Tag: cartesian product of sets

Questions Related to cartesian product of sets

Cartesian product of sets $A$ and $B$ is denoted by _______.

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $A \times B$

  2. $B \times A$

  3. $A \times A$

  4. $B \times B$

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Correct Option: A
Explanation:

Cartesian product of Set $A$ and $B$ is denoted by $A\times B$.

Identify the first component of an ordered pair $(0, -1) $.

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $0$

  2. $-1$

  3. $2$

  4. $1$

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Correct Option: A
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(0,-1)$, the first component is $0$.

Find the second component of an ordered pair $(2, -3)$

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $2$

  2. $3$

  3. $0$

  4. $-3$

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Correct Option: D
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(2,-3)$, the second component is $-3$.

The ______ product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. cartesian

  2. coordinate

  3. simple

  4. discrete

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Correct Option: A
Explanation:

The cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set. 
Example:$ A = {1, 2} \quad B = {2}$
cartesian product, $A \times B = {(1,2), (2, 2)}$

If $A = {a, b}, B={1, 2, 3}$, find B $\times$ A

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $B$ $\times$ $A$$ = {(1, a), (2, a), (3, a), (1, b) (2, b), (3, b)}$

  2. $B$ $\times$ $A$$ = { (2, a), (3, a), (1, b) (2, b), (3, b)}$

  3. $B$ $\times$ $A$$ = {(1, a), (2, a), (3, a), (1, b) (2, b)}$

  4. None of these

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Correct Option: A
Explanation:

To find B × A multiply each element of B with that of A & form an ordered pair.

 i.e. ordered pairs are (1,a); (2,a); (3,a); (1,b); (2,b); (3,b)
Therefore B × A = {  (1,a), (2,a), (3,a), (1,b), (2,b), (3,b)}

If A= {0, 1} and B ={1, 0}, then what is A x B equal to ?

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. {(0, 1), (1, 0)}

  2. {(0, 0), (1, 1)}

  3. {(0, 1), (1, 0), (1, I)}

  4. A X A

Show answer
Correct Option: D
Explanation:

$\left{ { 0,1 } \right} \times \left{ 1,0 \right} ={ \left{ (0,1),(0,0),(1,1),(1,0) \right}  }$

$\left{ { 0,1 } \right} \times \left{ 0,1 \right} ={ \left{ (0,0),(0,1),(1,0),(1,1) \right}  }$

So, $A\times B=A\times A$
Hence, D is correct.

If $A = {2, 3, 5}$ and $B = {5, 7}$, find the set with highest number of elements:

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $A \times B$

  2. $ B \times A$

  3. $A \times A$

  4. $B \times B$

Show answer
Correct Option: C
Explanation:
$A=\left \{ 2,3,5 \right \}$

$B=\left \{ 5,7 \right \}$

$A\times B=\left \{ (2,5),(2,7),(3,5),(3,7),(5,5),(5,7) \right \}$

$B\times A=\left \{ (5,2),(5,3),(5,5),(7,2),(7,3),(7,5) \right \}$

$A\times A=\left \{ (2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5) \right \}$

$B\times B=\left \{ (5,5),(5,7),(7,5),(7,7) \right \}$

$\therefore A\times A$ has the highest number of elements

For two sets $A$ and $B$, $A\times B=B\times A$.

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. True

  2. False

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Correct Option: B
Explanation:

The given statement is false.

Example:-
Let us consider $A={1,2}$ and $B={3,4}$.

Now $A\times B={(1,3),(1,4),(2,3),(2,4)}$......(1).

And $B\times A={(3,1),(3,2),(4,1),(4,2)}$..........(2).

Form (1) and (2) it's evident that $A\times B\ne B\times A$.

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets $A \times B$ having 3 or more elements is 

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $219$

  2. $211$

  3. $256$

  4. $220$

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Correct Option: A
Explanation:
Let $A=\left\{x,y\right\}$
$B=\left\{a,bc,d\right\}$
$A\times B$ has $2\times 4=8$ elements
Total substance of $A\times B={2}^{8}=256$
$\therefore\,$Total number of subsets of $A\times B$ having $3$ or more elements
$=256-\left(1\,null \,set+8\,single\,ton\,set-^{8}C _{2}\,having\,2\,elements\right)$
$=256-1-8-\dfrac{8!}{6!2!}$
$=256-1-8-\dfrac{8\times 7\times 6!}{6!2!}$
$=256-1-8-28=219$

If $A$ and $B$ are independent event such that $P(A \cap B')=\dfrac {3}{25}$ and $P(A' \cap B)=\dfrac {8}{25}$, then $P(A)=$

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $1/5$

  2. $3/8$

  3. $2/5$

  4. $4/5$

Show answer
Correct Option: C
Explanation:
$A$ & $B$ are independent
$P(A\cap B)=\dfrac {3}{25}$ and $P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$
$\rightarrow \ P(A)+P(B)=1---(i)$ ($A$& $B$ are independent )
$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$
$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$
$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$
$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$
$\rightarrow \ $ solving equation $(ii)$ and $(iii)$
$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 }  } $
                           $=\dfrac {-5}{25}$
$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$
$P(A)-(1-P(A))=\dfrac {-1}{5}$
$P(A)-I+P(A)=\dfrac {-1}{5}$
$\therefore \ 2P(A)=\dfrac {-1}{5}+1$
$\therefore \ 2P(A)=\dfrac {4}{5}$
$\therefore \ P(A)=\dfrac {4}{5\times 2}$
$\therefore \ P(A)=\dfrac {2}{5}$