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Model Paper 8 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 84 Marks
Question
Method to Find the Sets When Cartesian Product is Given
For finding these two sets, we write first element of each ordered pair in first set say $A$ and corresponding second element in second set $B\ ($say$)$.
Number of Elements in Cartesian Product of Two Sets
If there are $p$ elements in set $A$ and $q$ elements in set $B$, then there will be pq elements in $A \times B$ i.e. if $n(A)=p$ and $n(B)=q$, then $n(A \times B)=p q$.
$i$. The Cartesian product $A \times A$ has $9$ elements among which are found $(-1,0)$ and $(0,1)$. Find the set $A$ and the remaining elements of $A \times A. (1)$
$ii. A$ and $B$ are two sets given in such a way that $A \times B$ contains $6$ elements. If three elements of $A \times B$ are $(1, 3 ), (2,5)$ and $(3, 3),$ then find the remaining elements of $A \times B. (1)$
$iii$. If the set $A$ has $3$ elements and set $B$ has $4$ elements, then find the number of elements in $A \times B$. $(2)$
OR
If $A \times B=\{(a, 1),(b, 3),(a, 3),(b, 1),(a, 2),(b, 2)\}$. Find $A$ and $B .(2)$
✓

Answer

$i.$ Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4 !$ i.e. $24$
$\therefore A=\{a, b\}$ and $B=\{1,2,3\}$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _1$ be the event that Priyanka visits $A$ before $B$ .
Then,
$E_1=\{A B C D, A B D C, A C B D, A C D B, A D B C, A D C B,$
$ C A B D, C A D B, C D A B, D A B C, D A C B, D C A B\}$
$\Rightarrow n \left( E _1\right)=12$
$\therefore P ($ she visits $A$ before $B )= P \left( E _1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$
$ii$. Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
$ E _1=\{ ABCD , ABDC , ACBD , ACDB , ADBC , ADCB ,$
$ CABD , CADB , CDAB , DABC , DACB , DCAB \}$
$\Rightarrow n \left( E _1\right)=12$
$\therefore P ($she visits $A$ before $ B )=P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$
$iii$. Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is 
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _3$ be the event that she visits $A$ first and $B$ last.
Then,
$E_3=\{ ACDB , ADCB \}$
$n\left(E_3\right)=2$
$\because P($ she visits $A$ first and $B$ last $)=P\left(E_3\right)$
$=\frac{n\left(E_3\right)}{n(S)}=\frac{2}{24}=\frac{1}{12}$
OR
Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _4$ be the event that she visits $A$ either first or second.
Then, $E_4=\{A B C D, A B D C, A C B D, A C D B, A D B C, A D C B,$
$ B A C D, B A D C, C A B D, C A D B, D A B C, D A C B\}$
$\Rightarrow n\left(E_4\right)=12$
Hence $, P ($she visits $A$ either first or second$)$
$=P\left(E_4\right)=\frac{n\left(E_4\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$

 

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