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If A = {2, 3}, B = {4, 5}, C = {5, 6}, then what is the number of elements in A × (B ∩ C)?
1. 2
2. 4
3. 6
4. 8
Correct Answer - Option 1 : 2Concept:Let A, and B be any two sets(A ∩ B) = {x, \(\rm x \in A \;\;and \;\; x \in B\)}A × B = {(x, y), \(\rm x \in A \;\;and \;\; y \in B\)} Calculations:Given, A = {2, 3}, B = {4, 5}, C = {5, 6}⇒(B ∩ C) = {5}A × (B ∩ C) = {(2, 5), (3, 5)}⇒ The number of elements in A ×Read more
Correct Answer – Option 1 : 2
Concept:
Let A, and B be any two sets
(A ∩ B) = {x, \(\rm x \in A \;\;and \;\; x \in B\)}
A × B = {(x, y), \(\rm x \in A \;\;and \;\; y \in B\)}
Calculations:
Given, A = {2, 3}, B = {4, 5}, C = {5, 6}
⇒(B ∩ C) = {5}
A × (B ∩ C) = {(2, 5), (3, 5)}
⇒ The number of elements in A × (B ∩ C) = 2
See lessHow many elements has P(A), if A = Φ?
We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2m. If A = Φ, then n(A) = 0. ∴ n[P(A)] = 20 = 1 Hence, P(A) has one element.
We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2m.
If A = Φ, then n(A) = 0.
∴ n[P(A)] = 20 = 1
Hence, P(A) has one element.
See lessIf X = {a, b, c, d} and Y = {f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
(i) X – Y = {a, c} (ii) Y – X = {f, g} (iii) X ∩ Y = {b, d}
(i) X – Y = {a, c}
(ii) Y – X = {f, g}
(iii) X ∩ Y = {b, d}
See lessIf A and B are two sets such that A ⊂ B, then what is A ∪ B?
If A and B are two sets such that A ⊂ B, then A ∪ B = B.
If A and B are two sets such that A ⊂ B, then A ∪ B = B.
See lessWhich of the following relations are functions? Give reasons. If it is a function,
determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the givenrelation having their unique images, this relation is a function. Here,domain = {2, 5, 8, 11, 14, 17} and range = {1}(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6),Read more
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
See lessSince 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given
relation having their unique images, this relation is a function. Here,
domain = {2, 5, 8, 11, 14, 17} and range = {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given
relation having their unique images, this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
Since the same first element i.e., 1 corresponds to two different images
i.e., 3 and 5, this relation is not a function.
In the following, state whether A = B or not:
(i) A = {a, b, c, d}; B = {d, c, b, a}
(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}; B = {x: x is positive even integer and x ≤ 10}
(iv) A = {x: x is a multiple of 10}; B = {10, 15, 20, 25, 30 …}
(i) A = {a, b, c, d}; B = {d, c, b, a} The order in which the elements of a set are listed is not significant. ∴ A = B (ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18} It can be seen that 12 ∈ A but 12 ∉ B. ∴ A ≠ B (iii) A = {2, 4, 6, 8, 10} B = {x: x is a positive even integer and x ≤ 10} = {2, 4, 6, 8,Read more
(i) A = {a, b, c, d}; B = {d, c, b, a} The order in which the elements of a set are listed is not significant. ∴ A = B
(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18} It can be seen that 12 ∈ A but 12 ∉ B. ∴ A ≠ B
(iii) A = {2, 4, 6, 8, 10} B = {x: x is a positive even integer and x ≤ 10} = {2, 4, 6, 8, 10} ∴ A = B
(iv) A = {x: x is a multiple of 10} B = {10, 15, 20, 25, 30 …} It can be seen that 15 ∈ B but 15 ∉ A. ∴ A ≠ B
See lessLet A and B be two events. If \(P(A) = \dfrac{1}{2}, P(B) = \dfrac{1}{4}, P(A \cap B) = \dfrac{1}{5}\) then \(P({A’\over B’})\) =
1. 0.8
2. 0.4
3. 0.3
4. 0.6
Correct Answer - Option 4 : 0.6Concept:The complement of an event:The complement of an event is the subset of outcomes in the sample space that are not in the event.The probability of the complement of an event is one minus the probability of the event.P (A') = 1 - P (A)For two events A and B we havRead more
Correct Answer – Option 4 : 0.6
Concept:
The complement of an event:
The complement of an event is the subset of outcomes in the sample space that are not in the event.
The probability of the complement of an event is one minus the probability of the event.
P (A’) = 1 – P (A)
For two events A and B we have
1. P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
2. De morgan’s Law P (A ∪ B)’ = P (A’ ∩ B’)
3. \(P({A’\over B’})={{P(A’\cap B’)\over P(B’)}}={{P(A\cup B)’\over P(B’)}}\)
Calculation:
Given:
\(P(A) = \dfrac{1}{2}, P(B) = \dfrac{1}{4}, P(A \cap B) = \dfrac{1}{5}\)
we have,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(P(A ∪ B) = {1\over 2}+{1\over4}-{1\over5}={11\over 20}\)
The complement of events:
\(P(B’)=1-P(B)=1-{1\over 4}={3\over 4}\)
\(P(A\cup B)’=1-P(A\cup B)=1-{11\over 20}={9\over 20}\)
Now,
\(P({A’\over B’})={{P(A’\cap B’)\over P(B’)}}={{P(A\cup B)’\over P(B’)}}\)
\(P({A’\over B’})={{{9\over20}\over {3\over 4}}}={9\over 20}\times {4\over 3}={3\over 5}=0.6\)
See lessIf A = {x : x is square of natural numbers ≤ 8}, and B = {2x + 1 : x ∈ N}, the what is (A ∩ B)?
1. {0, 1, 4, 9, 25, 49, 121}
2. {1, 4, 16, 36, 64}
3. {9, 25,49}
4. {1, 9, 25, 49}
Correct Answer - Option 3 : {9, 25,49}Concept:Set theory:A ∪ B means set of all the values in the set A and B.A ∩ B is the set of common elements of A and B.Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from. CalculationGiven: B = {2x +Read more
Correct Answer – Option 3 : {9, 25,49}
Concept:
Set theory:
Calculation
Given: B = {2x + 1 : x ∈ N}
So, Set B = {3, 5, 7, 9, 11, 13, ……} (all the odd positive integers excluding 1)
Given: A = {x : x is square of natural numbers ≤ 8}
So, Set A = {1, 4, 9, 16, 25, 36, 49, 64}
∴ A ∩ B = {9, 25, 49}
See lessGiven the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of
the following may be considered as universals set (s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) Φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}
(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6} B ⊂ {0, 1, 2, 3, 4, 5, 6} However, C ⊄ {0, 1, 2, 3, 4, 5, 6} Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C. (ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ Therefore, Φ cannot be the universal set for the sets A, B, and C. (iRead more
(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6} B ⊂ {0, 1, 2, 3, 4, 5, 6}
However, C ⊄ {0, 1, 2, 3, 4, 5, 6}
Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.
(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ Therefore, Φ cannot be the universal set for the sets A, B, and C.
(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.
(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8} B ⊂ {1, 2, 3, 4, 5, 6, 7, 8} However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8} Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.
See lessThe universal set U is the set of all odd numbers less than 20 and set A = {x : x is an odd multiple of 3 and x < 20}. Find complementary set A'.
1. A’ = {1, 5, 11, 13, 17, 19}
2. A’ = {1, 5, 7, 11, 13, 17, 19}
3. A’ = {1, 19}
4. A’ = {1, 2, 5, 7, 8, 11, 13, 16, 17, 19}
Correct Answer - Option 2 : A' = {1, 5, 7, 11, 13, 17, 19}Concept:If U is a universal set and A be any subset of U then the complement of A is the set of all members of the universal set U which are not the elements of A.It is denoted by A'A' = U - ACalculation:U = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19Read more
Correct Answer – Option 2 : A’ = {1, 5, 7, 11, 13, 17, 19}
Concept:
If U is a universal set and A be any subset of U then the complement of A is the set of all members of the universal set U which are not the elements of A.
It is denoted by A’
A’ = U – A
Calculation:
U = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
A = {3, 9, 15}
A’ = U – A = {1, 5, 7, 11, 13, 17, 19}
See less