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?
A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?
a)
b)
c)
d)
|
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
| 1 Crore+ students have signed up on EduRev. Have you? Download the App |
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?
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
|
|
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
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
Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributes exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is - a)0.3
- b)0.5
- c)0.7
- d)0.9
Correct answer is 'A'. Can you explain this answer?
Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributes exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is
a)
0.3
b)
0.5
c)
0.7
d)
0.9
|
Aniket Saini answered |
Solution:
Given:
The time between successive arrivals follows poison distribution with an average time of 10 minutes.
The length of a phone call follows an exponential distribution with a mean of 3 minutes.
To find:
The probability that an arrival does not have to wait before service.
Method:
We can use the M/M/1 queuing model to solve this problem. M/M/1 stands for Markovian Arrival Process/Markovian Service Process/Single Server Queue.
Assumptions of M/M/1 queuing model:
1. Arrival process follows poison distribution.
2. Service time follows exponential distribution.
3. There is only one server.
4. Service is provided on a first-come-first-serve basis.
5. The queue is of infinite capacity.
Steps to solve the problem using M/M/1 queuing model:
1. Calculate the arrival rate (λ) and service rate (μ).
2. Calculate the utilization factor (ρ) using the formula ρ = λ/μ.
3. Calculate the probability that the server is busy (P0) using the formula P0 = 1 - ρ.
4. Calculate the average number of customers in the system (L) using the formula L = λ/(μ-λ).
5. Calculate the average time spent in the system (W) using the formula W = L/λ.
6. Calculate the probability that an arrival does not have to wait before service (P) using the formula P = e^(-μW).
Calculation:
1. Arrival rate (λ) = 1/10 per minute
Service rate (μ) = 1/3 per minute
2. Utilization factor (ρ) = λ/μ = (1/10)/(1/3) = 0.3
3. Probability that the server is busy (P0) = 1 - ρ = 1 - 0.3 = 0.7
4. Average number of customers in the system (L) = λ/(μ-λ) = (1/10)/[(1/3)-(1/10)] = 0.375
5. Average time spent in the system (W) = L/λ = 0.375/(1/10) = 3.75 minutes
6. Probability that an arrival does not have to wait before service (P) = e^(-μW) = e^(-1/3*3.75) = 0.3
Therefore, the probability that an arrival does not have to wait before service is 0.3, option (a) is correct.
Given:
The time between successive arrivals follows poison distribution with an average time of 10 minutes.
The length of a phone call follows an exponential distribution with a mean of 3 minutes.
To find:
The probability that an arrival does not have to wait before service.
Method:
We can use the M/M/1 queuing model to solve this problem. M/M/1 stands for Markovian Arrival Process/Markovian Service Process/Single Server Queue.
Assumptions of M/M/1 queuing model:
1. Arrival process follows poison distribution.
2. Service time follows exponential distribution.
3. There is only one server.
4. Service is provided on a first-come-first-serve basis.
5. The queue is of infinite capacity.
Steps to solve the problem using M/M/1 queuing model:
1. Calculate the arrival rate (λ) and service rate (μ).
2. Calculate the utilization factor (ρ) using the formula ρ = λ/μ.
3. Calculate the probability that the server is busy (P0) using the formula P0 = 1 - ρ.
4. Calculate the average number of customers in the system (L) using the formula L = λ/(μ-λ).
5. Calculate the average time spent in the system (W) using the formula W = L/λ.
6. Calculate the probability that an arrival does not have to wait before service (P) using the formula P = e^(-μW).
Calculation:
1. Arrival rate (λ) = 1/10 per minute
Service rate (μ) = 1/3 per minute
2. Utilization factor (ρ) = λ/μ = (1/10)/(1/3) = 0.3
3. Probability that the server is busy (P0) = 1 - ρ = 1 - 0.3 = 0.7
4. Average number of customers in the system (L) = λ/(μ-λ) = (1/10)/[(1/3)-(1/10)] = 0.375
5. Average time spent in the system (W) = L/λ = 0.375/(1/10) = 3.75 minutes
6. Probability that an arrival does not have to wait before service (P) = e^(-μW) = e^(-1/3*3.75) = 0.3
Therefore, the probability that an arrival does not have to wait before service is 0.3, option (a) is correct.
