Sets-And Relation
Question 1 |
Let R be the set of all binary relations on the set {1,2,3}. Suppose a relation is chosen from R at random. The probability that the chosen relation is reflexive (round off to 3 decimal places) is _____.
0.125 |
The number of reflexive relations is 2^(n^2-n).
The total number of relations on a set with n elements is 2^ (n^2).
The probability of choosing the reflexive relation out of set of relations is
= 2^(n^2-n) /2^ (n^2)
= 2^( n^2-n- n^2)
= 2^(-n)
Given n=3, the probability will be 2-n = ⅛ = 0.125
Question 2 |
Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x ≤ y, where ⇒ stands for logical implication. Which of the following statements CANNOT be true?
P(x) = True for all x ∈ S such that x ≠ b
| |
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c | |
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c | |
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x |
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
Question 3 |
Let A, B and C be non-empty sets and let X = (A - B) - C and Y = (A - C) - (B - C). Which one of the following is TRUE?
X = Y | |
X ⊂ Y | |
Y ⊂ X | |
None of these |
B = {1, 3, 4, 5}
C = {2, 4, 5, 6}
X = (A - B) - C
X = {2, 6} - {2, 4, 5, 6}
= ∅
Y = (A - C) - (B - C)
= {1, 3} - { 1, 3}
= ∅
X = Y
X = (A - B) - C
= (1, 5) - (5, 7, 4, 3)
= (1)
Y = (A - C) - (B - C)
= (1, 4) - (2, 4)
= (1)
X = Y
Question 4 |
The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively:
3 and 13 | |
2 and 11 | |
4 and 13 | |
8 and 14 |
Inverse of 4 = m; Inverse of 7 = n
(4×m)%15=1; (7*n)%15=1
Option A: m=3 n=13
12%15≠1 (✖️) 91%15=1 (✔️)
Option B: m=2 n=11
8%15≠1 (✖️) 11%15≠1 (✖️)
Option C: m=4 n=13
16%15=1(✔️) 91%15=1 (✔️)
Option D: m=8 n=14
120%15≠1(✖️) 98%15≠1(✖️)
Question 5 |
The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?
It is not closed | |
2 does not have an inverse | |
3 does not have an inverse | |
8 does not have an inverse |
Option A:
It is not closed under multiplication. After multiplication modulo (10) we get ‘0’. The ‘0’ is not present in the set.
(2*5)%10 ⇒ 10%10 = 0
Option B:
2 does not have an inverse such as
(2*x)%10 ≠ 1
Option C:
3 have an inverse such that
(3*7)%10 = 1
Option D:
8 does not have an inverse such that
(8*x)%10 ≠ 1
Question 6 |
Consider the set S = {a,b,c,d}. Consider the following 4 partitions π1, π2, π3, π4 on 
Let p be the partial order on the set of partitions S' = {π1, π2, π3, π4} defined as follows: πi p πj if and only if πi refines πj. The poset diagram for (S', p) is:
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And, neither π2 refines π3, nor π3 refines π2.
Here, only π1 refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.