_{3 9} * ( ^{1 3}⁄_{3 9}) = ^{10 30}⁄_{30 90}

To find the eigenvalues, solve the characteristic equation det (MM^T – λI) = 0: det(^{10 - λ 30}⁄_{30 90-λ}) = (10 - λ)(90 - λ) – 30² = 0
λ² – 100λ = 0
λ(λ – 100) = 0

Thus, the eigenvalues are λ1 = 0 and λ2 = 100. The singular values of M are the square roots of the eigenvalues:
σ₁ = √100 = 10, σ₂= 0

3. Final Answer:
Since the product of singular values is σ1·σ2, we have:
σ₁σ₂ = 10 x 0 = 0.

Final Answer: 0

Solution: The data in Table C is as follows:

Solution: We need to find the minimum odd value of k for which the k-nearest neighbor algorithm assigns the label "square" to the diamond (◊) shaped data point.

1. Understanding the k-Nearest Neighbor Algorithm: - In k-nearest neighbor (k-NN) classification, the label of a data point is assigned based on the majority class among the k nearest neighbors (based on Euclidean distance in this case). - We are given a dataset with 5 data points from two classes: circles and squares.

2. Properties of the k-Nearest Neighbor Algorithm: - To make sure that the classification is based on a majority vote, the value of k must be odd. This avoids ties when there is an equal number of data points from each class among the nearest neighbors.

3. Assigning the Label "Square": - The task asks for the minimum odd value of k that results in the "diamond" shaped data point being classified as a "square." - The value of k must be chosen such that the number of "square" labels among the k nearest neighbors is greater than the number of "circle" labels.

4. Conclusion: - Since we are looking for the smallest odd value of k, and there are 5 data points, the smallest odd value for k would be k = 3. For k = 3, the 3 nearest neighbors are likely to influence the classification based on the majority label among them.

k = 3

Solution: From the structure of the Bayesian Network and the conditional probability tables:
- P(U = 1) = 0.5 - P(V = 0|U = 0) = 0.5, P(V = 1|U = 0) = 0.5 - P(V = 0|U = 1) = 0.5, P(V = 1|U = 1) = 0.5 - P(W = 0|U = 0) = 1, P(W = 1|U = 0) = 0 - P(W = 0|U = 1) = 0, P(W = 1|U = 1) = 1 - P(Z = 0|V = 0, W = 0) = 0.5, P(Z = 1|V = 0,W = 0) = 0.5 - P(Z = 0|V = 0, W = 1) = 1, P(Z = 1|V = 0, W = 1) = 0 - P(Z = 0|V = 1, W = 0) = 1, P(Z = 1|V = 1, W = 0) = 0 - P(Z = 0|V = 1, W = 1) = 0.5, P(Z = 1|V = 1, W = 1) = 0.5

The goal is to compute P(U = 1, V = 1, W = 1, Z = 1).

We can break this down using the chain rule for Bayesian networks:
P(U = 1, V = 1, W = 1, Z = 1) = P(U = 1)⋅P(V = 1|U = 1)⋅P(W = 1|U = 1)⋅P(Z = 1|V = 1, W = 1)

Substitute the values from the given probabilities:
P(U = 1) = 0.5, P(V = 1|U = 1) = 0.5, P(W = 1|U = 1) = 1, P(Z = 1|V = 1, W = 1) = 0.5

Now, multiplying these together:
P(U = 1, V = 1, W = 1, Z = 1) = 0.5 × 0.5 × 1 × 0.

Solution: Given the matrix M = uu^T, where u = ¹⁄₃, we can calculate the singular values of the matrix. The singular values of a matrix M are the square roots of the eigenvalues of the matrix M^TM or MM^T.

1. Matrix Calculation:
First, calculate M = uu^T:
M = ¹⁄₃(1 3) = ^{1 3}⁄_{3 9}

2. Singular Values:
The matrix M is a 2 × 2 matrix. To find its singular values, we calculate the eigenvalues of MM^T.

MM^T = ^{1 3}⁄_{3 9} * (^{1 3}⁄_{3 9}) = ^{10 30}⁄_{30 90}

To find the eigenvalues, solve the characteristic equation det (MM^T – λI) = 0:
det(^{10 - λ 30}⁄_{30 90-λ}) = (10 - λ)(90 - λ) – 30² = 0
λ² – 100λ = 0
λ(λ – 100) = 0
Thus, the eigenvalues are λ₁ = 0 and λ₂ = 100. The singular values of M are the square roots of the eigenvalues:
σ₁ = √100 = 10, σ₂ = 0

3. Final Answer:
Since the product of singular values is σ₁·σ₂, we have:
σ₁σ₂ = 10 × 0 = 0

Final Answer: 0

Solution: The data in Table C is as follows:

Solution: We need to find the minimum odd value of k for which the k-nearest neighbor algorithm assigns the label "square" to the diamond (◊) shaped data point.

