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All questions of Discrete Mathematics for Computer Science Engineering (CSE) Exam

A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Rhea Reddy answered
Let A be the event that first toss is head  
And B be the event that second toss is head. 
By the given condition rest all 8 tosses should be tail
∴ The probability of getting head in first two cases 
Clear all your doubts with EduRev

The 12 houses on one side of a street are numbered with even numbers starting at 2 and going up to 24. A free newspaper is delivered on Monday to 3 different houses chosen at random from these 12. Find the probability that at least 2 of these newspapers are delivered to houses with numbers strictly greater than 14.
  • a)
    7/11
  • b)
    5/12
  • c)
    4/11
  • d)
    5/22
Correct answer is option 'C'. Can you explain this answer?

Ravi Singh answered
There are  12 houses on one side of a street are numbered with even numbers.
In which 5 houses are strictly greater than Number 14.
And remaining 7 houses are numbered smaller than 14 (i.e. including 14)
No of way of choosing at least 2 of these newspapers are delivered to houses with numbers strictly greater than 14.
5C3  +5C2 * 7C1 =80
Total way of choosing 3 houses=12C3 =220
So Required probability=80/220=4/11

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
  • a)
    10/21
  • b)
    5/12
  • c)
    2/3
  • d)
    1/6
Correct answer is option 'B'. Can you explain this answer?

Cstoppers Instructors answered
Here our sample space consists of 3 + 3 * 6 = 21 events- (4), (5), (6), (1,1), (1,2) ... (3,6).
Favorable cases = (6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6)
Required Probability = No. of favorable cases/Total cases = 10/21
But this is wrong way of doing. Because due to 2 tosses for some and 1 for some, individual probabilities are not the same. i.e., while (6) has 1/6 probability of occurrence, (1,5) has only 1/36 probability. So, our required probability 
= 1/6 + (9 * 1/36) = 5/12

F is an n*n real matrix. b is an n*1 real vector. Suppose there are two n*1 vectors, u and v such that, u ≠ v and Fu = b, Fv = b. Which one of the following statements is false?
  • a)
    Determinant of F is zero.  
  • b)
    There are an infinite number of solutions to Fx = b 
  • c)
    There is an x≠0 such that Fx = 0
  • d)
    F must have two identical rows
Correct answer is option 'D'. Can you explain this answer?

Ravi Singh answered
(A) : Correct. We are given


Since  so we have a non-zero solution  to homogeneous equation  Now any vector  is also a solution of and so we have infinitely many solutions of and so determinant of F is zero.
(B) : Correct. Consider a vector


So there are infinitely many vectors of the form   which are solutions to equation Fx = b.
(C) : Correct. In option (a), we proved that vector
(D) : False. This is not necessary.
So option (D) is the answer.

The larger of the two eigenvalues of the matrix 
    Correct answer is '6'. Can you explain this answer?

    Gate Gurus answered
    For finding the Eigen Values of a Matrix we need to build the Characteristic equation which is of the form,
    A - λI
    Where A is the given Matrix.
    λ is a constant
    I is the identity matrix.
    We'll have a Linear equation after solving A - λI. Which will give us 2 roots for λ.

    6 is larger and hence is the Answer.

    Consider the DAG with Consider V = {1, 2, 3, 4, 5, 6}, shown below. Which of the following is NOT a topological ordering? 
    • a)
      1 2 3 4 5 6
    • b)
      1 3 2 4 5 6
    • c)
      1 3 2 4 6 5
    • d)
      3 2 4 1 6 5
    Correct answer is option 'D'. Can you explain this answer?

    Ravi Singh answered
    In option D, 1 appears after 2 and 3 which is not possible in Topological Sorting. In the given DAG it is directly visible that there is an outgoing edge from vertex 1 to vertex 2 and 3 hence 2 and 3 cannot come before vertex 1 so clearly option D is incorrect topological sort. But for questions in which it is not directly visible we should know how to find topological sort of a DAG.

    Can you explain the answer of this question below:

    What is the logical translation of the following statement?