An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is- a)0.5
- b)0.18
- c) 0.12
- d) 0.06
Correct answer is option 'C'. Can you explain this answer?
An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is
a)
0.5
b)
0.18
c)
0.12
d)
0.06
|
Aditya Chavan answered |
Let A be the event that ‘failed in paper 1’.
B be the event that ‘failed in paper 2’.
Given P(A) = 0.3, P(B) = 0.2.
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?
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
|
|
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,
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.
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.
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?
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
|
Yash Patel 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
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
A fair dice is rolled twice. The probability that an odd number will follow an even number is - a)1/2
- b)1/6
- c)1/3
- d)1/4
Correct answer is option 'D'. Can you explain this answer?
A fair dice is rolled twice. The probability that an odd number will follow an even number is
a)
1/2
b)
1/6
c)
1/3
d)
1/4
|
Raghavendra Sharma answered |
Probability of even number =3/6 =1/2
Probability of odd number =3/6 =1/2
Both are independent so probability=1/2.1/2 =1/4
The larger of the two eigenvalues of the matrix 
Correct answer is '6'. Can you explain this answer?
The larger of the two eigenvalues of the matrix
|
Kabir Verma 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
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.
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.
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
a)
b)
c)
d)
|
Sanya Agarwal answered |
Poisson Probability Density Function (with mean 
We have to sum the probability density function for
(thus finding the cumulative mass function)
We have to sum the probability density function for
(thus finding the cumulative mass function) 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?
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
|
|
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.
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.
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.
What is the logical translation of the following statement?
"None of my friends are perfect."
|
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
A club with x members is organized into tour committees such that(a) each member is in exactly two committees,
(b) any two committees have exactly one member in common.Then x has- a)exactly two values both between 4 and 8
- b)exactly one value and this lies between 4 and 8
- c)exactly two values both between 8 and 16
- d)exactly one value and this lies between 8 and 16.
Correct answer is option 'B'. Can you explain this answer?
A club with x members is organized into tour committees such that
(a) each member is in exactly two committees,
(b) any two committees have exactly one member in common.
(b) any two committees have exactly one member in common.
Then x has
a)
exactly two values both between 4 and 8
b)
exactly one value and this lies between 4 and 8
c)
exactly two values both between 8 and 16
d)
exactly one value and this lies between 8 and 16.
|
|
Ravi Singh answered |
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?
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)
|
|
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
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
An examination paper has 150 multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 marks. Suppose 1000 students choose all their answers randomly with uniform probability.The sum total of the expected marks obtained by all these students is- a)0
- b)2550
- c)7525
- d)9375
Correct answer is option 'D'. Can you explain this answer?
An examination paper has 150 multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 marks. Suppose 1000 students choose all their answers randomly with uniform probability.The sum total of the expected marks obtained by all these students is
a)
0
b)
2550
c)
7525
d)
9375
|
|
Yash Patel answered |
Probability of choosing the correct option = 1/4
Probability of choosing a wrong option = 1/3
So, expected mark for a question for a student
Expected mark for a student for 150 questions
So, sum total of the expected marks obtained by all 1000 students
Probability of choosing a wrong option = 1/3
So, expected mark for a question for a student
Expected mark for a student for 150 questions
So, sum total of the expected marks obtained by all 1000 students
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?
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
|
|
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.
Let f(x) be a polynomial and 
be its derivative. If the degree of 
is 10, then the degree of 
Correct answer is '9'. Can you explain this answer?
