All questions of Discrete & Engineering 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
The CORRECT formula for the sentence, "not all rainy days are cold” is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The CORRECT formula for the sentence, "not all rainy days are cold” is
a)
b)
c)
d)
|
Sanvi Kapoor answered |
In other words it says ``Some rainy days are not cold"
Given statement is
¬∀d[R(d)→C(d)]
¬∀d[R(d)→C(d)]
≡¬∀d[¬R(d)∨C(d)
≡∃d[R(d)∧¬C(d)]
≡∃d[R(d)∧¬C(d)]
Minimum of the real valued function f(x) = (x-1)2/3 occurs at x equal to- a)- ∝
- b)0
- c)1
- d)∝
Correct answer is option 'C'. Can you explain this answer?
Minimum of the real valued function f(x) = (x-1)2/3 occurs at x equal to
a)
- ∝
b)
0
c)
1
d)
∝
|
Kunal Gupta answered |
As f(x) is square of hence its minimum value be 0 where at x = 1.
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
Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is - a)1/36
- b)1/3
- c)25/36
- d)11/36
Correct answer is option 'D'. Can you explain this answer?
Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is
a)
1/36
b)
1/3
c)
25/36
d)
11/36
|
Ravi Singh answered |
P(atleast one of dice will have 6 facing
= 1 - P(none of dice have 6 facing up)
= 1 - P(none of dice have 6 facing up)
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
|
Machine Experts answered |
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
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)
|
Pathways Academy answered |
Given Px (x ) is the probability density function for the random variable X.
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
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
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
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?
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
|
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
Condition for monoid is __________
- a)(a+e)=a
- b)(a*e)=(a+e)
- c)a=(a*(a+e)
- d)(a*e) = (e*a) = a
Correct answer is option 'D'. Can you explain this answer?
Condition for monoid is __________
a)
(a+e)=a
b)
(a*e)=(a+e)
c)
a=(a*(a+e)
d)
(a*e) = (e*a) = a
|
|
Sudhir Patel answered |
A Semigroup (S,*) is defined as a monoid if there exists an element e in S such that (a*e) = (e*a) = a for all a in S. This element is called identity element of S w.r.t *.
Consider the binary relation: R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?- a)R is symmetric but NOT antisymmetric
- b)R is NOT symmetric but antisymmetric
- c)R is both symmetric and antisymmetric
- d)R is neither symmetric nor antisymmetric
Correct answer is option 'D'. Can you explain this answer?
Consider the binary relation: R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?
a)
R is symmetric but NOT antisymmetric
b)
R is NOT symmetric but antisymmetric
c)
R is both symmetric and antisymmetric
d)
R is neither symmetric nor antisymmetric
|
Rajdeep Mehra answered |
∴ R is not antisymmetric.
R is neither-symmetric nor anti-symmetric.
If M is a square matrix with a zero determinant, which of the following assertion(S) is (are) correct?
S1: Each row of M can be represented as a linear combination of the other rows.
S2: Each column of M can be represented as a linear combination of the other columns.
S3: MX = O has a nontrivial solution.
S4: M has an inverse.- a)S3 and S2 only
- b)S1 and S4 only
- c)S1 and S3 only
- d)S1, S2 and S3 only
Correct answer is option 'D'. Can you explain this answer?
If M is a square matrix with a zero determinant, which of the following assertion(S) is (are) correct?
S1: Each row of M can be represented as a linear combination of the other rows.
S2: Each column of M can be represented as a linear combination of the other columns.
S3: MX = O has a nontrivial solution.
S4: M has an inverse.
S1: Each row of M can be represented as a linear combination of the other rows.
S2: Each column of M can be represented as a linear combination of the other columns.
S3: MX = O has a nontrivial solution.
S4: M has an inverse.
a)
S3 and S2 only
b)
S1 and S4 only
c)
S1 and S3 only
d)
S1, S2 and S3 only
|
Sreemoyee Singh answered |
S1 and S2:
Since M has zero determinant, its rank is not full i.e. if M is of size 3 x 3 , then its rank is not 3. So there is a linear combination of rows which evaluates to 0 i.e. k1R1 + k2R2 +...+knRn = 0 and there is a linear combination of columns which evaluates to 0 i.e.
Now any row Ri can be written as linear combination of other rows as:
Similar is the case for columns.
So S1 and S2 are true.
Since M has zero determinant, its rank is not full i.e. if M is of size 3 x 3 , then its rank is not 3. So there is a linear combination of rows which evaluates to 0 i.e. k1R1 + k2R2 +...+knRn = 0 and there is a linear combination of columns which evaluates to 0 i.e.