    "None of my friends are perfect."

    • A:

    • B:

    • C:

    • D:

    The answer is d.

    Aditya Deshmukh answered
    F(x) ==> x is my friend
    P(x) ==> x is perfect

    D is the correct answer.

    A. There exist some friends which are not perfect

    B. There are some people who are not my friend and are perfect

    C. There exist some people who are not my friend and are not perfect.

    D. There doesn't exist any person who is my friend and perfect 

    How many sub strings of different lengths (non-zero) can be found formed from a character string of length n ?
    • a)
      n
    • b)
      n2
    • c)
      2n
    • d)
    Correct answer is option 'D'. Can you explain this answer?

    Yash Patel answered
    assuming an string of length n provided all alphabets are distinct..
    no of strings of length 1 = n
    no of strings of length 2 = n-1
    no of strings of length 3 = n-2 .
    .
    .
    no of string of length n = 1
    total = n + (n -1) + (n - 2) + (n - 2) + ..... + 1
    = n(n+1)/2

    The following is the Hasse diagram of the poset [{a, b, c, d, e}, ≤]
    The poset is
    • a)
      not a lattice
    • b)
      a lattice but not a distributive lattice
    • c)
      a distributive lattice but not a Boolean algebra
    • d)
      a Boolean algebra
    Correct answer is option 'B'. Can you explain this answer?

    Ravi Singh answered
    It is a lattice but not a distributive lattice.
    Table for Join Operation of above Hesse diagram
    Table for Meet Operation of above Hesse diagram
    Therefore for any two element p, q in the lattice (A,<=) p <= p V q ; p^q <= p This satisfies for all element (a,b,c,d,e).
    which has 'a' as unique least upper bound and 'e' as unique greatest lower bound.
    The given lattice doesn't obey distributive law, so it is not distributive lattice,
    Note that for b,c,d we have distributive law
    b^(cVd) = (b^c) V (b^d). From the diagram / tables given above we can verify as follows,
    (i) L.H.S. = b ^ (c V d) = b ^ a = b
    (ii) R.H.S. = (b^c) V (b^d) = e v e = e
    b != e which contradict the distributive law. Hence it is not distributive lattice. so, option (B) is correct.

    The minimum number of colours required to colour the following graph, such that no two adjacent vertices are assigned the same colour, is
    • a)
      2
    • b)
      3
    • c)
      4
    • d)
      5
    Correct answer is option 'C'. Can you explain this answer?

    Yash Patel answered

    Two vertices are said to be adjacent if they are directly connected, i.e., there is a direct edge between them. So, here, we can assign same color to 1 & 2 (red), 3 & 4 (grey), 5 & 7 (blue) and 6 & 8 (brown). Therefore, we need a total of 4 distinct colors. Thus, C is the correct choice. 
      
    Please comment below if you find anything wrong in the above post.

    If matrix  and X2 - X + I = 0 (I is the identity matrix and  is the zero matrix), then the inverse of X is
    • a)
    • b)
    • c)
    • d)
    Correct answer is option 'B'. Can you explain this answer?

    Yash Patel answered
    Given, X2 - X + I  = O
    =>   X2 = X - I
    => X-1 (X2)  = X-1(X - I)    {Multiplying X-1 both sides..}
    => X = I - X-1 
     => X-1 = I - X   
    Which gives Option (B)..

    Which of the following is TRUE about formulae in Conjunctive Normal Form?
    • a)
      For any formula, there is a truth assignment for which at least half the clauses evaluate to true.
    • b)
      For any formula, there is a truth assignment for which all the clauses evaluate to true
    • c)
      There is a formula such that for each truth assignment, at most one-fourth of the clauses evaluate to true.
    • d)
      None of the above
    Correct answer is option 'A'. Can you explain this answer?