Let f(x) be a polynomial and be its derivative. If the degree of is 10, then the degree of
be its derivative. If the degree of is 10, then the degree of |
|
Yash Patel answered |
F is some function where the largest even degree term is having degree 10. no restiction on odd degree terms.
since f(x)+f(-x)= degree 10
even power gets converted to odd in derivative.
then the the degrre of required expression =9.
the odd powers in F will become even in derivative and G(X)-G(-X) retains only odd powers.
since f(x)+f(-x)= degree 10
even power gets converted to odd in derivative.
then the the degrre of required expression =9.
the odd powers in F will become even in derivative and G(X)-G(-X) retains only odd powers.
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?
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
|
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:
- 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.
- 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.
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?
How many different non-isomorphic Abelian groups of order 4 are there
a)
2
b)
3
c)
4
d)
5
|
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)}
Number of partitioning of 2 = no. of non isomorphic abelian groups
2 can be partitioned as {(2),(1,1)}
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?
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
|
|
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
= 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 104Let 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: - An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
- 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?
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:
- An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
- 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)}
|
|
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, 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.
Thus, option (C) is correct.
Please comment below if you find anything wrong in the above post.
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?
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
|
Shounak Sengupta answered |
The chromatic number of a graph is the smallest number of colours needed to colour the vertices of so that no two adjacent vertices share the same colour. These types of questions can be solved by substitution with different values of n. 1) n = 2
This simple graph can be coloured with 2 colours. 2) n = 3
Here, in this graph let us suppose vertex A is coloured with C1 and vertices B, C can be coloured with colour C2 => chromatic number is 2 In the same way, you can check with other values, Chromatic number is equals to 2
A simple graph with no odd cycles is bipartite graph and a Bipartite graph can be colored using 2 colors (See this)
The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:- a)16/25
- b)(9/10)3
- c)27/75
- d)18/25
Correct answer is option 'D'. Can you explain this answer?
The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is:
a)
16/25
b)
(9/10)3
c)
27/75
d)
18/25
|
|
Yash Patel answered |
First digit can be chosen in 8 ways from 1−9 excluding 7
Second digit can be chosen in 9 ways from 0−9 excluding 7 and similarly the third digit in 9 ways.
So, total no. of ways excluding 7 = 8∗9∗9
Total no. of ways including 7 = 9∗10∗10
So, ans = (8∗9∗9)/(9∗10∗10)=18/25
Second digit can be chosen in 9 ways from 0−9 excluding 7 and similarly the third digit in 9 ways.
So, total no. of ways excluding 7 = 8∗9∗9
Total no. of ways including 7 = 9∗10∗10
So, ans = (8∗9∗9)/(9∗10∗10)=18/25
A cycle on n vertices is isomorphic to its complement. The value of n is _____- a)2
- b)4
- c)6
- d)5
Correct answer is option 'D'. Can you explain this answer?
A cycle on n vertices is isomorphic to its complement. The value of n is _____
a)
2
b)
4
c)
6
d)
5
|
|
Subhankar Chawla answered |
Below is a cyclic graph with 5 vertices and its complement graph.
The complement graph is also isomorphic (same number of vertices connected in same way) to given graph.
Manish has to travel from A to D changing buses at stops B and C enroute. The maximum waiting time at either stop can be 8 minutes each, but any time of waiting up to 8 minutes is equally likely at both places. He can afford up to 13 minutes of total waiting time if he is to arrive at D on time. What is the probability that Manish will arrive late at D?
- a)9/128
- b)13/64
- c)119/128
- d)8/13
Correct answer is option 'A'. Can you explain this answer?
