If $$P(A) = 0.8 , P(B) = 0.5 $$ & $$P(B/A) =0.4 $$ find (i) $$P(A \cap B) $$ (ii) $$P(A/B)$$ (iii) $$P(A\cup B)$$.

$$i) 0.32, ii)0.64 , iii)0.98$$

$$i) 0.62, ii)0.34 , iii)0.88$$

$$i) 0.66, ii)0.32 , iii)0.98$$

MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 3

Let

$A_1, A_2$ and

$A_3$ be subsets of a set

$X$ . Which one of the following is correct?

0%
$A_1\cup{A_2}\cup{A_3}$ is the largest subset of $X$ containing elements of each of $A_1, A_2$ and $A_3$
0%
$A_1\cup{A_2}\cup{A_3}$ is the smallest subset of $X$ containing either $A_1$ or $A_2\cup{A_3}$ but not both
0%
The smallest subset of $X$ containing $A_1\cup{A_2}$ and $A_3$ equals the smallest subset of $X$ containing both $A_1$ and $A_2\cup{A_3}$ only if $A_2=A_3$
0%
None of these

Statements:
I. Some Politicians are social workers.
II. All Doctors are social workers.
Conclusions:
I. Some Doctors are Politicians.
II. Some social workers are Doctors as well as Politicians.

0%
Both conclusion (I) and (II) follow
0%
Only conclusion (II) follow
0%
Neither conclusion (I) nor (II) follow
0%
Only conclusion (I) follow

Explanation

R.E.F image

Thus, option

$(3)$ is correct.

How many rural uneducated people are employed?

0%
$10$
0%
$6$
0%
$12$
0%
$14$

Explanation

Rural uneducated people has been represented by

$'10'$ .

$A=\left\{ 2,4\left\{ 5,6 \right\} ,8 \right\}$ , then which one of the following statements is not correct?

0%
$\left\{ 5,6 \right\} \subseteq A$
0%
$\left\{ 5,6 \right\} \in A$
0%
$\left\{ 2,4,8 \right\} \subseteq A$
0%
$2,4,8\in A\quad$

Explanation

$\left\{ 5,6 \right\}$ is not the subset but is an element of

$A$ ,

$\therefore$ a is false.

For two sets

$A$ and

$B$ ,

$A\cap \left( A\cup B \right)=$

0%
$A$
0%
$B$
0%
$\phi$
0%
$None\ of\ these$

Explanation

$\left( A\cap A \right) \cup \left( A\cap B \right) =A\cup \left( A\cap B \right) =A$

$\because A\cap B\subset A$

$A=\left\{a,b,c\right\},B=\left\{c,d,e\right\},C\left\{a,d,f\right\},$ then

$A\times \left( B\cup C \right)$ is

0%
$\left\{(a,d),(a,e),(a,c)\right\}$
0%
$\left\{(a,d),(b,d),(c,d)\right\}$
0%
$\left\{(d,a),(d,b),(d,c)\right\}$
0%
$none\ of\ these$

Explanation

$(d)$ .

$A\times \left( B\cup C \right) =\{ a,b,c\} \times \{ a,c,d,e,f\}$
The above set will consist of

$15$ ordered pairs and not

$3$ .

$A$ and

$B$ are two sets having

$3$ and

$5$ elements respectively and having

$2$ elements in common. Then the number of elements in

$A\times B$ is

0%
$6$
0%
$36$
0%
$15$
0%
$None\ of\ these$

Explanation

Total ordered pairs

$= n(A) \times n(B) = 3\times 5 = 15$

Let

$A$ and

$B$ have

$3$ and

$6$ elements respectively. What can be the minimum number of elements in

$A\cup B$ ?

0%
$3$
0%
$6$
0%
$9$
0%
$18$

Explanation

Ans.

$(b)$ .

$n\left( A\cup B \right) =n(A)+n(B)-n\left( A\cap B \right)$
Now

$A$ has

$3$ elements and

$B$ has

$6$ elements. If they are disjoint, then

$n\left( A\cap B \right) =0$ .

$\therefore n\left( A\cup B \right) =6+3=9$
If

$A\subset B$ then

$A\cup B=B$

$\therefore \left( A\cup B \right) =n(B)=6$

$B$ cannot be a subset of

$A$ and hence the other possibility of

$A\cup B=A$ is ruled out.

$25$ people for applied for programme

$A$ ,

$50$ people for programme

$B$ ,

$10$ people for both. So number of employee applied only for

$A$ is

0%
$15$
0%
$20$
0%
$35$
0%
$40$

Explanation

$n(A - B) = n(A) - n(A \cap B) = 25 -10 -15$

State the whether given statement is true or false

$A$ is any set, prove that:

$A\subseteq \phi \Leftrightarrow A=\phi$ .