    Rounak Choudhary answered
    We can easily prove that for any formula, there is a truth assignment for which at least half the clauses evaluate to true . Proof : Consider an arbitrary truth assignment. For each of its clause ‘j’ , introduce a random variable. Xj = 1 if clause ‘j’ is satisfied Xj = 0 otherwise Then, X = summation of (j * Xj) is the number of satisfied clauses. Given any clause ’c’ , it is unsatisfied only if all of its ‘k’ constituent literals evaluates to false as they are joined by OR operator. Now, because each literal within a clause has a 1/2 chance of evaluating to true independently of any of the truth value of any of the other literals, the probability that they are all false is (1 / 2)k . Thus, the probability that ‘c’ is satisfied = 1 − (1 / 2)k So, E(Xj) = 1 * (1 / 2)k = (1 / 2)k Therefore, E(Xj) >= 1/2 Summation on both sides to get E(X). Therefore, we have E(X) = summation of (j * Xj) >= m/2 where ‘m’ is the number of clauses. E(X) represents expected number of satisfied clauses. Thus, there must exist an assignment that satisfies at least half of the clauses. Please comment below if you find anything wrong in the above post.

    Consider the binary relation:
    S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}
    Q. The reflexive transitive closure of S is
    • a)
      {(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}
    • b)
      {(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}
    • c)
      {(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}
    • d)
      {(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}
    Correct answer is option 'B'. Can you explain this answer?

    Ravi Singh answered
    Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R. 
    If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}. 
    Now transitive closure is defined as smallest transitive relation which contains S. 
    We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now. 
     
    Thus, option (B) matches the final set S. 
     
    Please comment below if you find anything wrong in the above post.

    The number of binary strings of  zeros and  ones in which no two ones are adjacent is
    • a)
      n−1Ck
    • b)
      nCk
    • c)
      nCk+1
    • d)
      None of the above
    Correct answer is option 'D'. Can you explain this answer?

    Yash Patel answered
    first place n zeroes side by side _ 0 _ 0 _ 0 ... 0 _
    k 1's can be placed in any of the (n+1) available gaps hence number of ways  = n+1Ck

    Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K x 104  . The value of K is __________.
    • a)
      198 × 104
    • b)
      194 × 104
    • c)
      192 × 104
    • d)
      196 × 104
    Correct answer is option 'A'. Can you explain this answer?

    Ravi Singh answered
    an = 6n2 + 2n + an−1
    = 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + an−2
    = 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +a1
    = 6n2 + 2n + 6(n − 1)2 + 2(n − 1) + 6(n − 2)2 + 2(n − 2)+. . . . . . +6.12 + 2.1
    = 6(n2 + (n − 1)2 +. . . +22 + 12) + 2(n + (n − 1)+. . . +2 + 1)
    for n = 99 a99 = 2 x 99 x (99+1)2 = 198 x 104

    In how many ways can three person, each throwing a single die once, make a score of 11
    • a)
      22  
    • b)
      27  
    • c)
      24  
    • d)
      38
    Correct answer is option 'B'. Can you explain this answer?

    Akshay Singh answered
    The sum 11 can be broken as 6 + 5
    Let us name the players as A, B and C.
    Case 1: Fix 6 and break 5
    Suppose A throws 6. Players B and C can throw dice in following four ways to make sum 5
    (1,4), (2,3), (3,2) and (4,1)
    As any of the three players can throw value 6, so there are total 3 x 4 = 12
    Case 2: Fix 5 and break 6
    Suppose A throws 5. Players B and C can throw dice in following five ways to make sum 6
    (1,5), (2,4), (3,3), (4,2), (5,1)
    As any of the three players can throw value 5, so there are total 3 x 5 = 15
    Adding case 1 and case 2 there are total  12 + 15 = 27 ways

    Which of the following statements is true for every planar graph on n vertices?
    • a)
      The graph is connected
    • b)
      The graph is Eulerian
    • c)
      The graph has a vertex-cover of size at most 3n/4
    • d)
      The graph has an independent set of size at least n/3
    Correct answer is option 'C'. Can you explain this answer?