Manish has to travel from A to D changing buses at stops B and C enroute. The maximum waiting time at either stop can be 8 minutes each, but any time of waiting up to 8 minutes is equally likely at both places. He can afford up to 13 minutes of total waiting time if he is to arrive at D on time. What is the probability that Manish will arrive late at D?
a)
9/128
b)
13/64
c)
119/128
d)
8/13
|
Siddharth Menon answered |
Solution:
Total waiting time allowed = 13 minutes
Maximum waiting time at each stop = 8 minutes
Probability of waiting time up to 8 minutes at each stop = 1/2
To arrive on time, the total waiting time should be less than or equal to 13 minutes.
Let's consider different cases of waiting times at stop B and stop C:
Case 1: No waiting time at B or C
In this case, Manish will spend zero waiting time at both stops. Hence, the total waiting time will be zero and he will arrive on time at D. Probability of this case = 1/4.
Case 2: Waiting time of 8 minutes at one stop and no waiting time at the other stop
In this case, Manish will spend 8 minutes waiting time at one stop and zero waiting time at the other stop. Hence, the total waiting time will be 8 minutes and he will arrive on time at D. Probability of this case = 2 x (1/2) x (1/4) = 1/4.
Case 3: Waiting time of 8 minutes at both stops
In this case, Manish will spend 8 minutes waiting time at both stops. Hence, the total waiting time will be 16 minutes and he will arrive late at D. Probability of this case = (1/2) x (1/2) = 1/4.
Case 4: Waiting time of less than 8 minutes at both stops
In this case, Manish will spend some waiting time at both stops, but the total waiting time will be less than 13 minutes. Hence, he will arrive on time at D. Probability of this case = 1 - (1/4) - (1/4) - (1/4) = 1/4.
Therefore, the probability of Manish arriving late at D = probability of Case 3 = 1/4.
Hence, the correct answer is option A) 8/13.
Total waiting time allowed = 13 minutes
Maximum waiting time at each stop = 8 minutes
Probability of waiting time up to 8 minutes at each stop = 1/2
To arrive on time, the total waiting time should be less than or equal to 13 minutes.
Let's consider different cases of waiting times at stop B and stop C:
Case 1: No waiting time at B or C
In this case, Manish will spend zero waiting time at both stops. Hence, the total waiting time will be zero and he will arrive on time at D. Probability of this case = 1/4.
Case 2: Waiting time of 8 minutes at one stop and no waiting time at the other stop
In this case, Manish will spend 8 minutes waiting time at one stop and zero waiting time at the other stop. Hence, the total waiting time will be 8 minutes and he will arrive on time at D. Probability of this case = 2 x (1/2) x (1/4) = 1/4.
Case 3: Waiting time of 8 minutes at both stops
In this case, Manish will spend 8 minutes waiting time at both stops. Hence, the total waiting time will be 16 minutes and he will arrive late at D. Probability of this case = (1/2) x (1/2) = 1/4.
Case 4: Waiting time of less than 8 minutes at both stops
In this case, Manish will spend some waiting time at both stops, but the total waiting time will be less than 13 minutes. Hence, he will arrive on time at D. Probability of this case = 1 - (1/4) - (1/4) - (1/4) = 1/4.
Therefore, the probability of Manish arriving late at D = probability of Case 3 = 1/4.
Hence, the correct answer is option A) 8/13.
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?
Consider the binary relation:
S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}
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, ... }}
|
|
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.
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 matrices 
commute under multiplication- a)
- b)always
- c)never
- d)
Correct answer is option 'A'. Can you explain this answer?
The matrices commute under multiplication
commute under multiplicationa)
b)
always
c)
never
d)
|
|
Yash Patel answered |
The multiplication will commute if
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?
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
|
|
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
k 1's can be placed in any of the (n+1) available gaps hence number of ways = n+1Ck
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?
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
|
|
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
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?
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
|
|
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
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
Px(x) = M exp(–2|x|) – N exp(–3 |x|) is the probability density function for the real random variable X, over the entire x axis. M and N are both positive real numbers. The equation relating M and N is - a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Px(x) = M exp(–2|x|) – N exp(–3 |x|) is the probability density function for the real random variable X, over the entire x axis. M and N are both positive real numbers. The equation relating M and N is
a)
b)
c)
d)
|
|
Avinash Sharma answered |
Given Px (x ) is the probability density function for the random variable X.