Now any row Ri can be written as linear combination of other rows as:
Similar is the case for columns.
So S1 and S2 are true.
Consider the set ∑* of all strings over the alphabet ∑ = {0,1}. ∑* with the concatenation operator for strings- a)Does not form a group
- b)Forms a non-commutative group
- c)Does not have a right identity element
- d)Forms a group if the empty string is removed from ∑*
Correct answer is option 'A'. Can you explain this answer?
Consider the set ∑* of all strings over the alphabet ∑ = {0,1}. ∑* with the concatenation operator for strings
a)
Does not form a group
b)
Forms a non-commutative group
c)
Does not have a right identity element
d)
Forms a group if the empty string is removed from ∑*
|
|
Gopal Gupta answered |
Let A and B be sets and let Ac and Bc denote the complements of the sets A and B. The set (A - B) ∪ (B - A) ∪ (A ∩ B) is equal to- a)A ∪ B
- b)Ac ∪ Bc
- c)A ∩ B
- d)Ac ∩ Bc
Correct answer is option 'A'. Can you explain this answer?
Let A and B be sets and let Ac and Bc denote the complements of the sets A and B. The set (A - B) ∪ (B - A) ∪ (A ∩ B) is equal to
a)
A ∪ B
b)
Ac ∪ Bc
c)
A ∩ B
d)
Ac ∩ Bc
|
Rajdeep Nambiar answered |
Representing above set using Venn diagram as follows:
∴ By Venn diagram,
Suppose X and V are sets and |X| and |Y| are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. From this one can conclude that- a)|X| = 1, |Y| = 97
- b)|X| = 97, |Y| = 1
- c)|X| = 97, |Y| = 97
- d) None of the above
Correct answer is option 'A'. Can you explain this answer?
Suppose X and V are sets and |X| and |Y| are their respective cardinalities. It is given that there are exactly 97 functions from X to Y. From this one can conclude that
a)
|X| = 1, |Y| = 97
b)
|X| = 97, |Y| = 1
c)
|X| = 97, |Y| = 97
d)
None of the above
|
|
Nayanika Deshpande answered |
X and Y are sets.
The card in a lities of X and Y are |X| and |Y| respectively.
The card in a lities of X and Y are |X| and |Y| respectively.
Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even, is - a)0.5
- b)0.25
- c)0.167
- d)0.125
Correct answer is option 'A'. Can you explain this answer?
Two dices are rolled simultaneously. The probability that the sum of digits on the top surface of the two dices is even, is
a)
0.5
b)
0.25
c)
0.167
d)
0.125
|
Dipika Bose answered |
Here sample space S= 6 × 6 = 36
Total no. of way in which sum of digits on the top surface of the two dice is is even is 18.
∴ The require probability = 0.5
If three coins are tossed simultaneously, the probability of getting at least one head is - a)1/8
- b)3/8
- c)1/2
- d)7/8
Correct answer is option 'D'. Can you explain this answer?
If three coins are tossed simultaneously, the probability of getting at least one head is
a)
1/8
b)
3/8
c)
1/2
d)
7/8
|
|
Pallabi Kulkarni answered |
Understanding the Problem
When tossing three coins, we want to find the probability of getting at least one head.
Sample Space
The sample space (all possible outcomes) when tossing three coins is:
- HHH
- HHT
- HTH
- HTT
- THH
- THT
- TTH
- TTT
This gives us a total of 2^3 = 8 outcomes.
Calculating Desired Outcomes
To find the probability of getting at least one head, it's often easier to calculate the probability of the complementary event — getting no heads (i.e., all tails).
- The only outcome for no heads is: TTT
Thus, there is 1 outcome where we get no heads.
Probability of No Heads
The probability of getting no heads (all tails) is:
- P(no heads) = Number of favorable outcomes for no heads / Total outcomes = 1/8
Calculating Probability of At Least One Head
Now, we can find the probability of getting at least one head:
- P(at least one head) = 1 - P(no heads) = 1 - 1/8 = 7/8
Final Answer
Thus, the probability of getting at least one head when tossing three coins is:
- 7/8
Therefore, the correct answer is option 'D'.
When tossing three coins, we want to find the probability of getting at least one head.
Sample Space
The sample space (all possible outcomes) when tossing three coins is:
- HHH
- HHT
- HTH
- HTT
- THH
- THT
- TTH
- TTT
This gives us a total of 2^3 = 8 outcomes.
Calculating Desired Outcomes
To find the probability of getting at least one head, it's often easier to calculate the probability of the complementary event — getting no heads (i.e., all tails).
- The only outcome for no heads is: TTT
Thus, there is 1 outcome where we get no heads.