    Sanya Agarwal answered
    A planar graph is a graph which can drawn on a plan without any pair of edges crossing each other. A) FALSE: A disconnected graph can be planar as it can be drawn on a plane without crossing edges. B) FALSE: An Eulerian Graph may or may not be planar.An undirected graph is eulerian if all vertices have even degree. For example, the following graph is Eulerian, but not planar C) TRUE: D) FALSE:

    What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? Assume n >= 2.
    • a)
      2
    • b)
      3
    • c)
      n-1
    • d)
      n
    Correct answer is option 'A'. Can you explain this answer?

    Sanaya Gupta answered
    The Chromatic Number of a Graph

    The chromatic number of a graph is the minimum number of colors needed to color the vertices of the graph in such a way that no two adjacent vertices have the same color. In other words, it is the minimum number of colors needed to color the graph such that no two adjacent vertices have the same color.

    Graphs Without Odd Length Cycles

    An odd length cycle in a graph is a cycle (a path that starts and ends at the same vertex) with an odd number of edges. A graph that does not contain any odd length cycle is called bipartite. A bipartite graph is a graph whose vertices can be divided into two disjoint sets such that no two vertices within the same set are adjacent.

    Chromatic Number of a Graph Without Odd Length Cycles

    For a graph without any odd length cycle, it is always possible to color the vertices using only two colors. This is because we can divide the vertices into two sets, such that no two vertices within the same set are adjacent. Therefore, the chromatic number of the graph is 2.

    Explanation

    To understand why the chromatic number of a graph without any odd length cycle is 2, let's consider the following steps:

    1. Assume that the graph has n vertices.
    2. Since the graph is simple and connected, it must be a tree or a forest (a collection of disjoint trees).
    3. If the graph is a tree, it is trivially bipartite, as we can color the vertices alternatively using two colors.
    4. If the graph is a forest, we can still color the vertices using two colors. We can start with any vertex and assign it color 1. Then, for each adjacent vertex, assign it the opposite color. Continue this process until all vertices are colored.
    5. Since the graph does not contain any odd length cycle, there will be no adjacent vertices with the same color, as the colors are assigned alternately.
    6. Therefore, we have successfully colored the graph using only two colors.
    7. Hence, the chromatic number of the graph is 2.

    Conclusion

    The chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle is 2. This means that the graph can be colored using only two colors such that no two adjacent vertices have the same color.

    E1 and E2 are events in a probability space satisfying the following constraints:
    • Pr(E1) = Pr(E2)
    • Pr(E1 ∪ E2) = 1
    • E1 and E2 are independent
    The value of Pr(E1), the probability of the event E1, is
    • a)
      0
    • b)
      ¼
    • c)
      ½
    • d)
      1
    Correct answer is option 'D'. Can you explain this answer?

    Yash Patel answered
    let probability of Event E1 = x = prob of E2
    prob(E1 union E2) = prob(E1) + prob(E2) - prob(E1 intersect E2)
    1 = x + x -x2 (prob(E1 intersect E2) = prob(E1) * prob(E2) as events are independent)
    x = 1

    Let R1 be a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} and R2 be another relation from B to C = {1, 2, 3, 4} as defined below:
    1. An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
    2. An element x in B is related to an element y in C (under R2) if x + y is even but not divisible by 3.
     
    Q. Which is the composite relation R1R2 from A to C?  
    • a)
      R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)}
    • b)
      R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)}
    • c)
      R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)}
    • d)
      R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)}
    Correct answer is option 'C'. Can you explain this answer?

    Yash Patel answered
    R1 is a relation from A = {1, 3, 5, 7} to B = {2, 4, 6, 8} . Under R1, an element x in A is related to an element y in B if x + y is divisible by 3. 
    Thus, R1 = {(1, 2), (1, 8), (3, 6), (5, 4), (7, 2), (7, 8)} 
    R2 is a relation from B = {2, 4, 6, 8} to C = {1, 2, 3, 4} Under R2, an element y in B is related to an element z in C if y + z is even but not divisible by 3. 
    Thus, R2 = {(2, 2), (4, 4), (6, 2), (6, 4), (8, 2)} 
    Thus, R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)} 
     
    Thus, option (C) is correct. 
     