A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is- a)2/315
- b)1/630
- c)1/1260
- d)1/2520
Correct answer is option 'C'. Can you explain this answer?
A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is
a)
2/315
b)
1/630
c)
1/1260
d)
1/2520
|
|
Jaideep Dasgupta answered |
Here sample space = 9
The required probability of drawing 2 washers, 3 nuts and 4 bolts respectively without replac ement
In questions 1.1 to 1.7 below, one or more of the alternatives are correct. Write the code letter(s) a, b, c, d corresponding to the correct alternative(s) in the answer book. Marks will be given only if all the correct alternatives have been selected and no incorrect alternative is picked up. 1.1). The eigen vector (s) of the matrix
is (are)- a)
- b)
- c)
- d)
Correct answer is option 'B,D'. Can you explain this answer?
In questions 1.1 to 1.7 below, one or more of the alternatives are correct. Write the code letter(s) a, b, c, d corresponding to the correct alternative(s) in the answer book. Marks will be given only if all the correct alternatives have been selected and no incorrect alternative is picked up. 1.1). The eigen vector (s) of the matrix
is (are)
a)
b)
c)
d)
|
|
Ravi Singh answered |
Eigen values are: 0,0,0
The eigen vector should satisfy the equation: αz = 0
The eigen vector should satisfy the equation: αz = 0
Make is a utility that automatically builds executable programs and libraries from source code by reading files called makefiles which specify how to derive the target program. Which of the following standard graph algorithms is used by Make.- a)Strongly Connected Components
- b)Topological Sorting
- c)Breadth First Search
- d)Dijkstra's Shortest Path
Correct answer is option 'B'. Can you explain this answer?
Make is a utility that automatically builds executable programs and libraries from source code by reading files called makefiles which specify how to derive the target program. Which of the following standard graph algorithms is used by Make.
a)
Strongly Connected Components
b)
Topological Sorting
c)
Breadth First Search
d)
Dijkstra's Shortest Path
|
Tarun Saini answered |
Make can decide the order of building a software using topological sorting. Topological sorting produces the order considering all dependencies provide by makefile. See following for details. Topological Sorting
A probability density function is of the form 
The value of K is - a)0.5
- b)1
- c)0.5α
- d)α
Correct answer is option 'C'. Can you explain this answer?
A probability density function is of the form
The value of K is
a)
0.5
b)
1
c)
0.5α
d)
α
|
|
Ravi Singh answered |
As (x) is a probability density function
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?
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
|
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
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
(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
(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
Can you explain the answer of this question below:Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below
Given that a computer is defective, the probability that it was supplied by Y is
- A:
0. 1
- B:
0.2
- C:
0.3
- D:
0.4
The answer is d.
Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below
Given that a computer is defective, the probability that it was supplied by Y is
0. 1
0.2
0.3
0.4
|
|
Kabir Verma answered |
Probability of defective computer supplied by Y =
(Case when Y produces defective)/(All cases of producing defective product)
Case when Y produces defective = (0.3)(0.02) = 0.006
All cases of producing defective product= (0.6x0.01)+(0.3x0.02)
(0.1x0.03)= 0.006+0.006+0.003=0.015
(Case when Y produces defective)/(All cases of producing defective product)
Case when Y produces defective = (0.3)(0.02) = 0.006
All cases of producing defective product= (0.6x0.01)+(0.3x0.02)
(0.1x0.03)= 0.006+0.006+0.003=0.015
Probability = 0.006/0.015=0.4
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?
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
|
|
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:
Can you explain the answer of this question below:Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE?
- A:
E (XY) = E (X) E (Y)
- B:
Cov (X, Y) = 0
- C:
Var (X + Y) = Var (X) + Var (Y)
- D:
E (X2 y2) = (E (X))2 (E (y))2
The answer is b.
Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE?
E (XY) = E (X) E (Y)
Cov (X, Y) = 0
Var (X + Y) = Var (X) + Var (Y)
E (X2 y2) = (E (X))2 (E (y))2
|
Sooraj Krishnan answered |
Option d is the correct one as all the other options hold true
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?
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.
|
|
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.
Which one of the following is the most appropriate logical formula to represent the statement? "Gold and silver ornaments are precious". The following notations are used: G(x): x is a gold ornament S(x): x is a silver ornament P(x): x is precious- a)∀x(P(x)→(G(x)∧S(x)))
- b)∀x((G(x)∧S(x))→P(x))
- c)∃x((G(x)∧S(x))→P(x)
- d)∀x((G(x)∨S(x))→P(x))
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is the most appropriate logical formula to represent the statement? "Gold and silver ornaments are precious". The following notations are used: G(x): x is a gold ornament S(x): x is a silver ornament P(x): x is precious
a)
∀x(P(x)→(G(x)∧S(x)))
b)
∀x((G(x)∧S(x))→P(x))
c)
∃x((G(x)∧S(x))→P(x)
d)
∀x((G(x)∨S(x))→P(x))
|
Anirban Khanna answered |
=> This statement can be expressed as
=> For all X, x can be either gold or silver then the ornament X is precious
=> For all X, (G(X) v S(x)) => P(X).
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?
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
|
|
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.
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?
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:
(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
|
|
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
= 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 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?
"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)
|
|
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.
So option (a) is (X∧¬Z)→Y = ¬((X∧¬Z))∨Y = (¬X∨Z)∨Y, which matches our expression. So option A is correct.
We need to choose a team of 11 from a pool of 15 players and also select a captain. The number of different ways this can be done is- a)
- b)
- c)15 . 14 . 13 . 12 . 11 .10 . 9 . 8 . 7 . 6 . 5
- d)(15 . 14 . 13 . 12 . 11 .10 . 9 . 8 . 7 . 6 . 5) . 11
Correct answer is option 'B'. Can you explain this answer?
We need to choose a team of 11 from a pool of 15 players and also select a captain. The number of different ways this can be done is
a)
b)
c)
15 . 14 . 13 . 12 . 11 .10 . 9 . 8 . 7 . 6 . 5
d)
(15 . 14 . 13 . 12 . 11 .10 . 9 . 8 . 7 . 6 . 5) . 11
|
|
Ajit Kumar answered |
11 players can be selected in 15C11 ways and then the captain can be selected in 11 ways. so total numbers of ways is 11*15C11
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?
Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
a)
b)
c)
d)
|
|
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
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
When a coin is tossed, the probability of getting a Head is 
. Let be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is- a)1/p
- b)1/(1 - p)
- c)1/p2
- d)1/(1 - p2)
Correct answer is option 'A'. Can you explain this answer?
When a coin is tossed, the probability of getting a Head is . Let be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is
. Let be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N isa)
1/p
b)
1/(1 - p)
c)
1/p2
d)
1/(1 - p2)
|
|
Yash Patel answered |
multiply both side with and subtract:
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?
If matrix and X2 - X + I = 0 (I is the identity matrix and is the zero matrix), then the inverse of X is
and X2 - X + I = 0 (I is the identity matrix and is the zero matrix), then the inverse of X isa)
b)
c)
d)
|
|
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
=> 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)..
Chapter doubts & questions for Discrete Mathematics - GATE Computer Science Engineering(CSE) 2025 Mock Test Series 2024 is part of Computer Science Engineering (CSE) exam preparation. The chapters have been prepared according to the Computer Science Engineering (CSE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Discrete Mathematics - GATE Computer Science Engineering(CSE) 2025 Mock Test Series in English & Hindi are available as part of Computer Science Engineering (CSE) exam.
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup