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Set Theory Test 4

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Set Theory Test 4
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  • Question 1
    1 / -0
    Let  $$U=\{x: \in, W: 3<x< 12\} $$, $$B=\{4,6,8,10\}$$ . $$B'$$
    Solution
    Given universal set is $$\{x\in W ,3<x<12\}$$
    It can be written as a set $$\{4,5,6,7,8,9,10,11\}$$
    $$B$$ is given as $$B=\left\{4,6,8,10\right\}$$
    $$\therefore B'=W-B=$$ All the elements in the universal set but not in set $$B$$
             $$ =\{5,7,9,11\}$$
    Hence, option B is correct
  • Question 2
    1 / -0

    Directions For Questions

    $$\mu = \left \{a, b, c, d, e, f, g, h, i, j\right \}$$
    $$P = \left \{a, b, c, e\right \}$$
    $$Q = \left \{b, c, d, f\right \}$$ and
    $$R = \left \{c, f, h, i, j\right \}$$
    Find the number of elements of the set

    ...view full instructions

    $$P\cap Q\cup R$$
    Solution
    $$ P = \left \{ a,b,c,e \right \} $$
    $$ Q = \left \{ b,c,d,f \right \} $$
    $$ \Rightarrow (P\cap Q) = \left \{ b,c \right \} $$
    As $$ R = \left \{ c,f,h,i,j\right \} $$
    $$ \Rightarrow \boxed {(P\cap Q)\cup R = \left \{ b,f,h,i,j \right \}} $$ 

  • Question 3
    1 / -0
    Given $$P(A \cup B)=0.6, P(A\cap B)=0.2$$, the probability of exactly one of the event occurs is
    Solution
    Given, $$P(A\cup B)=0.6, P(A\cap B)=0.2$$
    Probability of exactly one of the event occurs is $$P(\bar{A}\cap B)+P(A\cap \bar{B})$$
    $$=P(B)-P(A\cap B)+P(A)-P(A\cap B)$$
    $$=P(A\cup B)+P(A\cap B)-2P(A\cap B)$$
    $$[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]$$
    $$=P(A\cup B)-P(A\cap B)$$
    $$=0.6-0.2$$
    $$=0.4$$
  • Question 4
    1 / -0
    If $$A, B$$ and $$C$$ are any three set, then $$A \cup (B\cap C) =$$

    Solution
    Using distributive law of sets option B is correct
     Or it is the distributive law itself
  • Question 5
    1 / -0
    If $$A = \left \{1, 2, 3, 4\right \}$$, what is the number of subsets of A with at least three elements?
    Solution
    A subset containing $$3$$ elements $$= \left \{1, 2, 3\right \}; \left \{1, 3, 4\right \}; \left \{1, 2, 4\right \}$$ and $$\left \{2, 3, 4\right \}$$
    A subset containing $$4$$ elements $$= \left \{1, 2, 3, 4\right \}$$
    $$\therefore$$ there are five subsets containing at least $$3$$ elements.
  • Question 6
    1 / -0
    The set of integers is closed with respect to which one of the following?
    Solution
    From group theory, integers are closed w.r.t. both addition & multiplication
  • Question 7
    1 / -0
    In the Venn diagram, the numbers represent the number of elements in the subsets. Given that $$\xi = F\cup G\cup H$$ and $$n(\xi) = 42$$, find $$n(G'\cup H)$$

    Solution

    $$n(\xi) = 42$$

    $$n(G) = 12, n(F) = 17, n(H) = 8$$

    $$n(F \cap G) = 5, n(G \cap H) = 0, n(F \cap H) = 0$$

    $$n(G’)= 42-7-5 =30$$

    $$n(H)= 8, n(G’ \cup H) = n(G')+n(H)-n(G'\cap H)=30+8-8 = 30$$

  • Question 8
    1 / -0

    Directions For Questions

    $$\mu = \left \{a, b, c, d, e, f, g, h, i, j\right \}$$
    $$P = \left \{a, b, c, e\right \}$$
    $$Q = \left \{b, c, d, f\right \}$$ and
    $$R = \left \{c, f, h, i, j\right \}$$
    Find the number of elements of the set

    ...view full instructions

    $$(P\cap Q)\cup (Q\cap R)$$
    Solution
    $$ P = \left \{ a,b,c,e \right \};Q = \left \{ b_{n}c_{n}d_{n}f \right \} $$
    $$ \Rightarrow (P\cap Q) = \left \{ b,c \right \} $$
    As $$ R = \left \{ c,f,n,i,j \right \} \Rightarrow (Q\cap R) = \left \{ c,f \right \} $$
    $$ \boxed { (P\cap Q)\cup (Q\cap R) = \left \{ b,c,f \right \}} $$ 

  • Question 9
    1 / -0
    In a school with an envolment of $$950$$ students, each student must join either the lions club or the country club or both. Given that $$646$$ students are members of the lions club and $$532$$ are members of the country club, calculate the number of students who are members of both clubs
    Solution
    Total number of students $$=950$$
    Number of students who are members of lion club $$=646$$
    Number of students who are members of country club $$=532$$
    Number of students who are members of both club $$=(646+532)-950$$
                                                                                           $$=1178-950$$
                                                                                           $$=228$$
    $$\therefore$$  There are $$228$$ students who joined both clubs.
  • Question 10
    1 / -0
    If A and B be two sets containing $$4$$ and $$8$$ elements respectively, what can be the maximum number of elements in $$A\cup B$$? Find also, the minimum number of elements in $$(A\cup B)$$?
    Solution
    $${\textbf{Step -1: Find maximum number of elements.}}$$
                      $${\text{Given,}}$$
                      $$n(A) =4$$
                      $$n(B) =8$$
                      $${\text{Minimum number of elements in}}$$ $$A\cup B=8.$$
                      $${\text{A is subset of B.}}$$
                      $${\text{Maximum number of elements in}}$$ $$A\cup B=4+8=12.$$
                      $${\textbf{[All element taken in both set A and B and elements in both sets are different.]}}$$
    $${\textbf{Hence , maximum number of elements in}}$$ $$\mathbf{(A \cup B) =12.}$$
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