Probability of No Heads
The probability of getting no heads (all tails) is:
- P(no heads) = Number of favorable outcomes for no heads / Total outcomes = 1/8
Calculating Probability of At Least One Head
Now, we can find the probability of getting at least one head:
- P(at least one head) = 1 - P(no heads) = 1 - 1/8 = 7/8
Final Answer
Thus, the probability of getting at least one head when tossing three coins is:
- 7/8
Therefore, the correct answer is option 'D'.
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times? - a)1/4
- b)3/8
- c)1/2
- d)3/4
Correct answer is option 'A'. Can you explain this answer?
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?
a)
1/4
b)
3/8
c)
1/2
d)
3/4
|
|
Mrinalini Sharma answered |
Probability of getting exactly three heads
The minimum value of the function f(x) = x3-3x2 - 24x + 100 in the interval [-3, 3] is- a)20
- b)28
- c)16
- d)32
Correct answer is option 'B'. Can you explain this answer?
The minimum value of the function f(x) = x3-3x2 - 24x + 100 in the interval [-3, 3] is
a)
20
b)
28
c)
16
d)
32
|
Priyanka Bose answered |
Hence f(x) has minimum value at x = 3 which is 28.
The probability that two friends share the same birth-month is - a)1/6
- b)1/12
- c)1/144
- d)1/24
Correct answer is option 'B'. Can you explain this answer?
The probability that two friends share the same birth-month is
a)
1/6
b)
1/12
c)
1/144
d)
1/24
|
|
Atharva Majumdar answered |
Let A = the event that the birth month of first friend
And B= that of second friend.
∴ P( A )= 1, as 1st friend can born in any month
and P(B) = 1/12, by the condition.
∴ Probability of two friends share same birth-month
Let AX - B be a system of linear equations where A is an m x n matrix and b is a m x 1 column vector and X is an x 1 column vector of unknown. Which of the following is false?- a)The system has a solution if and only if, both A and the augmented matrix [A, B] have the same rank.
- b)If m < n and B is the zero vector, then the system has infinitely many solutions.
- c)If m = n and B is non-zero vector, then the system has a unique solution.
- d)The system will have only a trivial solution when m - n, B is the zero vector and rank (A) = n.
Correct answer is option 'C'. Can you explain this answer?
Let AX - B be a system of linear equations where A is an m x n matrix and b is a m x 1 column vector and X is an x 1 column vector of unknown. Which of the following is false?
a)
The system has a solution if and only if, both A and the augmented matrix [A, B] have the same rank.
b)
If m < n and B is the zero vector, then the system has infinitely many solutions.
c)
If m = n and B is non-zero vector, then the system has a unique solution.
d)
The system will have only a trivial solution when m - n, B is the zero vector and rank (A) = n.
|
|
Anushka Sengupta answered |
Following are the possibilities for a system of linear equations:
(i) If matrix A and augmented matrix [AB] have same rank, then the system has solution otherwise there is no solution.
(ii) If matrix A and augmented matrix [AB] have same rank which is equal to the number of variables, then the system has unique solution and if B is zero vector then the system have only a trivial solution.
(iii) If matrix A and matrix [AB] have same rank which is less than the number of variables, then the system has infinite solution.
Therefore, option (c) is false because if m - n and B is non-zero vector, then it is not necessary that system has a unique solution, because m is the number of equations (quantity) and not the number of linearly independent equations (quality).
(i) If matrix A and augmented matrix [AB] have same rank, then the system has solution otherwise there is no solution.
(ii) If matrix A and augmented matrix [AB] have same rank which is equal to the number of variables, then the system has unique solution and if B is zero vector then the system have only a trivial solution.
(iii) If matrix A and matrix [AB] have same rank which is less than the number of variables, then the system has infinite solution.
Therefore, option (c) is false because if m - n and B is non-zero vector, then it is not necessary that system has a unique solution, because m is the number of equations (quantity) and not the number of linearly independent equations (quality).
The binary relation S = f (empty set) on set A = {1,2, 3} is- a)Neither reflexive nor symmetric
- b)Symmetric and reflexive
- c)Transitive and reflexive
- d)Transitive and symmetric
Correct answer is option 'D'. Can you explain this answer?
The binary relation S = f (empty set) on set A = {1,2, 3} is
a)
Neither reflexive nor symmetric
b)
Symmetric and reflexive
c)
Transitive and reflexive
d)
Transitive and symmetric
|
|
Partho Yadav answered |
The empty relation on any set is always transitive and symmetric but not reflexive.
What is the maximum number of edges in an acyclic undirected graph with n vertices?- a)n - 1
- b)n
- c)n + 1
- d)2n - 1
Correct answer is option 'A'. Can you explain this answer?
What is the maximum number of edges in an acyclic undirected graph with n vertices?
a)
n - 1
b)
n
c)
n + 1
d)
2n - 1
|
Gaurav Verma answered |
Introduction:
An acyclic undirected graph is a graph that does not contain any cycles, which means there are no closed loops in the graph. In such a graph, the maximum number of edges can be determined based on the number of vertices.
Explanation:
To find the maximum number of edges in an acyclic undirected graph with n vertices, we need to consider that each vertex can be connected to all other vertices except itself and its adjacent vertices.
Maximum Number of Edges:
Let's consider a specific vertex in the graph. This vertex can be connected to all the remaining (n-1) vertices. However, it cannot be connected to itself or its adjacent vertices since it is an acyclic graph.
So, for each vertex, the maximum number of edges that can be added is (n-1) - 1 = n - 2.
Total Number of Edges:
Since there are n vertices in the graph, we need to sum up the number of edges for each vertex.
Total number of edges = n * (n - 2)
However, this counts each edge twice, once for each vertex it is connected to. Therefore, we need to divide the total number of edges by 2 to get the actual number of edges.
Total number of edges = n * (n - 2) / 2
Simplification:
Let's simplify the equation further:
n * (n - 2) / 2 = n * n / 2 - 2 * n / 2 = n^2 / 2 - n
Conclusion:
Hence, the maximum number of edges in an acyclic undirected graph with n vertices is n^2 / 2 - n, which can be approximated to n - 1.
Therefore, the correct answer is option 'A' (n - 1).
An acyclic undirected graph is a graph that does not contain any cycles, which means there are no closed loops in the graph. In such a graph, the maximum number of edges can be determined based on the number of vertices.
Explanation:
To find the maximum number of edges in an acyclic undirected graph with n vertices, we need to consider that each vertex can be connected to all other vertices except itself and its adjacent vertices.
Maximum Number of Edges:
Let's consider a specific vertex in the graph. This vertex can be connected to all the remaining (n-1) vertices. However, it cannot be connected to itself or its adjacent vertices since it is an acyclic graph.
So, for each vertex, the maximum number of edges that can be added is (n-1) - 1 = n - 2.
Total Number of Edges:
Since there are n vertices in the graph, we need to sum up the number of edges for each vertex.
Total number of edges = n * (n - 2)
However, this counts each edge twice, once for each vertex it is connected to. Therefore, we need to divide the total number of edges by 2 to get the actual number of edges.
Total number of edges = n * (n - 2) / 2
Simplification:
Let's simplify the equation further:
n * (n - 2) / 2 = n * n / 2 - 2 * n / 2 = n^2 / 2 - n
Conclusion:
Hence, the maximum number of edges in an acyclic undirected graph with n vertices is n^2 / 2 - n, which can be approximated to n - 1.
Therefore, the correct answer is option 'A' (n - 1).
Aishwarya studies either computer science or mathematics everyday. if the studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday? - a)0.24
- b)0.36
- c)0.4
- d)0.6
Correct answer is option 'C'. Can you explain this answer?
Aishwarya studies either computer science or mathematics everyday. if the studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
a)
0.24
b)
0.36
c)
0.4
d)
0.6
|
|
Bhavya Ahuja answered |
Given information:
- Aishwarya studies either computer science or mathematics every day.
- If Aishwarya studies computer science on a day, the probability that she studies mathematics the next day is 0.6.
- If Aishwarya studies mathematics on a day, the probability that she studies computer science the next day is 0.4.
We are asked to find the probability that Aishwarya studies computer science on Wednesday, given that she studies computer science on Monday.
To solve this problem, we can use conditional probability.
Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted by P(A|B), where A and B are events.
In this case, we are interested in finding the probability of Aishwarya studying computer science on Wednesday (event A), given that she studies computer science on Monday (event B). So, we need to find P(A|B).
Applying Conditional Probability:
We know that Aishwarya studies either computer science or mathematics every day. So, we can consider these two events as the sample space.
Let's define the events:
- C: Aishwarya studies computer science
- M: Aishwarya studies mathematics
We are given that Aishwarya studies computer science on Monday (event B), so we are interested in finding the probability of Aishwarya studying computer science on Wednesday (event A).
We need to find P(C|C), which is the probability of Aishwarya studying computer science on Wednesday, given that she studied computer science on Monday.
We can use the formula for conditional probability:
P(C|C) = P(C and C) / P(C)
Finding P(C and C):
Since Aishwarya studies either computer science or mathematics every day, the events C and M are mutually exclusive. So, P(C and C) is equal to P(C).
Finding P(C):
We are given that Aishwarya studies computer science on Monday. So, P(C) = 1.
Calculating the Probability:
Using the formula for conditional probability, we have:
P(C|C) = P(C and C) / P(C)
P(C|C) = P(C) / P(C)
P(C|C) = 1 / 1
P(C|C) = 1
Therefore, the probability that Aishwarya studies computer science on Wednesday, given that she studies computer science on Monday, is 1 or 100%.
Answer:
The probability that Aishwarya studies computer science on Wednesday is 0.4 (or 40%). Therefore, the correct answer is option C) 0.4.
- Aishwarya studies either computer science or mathematics every day.
- If Aishwarya studies computer science on a day, the probability that she studies mathematics the next day is 0.6.
- If Aishwarya studies mathematics on a day, the probability that she studies computer science the next day is 0.4.
We are asked to find the probability that Aishwarya studies computer science on Wednesday, given that she studies computer science on Monday.
To solve this problem, we can use conditional probability.
Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted by P(A|B), where A and B are events.
In this case, we are interested in finding the probability of Aishwarya studying computer science on Wednesday (event A), given that she studies computer science on Monday (event B). So, we need to find P(A|B).
Applying Conditional Probability:
We know that Aishwarya studies either computer science or mathematics every day. So, we can consider these two events as the sample space.
Let's define the events:
- C: Aishwarya studies computer science
- M: Aishwarya studies mathematics
We are given that Aishwarya studies computer science on Monday (event B), so we are interested in finding the probability of Aishwarya studying computer science on Wednesday (event A).
We need to find P(C|C), which is the probability of Aishwarya studying computer science on Wednesday, given that she studied computer science on Monday.
We can use the formula for conditional probability:
P(C|C) = P(C and C) / P(C)
Finding P(C and C):
Since Aishwarya studies either computer science or mathematics every day, the events C and M are mutually exclusive. So, P(C and C) is equal to P(C).
Finding P(C):
We are given that Aishwarya studies computer science on Monday. So, P(C) = 1.
Calculating the Probability:
Using the formula for conditional probability, we have:
P(C|C) = P(C and C) / P(C)
P(C|C) = P(C) / P(C)
P(C|C) = 1 / 1
P(C|C) = 1
Therefore, the probability that Aishwarya studies computer science on Wednesday, given that she studies computer science on Monday, is 1 or 100%.
Answer:
The probability that Aishwarya studies computer science on Wednesday is 0.4 (or 40%). Therefore, the correct answer is option C) 0.4.
An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches – 0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained 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 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained all these students is
a)
0
b)
2550
c)
7525
d)
9375
|
|
Nandini Basak answered |
Ans.
Method to Solve :
Probability of choosing the correct option = 1/4
Probability of choosing a wrong option = 3/4
So, expected mark for a question for a student = 1/4�1+3/4�(−0.25)=1/16
Expected mark for a student for 150questions =1/16�150=9.375
So, sum total of the expected marks obtained by all 1000students = 9.375�1000=9375
Probability of choosing a wrong option = 3/4
So, expected mark for a question for a student = 1/4�1+3/4�(−0.25)=1/16
Expected mark for a student for 150questions =1/16�150=9.375
So, sum total of the expected marks obtained by all 1000students = 9.375�1000=9375
A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is - a)1/90
- b)1/2
- c)19/90
- d)2/9
Correct answer is option 'D'. Can you explain this answer?
A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is
a)
1/90
b)
1/2
c)
19/90
d)
2/9
|
Avik Ghosh answered |
The probability of drawing two red balls without replacement
Consider the following set of equations:
x + 2y = 5
4x + 8y = 12
3x + 6y + 3z = 15
This set- a)Has a unique solution
- b)Has no solution
- c)Has finite number of solutions
- d)Has infinite number of solutions
Correct answer is option 'B'. Can you explain this answer?
Consider the following set of equations:
x + 2y = 5
4x + 8y = 12
3x + 6y + 3z = 15
This set
x + 2y = 5
4x + 8y = 12
3x + 6y + 3z = 15
This set
a)
Has a unique solution
b)
Has no solution
c)
Has finite number of solutions
d)
Has infinite number of solutions
|
|
Siddharth Saha answered |
Set of equations is
Above set of equations can be written as
Augmented matrix [AB] is given as
Performing gauss-Elimination on the above matrix
Above set of equations can be written as
Augmented matrix [AB] is given as
Performing gauss-Elimination on the above matrix
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between.- a)k and n
- b) k - 1 and k + 1
- c)k - 1 and n - 1
- d)k + 1 and n - k
Correct answer is option 'C'. Can you explain this answer?
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between.
a)
k and n
b)
k - 1 and k + 1
c)
k - 1 and n - 1
d)
k + 1 and n - k
|
|
Aashna Kaur answered |
Maximum components will result after removal of a node if graph G is a star graph as shown below.
or a null graph of n vertices as shown below:
In either case, if node vis removed, the number of components will be n - 1, where, n is the total number of nodes in the star graph.
∴ n - 1 is the maximum number of components possible. Minimum components will result if the node being removed is a lone vertex in which case, the number of components will be k - 1.
∴ The number of components must necessarily lie between k - 1 and n - 1.
or a null graph of n vertices as shown below:
In either case, if node vis removed, the number of components will be n - 1, where, n is the total number of nodes in the star graph.
∴ n - 1 is the maximum number of components possible. Minimum components will result if the node being removed is a lone vertex in which case, the number of components will be k - 1.
∴ The number of components must necessarily lie between k - 1 and n - 1.
The probability that there are 53 Sundays in a randomly chosen leap year is - a)1/7
- b)1/14
- c)1/28
- d)2/7
Correct answer is option 'D'. Can you explain this answer?
The probability that there are 53 Sundays in a randomly chosen leap year is
a)
1/7
b)
1/14
c)
1/28
d)
2/7
|
Gauri Roy answered |
Problem:
The probability that there are 53 Sundays in a randomly chosen leap year is:
a) 1/7
b) 1/14
c) 1/28
d) 2/7
The correct answer is option 'D'. Let's explain why it is the correct answer.
Leap Year:
A leap year is a year that is exactly divisible by 4, except for years that are divisible by 100. However, years that are divisible by 400 are also leap years. This means that every 4 years, there is an extra day, February 29th, in the calendar.
Number of Days in a Leap Year:
In a leap year, there are 366 days instead of the usual 365 days. This is because of the extra day, February 29th.
Number of Weeks in a Leap Year:
Since there are 7 days in a week, the total number of weeks in a leap year is given by:
Total number of weeks = Number of days / Number of days in a week
= 366 days / 7 days per week
= 52 weeks and 2 days
Number of Sundays in a Leap Year:
To find the probability of having 53 Sundays in a leap year, we need to determine the number of Sundays that can occur in a leap year.
In a leap year, the first day of the year can be any day of the week. Let's consider each possibility:
1) If the first day of the year is a Sunday, then there will be 53 Sundays in the year.
2) If the first day of the year is a Monday, then there will be 53 Sundays in the year.
3) If the first day of the year is a Tuesday, then there will be 52 Sundays in the year.
4) If the first day of the year is a Wednesday, then there will be 52 Sundays in the year.
5) If the first day of the year is a Thursday, then there will be 52 Sundays in the year.
6) If the first day of the year is a Friday, then there will be 52 Sundays in the year.
7) If the first day of the year is a Saturday, then there will be 52 Sundays in the year.
In options a), b), and c), the probability given is less than 1/7, which means it is incorrect. The only option left is option d), which is 2/7. This is the correct answer.
Conclusion:
The probability that there are 53 Sundays in a randomly chosen leap year is 2/7, which is the correct answer (option 'D').
The probability that there are 53 Sundays in a randomly chosen leap year is:
a) 1/7
b) 1/14
c) 1/28
d) 2/7
The correct answer is option 'D'. Let's explain why it is the correct answer.
Leap Year:
A leap year is a year that is exactly divisible by 4, except for years that are divisible by 100. However, years that are divisible by 400 are also leap years. This means that every 4 years, there is an extra day, February 29th, in the calendar.
Number of Days in a Leap Year:
In a leap year, there are 366 days instead of the usual 365 days. This is because of the extra day, February 29th.
Number of Weeks in a Leap Year:
Since there are 7 days in a week, the total number of weeks in a leap year is given by:
Total number of weeks = Number of days / Number of days in a week
= 366 days / 7 days per week
= 52 weeks and 2 days
Number of Sundays in a Leap Year:
To find the probability of having 53 Sundays in a leap year, we need to determine the number of Sundays that can occur in a leap year.
In a leap year, the first day of the year can be any day of the week. Let's consider each possibility:
1) If the first day of the year is a Sunday, then there will be 53 Sundays in the year.
2) If the first day of the year is a Monday, then there will be 53 Sundays in the year.
3) If the first day of the year is a Tuesday, then there will be 52 Sundays in the year.
4) If the first day of the year is a Wednesday, then there will be 52 Sundays in the year.
5) If the first day of the year is a Thursday, then there will be 52 Sundays in the year.
6) If the first day of the year is a Friday, then there will be 52 Sundays in the year.
7) If the first day of the year is a Saturday, then there will be 52 Sundays in the year.
In options a), b), and c), the probability given is less than 1/7, which means it is incorrect. The only option left is option d), which is 2/7. This is the correct answer.
Conclusion:
The probability that there are 53 Sundays in a randomly chosen leap year is 2/7, which is the correct answer (option 'D').
In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses. How many students have not taken any of the three courses?- a)15
- b)20
- c)25
- d)30
Correct answer is option 'C'. Can you explain this answer?
In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses. How many students have not taken any of the three courses?
a)
15
b)
20
c)
25
d)
30
|
Siddharth Basu answered |
According to inclusion - exclusion formula:
PL = students who have taken programming.
DS = Students who have taken Data structures.
CO = Students who have taken Computer Organization.
So, the number of students who have taken atleast 1 of the 3 courses is given by:
= 125 + 85 + 65 - 50 - 35 - 30 + 15
= 175
Therefore, the number of students who have not taken any of the 3 courses is
= Total students-students taken atleast 1 course
= 200-175 = 25
The following propositional statement is
(P → (Q v R)) → (( P ∧ Q) → R)- a)Satisfiable but not valid
- b)Valid
- c)A contraditiction
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The following propositional statement is
(P → (Q v R)) → (( P ∧ Q) → R)
(P → (Q v R)) → (( P ∧ Q) → R)
a)
Satisfiable but not valid
b)
Valid
c)
A contraditiction
d)
None of the above
|
|
Naina Banerjee answered |
(P → (Q v R)) → ((P ∧ Q) → R)
≡ (P→ Q+ R) → (PQ → R)
≡ [P' + Q + R] → [(PQ)' + R]
≡ [P' + Q + R] → [P' + Q' + R]
≡ (P' + Q + R)' + P' + Q' + R
≡ PQ' R' + P' + Q' + R
≡ Q' + Q' PR' + P' + R
≡ Q' + P' + R (by absorption law)
Which is a contingency (i.e. satisfiable but not valid).
≡ (P→ Q+ R) → (PQ → R)
≡ [P' + Q + R] → [(PQ)' + R]
≡ [P' + Q + R] → [P' + Q' + R]
≡ (P' + Q + R)' + P' + Q' + R
≡ PQ' R' + P' + Q' + R
≡ Q' + Q' PR' + P' + R
≡ Q' + P' + R (by absorption law)
Which is a contingency (i.e. satisfiable but not valid).
If the rank of a (5 x 6) matrix Q is 4, then which one of the following statements is correct?- a)Q will have four linearly independent rows and four linearly independent columns
- b)Q will have four linearly independent rows and five linearly independent columns
- c)QQT will be invertible
- d)QTQ will be invertible
Correct answer is option 'A'. Can you explain this answer?
If the rank of a (5 x 6) matrix Q is 4, then which one of the following statements is correct?
a)
Q will have four linearly independent rows and four linearly independent columns
b)
Q will have four linearly independent rows and five linearly independent columns
c)
QQT will be invertible
d)
QTQ will be invertible
|
|
Prisha Desai answered |
If rank of (5 x 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as will as 4 linearly in dependent columns.
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is - a)1/3
- b)3/7
- c)1/2
- d)4/7
Correct answer is option 'C'. Can you explain this answer?
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is
a)
1/3
b)
3/7
c)
1/2
d)
4/7
|
|
Prerna Menon answered |
Problem:
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is:
a) 1/3
b) 3/7
c) 1/2
d) 4/7
Solution:
To find the probability that the second removed ball is red, given that the first removed ball is white, we can use conditional probability.
Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), where A and B are events.
In this problem, we want to find P(Red ball is selected second|White ball is selected first), which can be written as P(R|W).
Step 1: Find the Probability of Selecting a White Ball First
The probability of selecting a white ball first can be calculated by dividing the number of white balls by the total number of balls:
P(White ball is selected first) = Number of white balls / Total number of balls = 4 / 7
Step 2: Find the Probability of Selecting a Red Ball Second, Given that a White Ball is Selected First
Since we know that a white ball is selected first, the total number of balls is reduced by 1. Therefore, the probability of selecting a red ball second, given that a white ball is selected first, can be calculated by dividing the number of red balls by the reduced total number of balls:
P(Red ball is selected second|White ball is selected first) = Number of red balls / (Total number of balls - 1) = 3 / 6 = 1 / 2
Conclusion:
The probability that the second removed ball is red, given that the first removed ball is white, is 1/2. Therefore, the correct answer is option C.
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is:
a) 1/3
b) 3/7
c) 1/2
d) 4/7
Solution:
To find the probability that the second removed ball is red, given that the first removed ball is white, we can use conditional probability.
Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), where A and B are events.
In this problem, we want to find P(Red ball is selected second|White ball is selected first), which can be written as P(R|W).
Step 1: Find the Probability of Selecting a White Ball First
The probability of selecting a white ball first can be calculated by dividing the number of white balls by the total number of balls:
P(White ball is selected first) = Number of white balls / Total number of balls = 4 / 7
Step 2: Find the Probability of Selecting a Red Ball Second, Given that a White Ball is Selected First
Since we know that a white ball is selected first, the total number of balls is reduced by 1. Therefore, the probability of selecting a red ball second, given that a white ball is selected first, can be calculated by dividing the number of red balls by the reduced total number of balls:
P(Red ball is selected second|White ball is selected first) = Number of red balls / (Total number of balls - 1) = 3 / 6 = 1 / 2
Conclusion:
The probability that the second removed ball is red, given that the first removed ball is white, is 1/2. Therefore, the correct answer is option C.
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?
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
|
Deepika Deshpande answered |
Sum of degree of all vertices = 2e (using Handshaking lemma)
So, 6 x 2 + 3 x 4 + x x 3 = 2 x 27
12 + 12 + 3x = 54
3x = 54 - 24
x = 30/3
x = 10
So, total number of vertices
⇒ x + 6 + 3 = 10 + 6 + 3 = 19
So, 6 x 2 + 3 x 4 + x x 3 = 2 x 27
12 + 12 + 3x = 54
3x = 54 - 24
x = 30/3
x = 10
So, total number of vertices
⇒ x + 6 + 3 = 10 + 6 + 3 = 19
The probability of a defective piece being produced in a manufacturing process is 0.01. The probability that out of 5 successive pieces, only one is defective, is - a)(0. 99)2 (0.01)
- b)(0.99)( 0.01)4
- c)5× (0.99)(0.01)4
- d)5× (0.99)4(0.01)
Correct answer is option 'D'. Can you explain this answer?
The probability of a defective piece being produced in a manufacturing process is 0.01. The probability that out of 5 successive pieces, only one is defective, is
a)
(0. 99)2 (0.01)
b)
(0.99)( 0.01)4
c)
5× (0.99)(0.01)4
d)
5× (0.99)4(0.01)
|
|
Keerthana Joshi answered |
The required probability 
A monoid is called a group if _______- a)(a*a)=a=(a+c)
- b)(a*c)=(a+c)
- c)(a+c)=a
- d)(a*c)=(c*a)=e
Correct answer is option 'D'. Can you explain this answer?
A monoid is called a group if _______
a)
(a*a)=a=(a+c)
b)
(a*c)=(a+c)
c)
(a+c)=a
d)
(a*c)=(c*a)=e
|
|
Sudhir Patel answered |
A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.
Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z= min (X, Y), then the mean of Z is given by- a)(1/(α + β))
- b)min (α, β)
- c)(αβ/(α + β))
- d)α + β
Correct answer is option 'C'. Can you explain this answer?
Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z= min (X, Y), then the mean of Z is given by
a)
(1/(α + β))
b)
min (α, β)
c)
(αβ/(α + β))
d)
α + β
|
|
Ashwini Ghosh answered |
X and Y are two independent exponentially distributed random variables. Let λ1 and λ2 parameters of X and Y respectively.
Given, Z = min (X, Y)
Since mean of exponential distribution = 1/Parameter
So,
∴ Z is random variable with parameter
Mean of Z =
How many solutions does the following system of linear equations have?
- x + 5y = - 1
x - y = 2
x + 3y = 3- a)Infinitely many
- b)Two distinct solutions
- c)Unique
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
How many solutions does the following system of linear equations have?
- x + 5y = - 1
x - y = 2
x + 3y = 3
- x + 5y = - 1
x - y = 2
x + 3y = 3
a)
Infinitely many
b)
Two distinct solutions
c)
Unique
d)
None of these
|
Srestha Rane answered |
The augmented matrix is
Using gauss-efimination on above matrix we get,
Rank [A | B] - 2 (number of non zero rows in [A | B])
Rank [A ] = 2 (number of non zero row s in [A ])
Rank [A | B] = Rank [A] = 2 = number of variables
∴ Unique solution exists.
A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is- a)1/6
- b)3/8
- c)1/8
- d)1/2
Correct answer is option 'B'. Can you explain this answer?
A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is
a)
1/6
b)
3/8
c)
1/8
d)
1/2
|
Stuti Datta answered |
Probability of getting an odd number in rolling of a die = 3/6 = 1/2.
Now using binomial distribution
P(Exactly one odd number among three outcomes)
Now using binomial distribution
P(Exactly one odd number among three outcomes)
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