    Please comment below if you find anything wrong in the above post.

    Consider an undirected random graph of eight vertices. The probability that there is an edge between a pair of vertices is 1/2. What is the expected number of unordered cycles of length three?
    • a)
      1/8
    • b)
      1
    • c)
      7
    • d)
      8
    Correct answer is option 'C'. Can you explain this answer?

    Rajeev Menon answered
    A cycle of length 3 can be formed with 3 vertices. There can be total 8C3 ways to pick 3 vertices from 8. The probability that there is an edge between two vertices is 1/2. So expected number of unordered cycles of length 3 = (8C3)*(1/2)^3 = 7

    Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
    • a)
    • b)
    • c)
    • d)
    Correct answer is option 'B'. Can you explain this answer?

    Yash Patel answered
    for every car accident we can pick a day in 7 ways
    total number of ways in which accidents can be assigned to days = 77
    probability of accidents happening on a particular day = 1/77
    we can choose a day in 7 ways
    hence probability = 7/77= 1/76

    Compute the value of:
      Correct answer is '-1'. Can you explain this answer?

      Naroj Boda answered
      For the integrand 

      This gives a new lower bound  and upper bound  Now, our integral becomes:

      Since the antiderivative of cos(u) is sin (u) , applying the fundamental theorem of calculus, we get:

      Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:
      (i) select a box
      (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are 1/3 and 2/3 respectively. Given that a ball selected in the above process is a red ball, the probability that it came from the box P is:
      • a)
        4/19
      • b)
        5/19
      • c)
        2/9
      • d)
        19/30
      Correct answer is option 'A'. Can you explain this answer?

      Ravi Singh answered
      The probability of selecting a red ball  = (1/3) * (2/5) + (2/3) * (3/4) = 2/15 + 1/2 
       = 19/30 
      Probability of selecting a red ball from box P = (1/3) * (2/5) = 2/15 
      Given that a ball selected in the above process is a red ball, the probability that it came from the box P is = (2/15) / (19/30) = 4/19

      If P and Q are two random events, then the following is TRUE:
      • a)
        Independence of P and Q implies that probability (P ∩ Q) = 0
      • b)
        Probability (P ∪ Q) ≥ Probability (P) + Probability (Q)
      • c)
        If P and Q are mutually exclusive, then they must be independent
      • d)
        Probability (P ∩ Q) ≤ Probability (P)
      Correct answer is option 'D'. Can you explain this answer?

      Yash Patel answered
      here i check option---> i let P = probability,  P and Q is replaced by A and B.
      A) it  is saying independent means P(A ^ B) =  P(A)* P(B)  so option A is false
      B) P(AÜ B) =  P(A) + P(B) - P(A ^ B)
      hence P(AÜB) >= P(A) + P(B)  its false option B is false
      C) if A and B is mutually exclusive means P(A ^ B) =  0 then its independent ........its false  there is no relation independent and mutually exclusive
      D)  P(A ^ B) <= P(A)
      its true
      hence option D is correct

      How many different non-isomorphic Abelian groups of order 4 are there
      • a)
        2
      • b)
        3
      • c)
        4
      • d)
        5
      Correct answer is option 'A'. Can you explain this answer?

      Pankaj Rane answered
      2 can be written as 2 power 2.
      Number of partitioning of 2 = no. of non isomorphic abelian groups
      2 can be partitioned as {(2),(1,1)}

      What is the number of vertices in an undirected connected graph with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and remaining of degree 3?
      • a)
        10
      • b)
        11
      • c)
        18
      • d)
        19
      Correct answer is option 'D'. Can you explain this answer?

      Telecom Tuners answered
      The idea is to use Handshaking Lemma :- In any graph, the sum of all the vertex-degree is equal to twice the number of edges.
      Let x = Number of vertices of degree 3.
      By Handshaking Lemma
      6*2 + 3*4 + (x-9)*3 = 27*2
      24 + (x-9)*3 = 54
      x-9 = 10
      x = 19

      A graph is self-complementary if it is isomorphic to its complement. For all self-complementary graphs on n vertices, n is
      • a)
        A multiple of 4
      • b)
        Even
      • c)
        Odd
      • d)
        Congruent to 0 mod 4, or 1 mod 4
      Correct answer is option 'D'. Can you explain this answer?

      Ujwal Nambiar answered
      Explanation:

      In order to understand why the correct answer is option 'D', let's break down the given options and analyze them one by one.


      Option A: A multiple of 4

      A graph being self-complementary doesn't have any direct relation to the number of vertices being a multiple of 4. There are self-complementary graphs with any number of vertices, not just multiples of 4. Therefore, option A is incorrect.


      Option B: Even

      This option is also incorrect because self-complementary graphs can have an odd number of vertices. There are self-complementary graphs with 3, 5, 7, etc. vertices.


      Option C: Odd

      Similar to the previous option, self-complementary graphs can have an odd number of vertices. Therefore, option C is incorrect.


      Option D: Congruent to 0 mod 4, or 1 mod 4

      This option is correct. A graph being self-complementary is related to its number of vertices modulo 4.

      Let's consider the two cases:


      1. If the number of vertices is congruent to 0 modulo 4 (i.e., a multiple of 4), then we can divide the vertices into two equal sets of size n/2. In the complement of the graph, each vertex in one set is adjacent to every vertex in the other set. When we swap the adjacency relationships, we get an isomorphic graph, making it self-complementary.

      2. If the number of vertices is congruent to 1 modulo 4, then we can divide the vertices into two sets, one with (n+1)/2 vertices and the other with (n-1)/2 vertices. In the complement of the graph, each vertex in the larger set is adjacent to every vertex in the smaller set. Again, swapping the adjacency relationships results in an isomorphic graph, making it self-complementary.


      Therefore, self-complementary graphs on n vertices are congruent to 0 mod 4 or 1 mod 4.

      A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective?  
      • a)
        1/5      
      • b)
        1/25  
      • c)
        20/99  
      • d)
        11/495 
      Correct answer is option 'D'. Can you explain this answer?

      Kaavya Sengupta answered
      Total number of items = 100

      Number of defective items = 20

      Number of Non-defective items = 80

      Then the probability that both items are defective, when 2 items are selected at random is,

       ⇒ P= (20C2x80C0)/(100C2) = 19/495

      "If X, then Y unless Z" is represented by which of the following formulae in propositional logic? ("¬" is negation "^" is conjunction, and "→" is implication)
      • a)
        (X ^ ¬ Z) → Y
      • b)
        (X ^ Y) → ¬ Z
      • c)
        (X → (Y ^ ¬ Z)
      • d)
        (X → Y(^ ¬ Z)
      Correct answer is option 'A'. Can you explain this answer?

      Ravi Singh answered
      The statement "If X then Y unless Z" means, if Z doesn't occur, X implies Y i.e. ¬Z→(X→Y), which is equivalent to Z∨(X→Y) (since P→Q ≡ ¬P∨Q), which is then equivalent to Z∨(¬X∨Y). Now we can look into options which one matches with this. 
      So option (a) is (X∧¬Z)→Y = ¬((X∧¬Z))∨Y = (¬X∨Z)∨Y, which matches our expression. So option A is correct.

      You have to play three games with opponents A and B in a specified sequence. You win the series if you win two consecutive games. A is a stronger player than B. Which sequence maximizes your chance of winning the series?
      • a)
        AAB
      • b)
        ABA
      • c)
        BAB
      • d)
        BAA
      • e)
        All are the same.
      Correct answer is option 'B'. Can you explain this answer?

      Yash Patel answered
      Let the probability of winning against player A be  and the probability of winning against player B be b.

       be the probability of winning the series in which the games played are against x,y and z in order.

      We can see that not all probabilities are equal, so option E is not correct.
      We can also see that options A and D result in the same value, so they are not correct either.
      Comparing option B and option C.

      Hence, option B is the correct answer.

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