1. Understanding the k-Nearest Neighbor Algorithm: - In k-nearest neighbor (k-NN) classification, the label of a data point is assigned based on the majority class among the k nearest neighbors (based on Euclidean distance in this case). - We are given a dataset with 5 data points from two classes: circles and squares.

2. Properties of the k-Nearest Neighbor Algorithm: - To make sure that the classification is based on a majority vote, the value of k must be odd. This avoids ties when there is an equal number of data points from each class among the nearest neighbors.

3. Assigning the Label "Square”: - The task asks for the minimum odd value of k that results in the "diamond" shaped data point being classified as a "square." - The value of k must be chosen such that the number of "square" labels among the k nearest neighbors is greater than the number of "circle" labels.

4. Conclusion: - Since we are looking for the smallest odd value of k, and there are 5 data points, the smallest odd value for k would be k = 3. For k = 3, the 3 nearest neighbors are likely to influence the classification based on the majority label among them.

k = 3

Solution: From the structure of the Bayesian Network and the conditional probability tables:
- P(U = 1) = 0.5 - P(V = 0|U = 0) = 0.5, P(V = 1|U = 0) = 0.5 - P(V = 0|U = 1) = 0.5, P(V = 1|U = 1) = 0.5 - P(W = 0|U = 0) = 1, P(W = 1|U = 0) = 0 - P(W = 0|U = 1) = 0, P(W = 1|U = 1) = 1 - P(Z = 0|V = 0, W = 0) = 0.5, P(Z = 1|V = 0,W = 0) = 0.5 - P(Z = 0|V = 0, W = 1) = 1, P(Z = 1|V = 0, W = 1) = 0 - P(Z = 0|V = 1, W = 0) = 1, P(Z = 1|V = 1, W = 0) = 0 - P(Z = 0|V = 1, W = 1) = 0.5, P(Z = 1|V = 1, W = 1) = 0.5

The goal is to compute P(U = 1, V = 1, W = 1, Z = 1).

We can break this down using the chain rule for Bayesian networks:
P(U = 1, V = 1, W = 1, Z = 1) = P(U = 1)⋅P(V = 1|U = 1)⋅P(W = 1|U = 1)⋅P(Z = 1|V = 1, W = 1)

Substitute the values from the given probabilities:
P(U = 1) = 0.5, P(V = 1|U = 1) = 0.5, P(W = 1|U = 1) = 1, P(Z = 1|V = 1, W = 1) = 0.5

Now, multiplying these together:
P(U = 1, V = 1, W = 1, Z = 1) = 0.5 × 0.5 × 1 × 0.5 = 0.125

Thus, the value of P(U = 1, V = 1, W = 1, Z = 1) is 0.125.

Solution: We need to compute the covariance between X and Y, which is given by the formula:
Cov(X, Y) = E[XY] – E[X]E[Y]

Step 1: Define the possible outcomes for the coin tosses
Since we are tossing two fair coins, the possible outcomes are:
{HH, HT, TH, TT}

Each outcome has a probability of ¹⁄₄.

Step 2: Compute the values of X and Y for each outcome
- For HH, both tosses are heads, so X = 1 and Y = 1. - For HT, X = 0 (since both tosses are not heads), and Y = 1 (since at least one toss is heads). - For TH, X = 0, and Y = 1. - For TT, X = 0, and Y = 0.

Step 3: Compute the expected values of X and Y
We calculate E[X] and E[Y]:
E[X] = ¹⁄₄(1+0+0+0) = ¹⁄₄
E[Y] = ¹⁄₄(1+1+1+0) = ³⁄₄

Step 4: Compute E[XY]
Next, we calculate E[XY]:
E[XY] = ¹⁄₄(1⋅1+0⋅1+0⋅1+0⋅0) = ¹⁄₄

Step 5: Compute the covariance
Now we can compute the covariance:
Cov(X, Y) = E[XY] – E[X]E[Y]

Substitute the values:
Cov(X,Y) = ¹⁄₄ - (¹⁄₄ x ³⁄₄) = ¹⁄₄ - ³⁄₁₆ = ¹⁄₁₆

Thus, the value of the covariance of X and Y is: ¹⁄₁₆