0%
True
0%
False

Explanation

Let

$A \subseteq \phi$

Now since

$\phi$ is subset of every set

therefore,

$\phi \subseteq A$ ---- (i)

also,

$A \subseteq \phi$ --- (ii)

From (i) ad (ii)

$A=\phi$

Conversely, let

$A=\phi$

Implies that

$A \subseteq \phi$

Therefore,

$A \subseteq \phi \Leftrightarrow A = \phi$

$P(A) = 0.8 , P(B) = 0.5$ &

$P(B/A) =0.4$ find (i)

$P(A \cap B)$ (ii)

$P(A/B)$ (iii)

$P(A\cup B)$ .

0%
$i) 0.32, ii)0.64 , iii)0.98$
0%
$i) 0.62, ii)0.34 , iii)0.88$
0%
$i) 0.66, ii)0.32 , iii)0.98$
0%
none

Explanation

$\\\>P(\frac{B}{A})=\>\frac{P(B\>\cap\>A)}{P(A)}\\\>0.4=\frac{P(B\cap\>A)}{0.8}\\$

$\therefore\>P(B\cap\>A)=0.32\\$ , $\implies \>P(A\cap\>B)=0.32\\$

$then\>\>P(\frac{A}{B})=\>\frac{P(A\>\cap\>B)}{P(B)}=(\frac{0.32}{0.5})=0.64\>\\$

$and\>P(A\cup\>B)=P(A)+P(B)-P(A\cap\>B)\\=0.8+0.5-0.32\\\>=0.98$

If two sets

$A$ and

$B$ are having

$80$ elements in common, then the number of element common to each of the sets

$A\times B$ and

$B\times A$ are

0%
${2}^{80}$
0%
${80}^{2}$
0%
$81$
0%
$79$

Given P(A)=

$0.5,$ P(B)=

$0.2$ and P(AB)=0.1;find

0%
$P\left( {A \cup B} \right)$
0%
$P\left( {\overline A \cap B} \right)$
0%
$P\left( {A \cup \overline B} \right)$
0%
$P\left( {\overline A \cap \overline B} \right)$

Explanation

$P(A)=0.5$

$P(B)=0.2$

$P(AB)=0.1$

$(a)$

$P(A\cup B)=P(A)+P(B)-P(A)P(B)$

$=[0.5+0.2-(0.5)(0.2)]$

$\Rightarrow 0.7-0.1$

$\Rightarrow 0.6$

$(b)$

$P(\bar A \cup B)=(P(B)-P(A\cap B))$

$P(A\cup B)=P(A)+P(B)-P((A\cap B))$

$\Rightarrow 0.3=0.5+0.2-P(A\cap B)$

$P(A\cap B)=0.1$

$\because P(\bar A \cap B)=0.2-0.1$

$=0.1$

$n(A)$ denotes the number of elements in set A and if

$n(A)=4, n(B)=5$ and

$n(A\cap B)=3$ then

$n\left[ \left( A\times B \right) \cap \left( B\times A \right) \right] =$

0%
$8$
0%
$9$
0%
$10$
0%
$11$

Explanation

For

$(A\times B)\cap (B\times A)$ we have to do the mapping of

$A\times B$ or

$B\times A$ between common elements.

no. of ways of mapping will be

$3\times 3=9$

$n[(A\times B)\cap(B\times A)]=9$

$n(A \cup B)=8, n(A)=6, n(B)=4$ , then

$n(A\cap B)$ =

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$n(A\cup B)=8\\ n(A)=6\quad ,\quad n(B)=4$

we know that

$n(A\cup B)=n(A)+n(B)-n(A\cap B)\\$

$8=6+4-n(A\cap B)\\$

$n(A\cap B)=10-8=2$

The set

$\left( {A \cap B'} \right)' \cup \left( {B \cap C} \right)$ is equal to :

0%
$A' \cup B \cup C$
0%
$A' \cup B$
0%
$A' \cup C'$
0%
$A' \cap B$

If A and B are any two sets, then

$A \cup B$ is equal to:

0%
$(A-B) \cup (B-A) \cup (A \cap B)$
0%
$(A^0 \cap B^0)^0$
0%
$(A-B) \cup (B-A)$
0%
$A \cup (B-A)$

Let

$A, B$ are two sets such that

$n(A)=6, n(B)=8$ then the maximum number of elements in

$n(A\cup B)$ is _________

0%
$7$
0%
$9$
0%
$14$
0%
$None$

Explanation

$n(A)=6,n(B)=8$

$n(A\cup B)$ will be maximum when

$A$ &

$B$ are disjoint sets.

$A\cap B=\phi \\ n(A\cap B)=0\\ n(A\cup B)=n(A)+n(B)-n(A\cap B)\\ =6+8-0\\ =14\\ n(A\cup B)=14$

$A=\left\{2, 4, 6, 8, 10\right\}, B=\left\{1, 3, 5, 7, 9\right\}$ , then

$A-B$ =____________

0%
$\left\{\right\}$
0%
$\left\{2, 4, 6, 8, 10\right\}$
0%
$\left\{1, 3, 5, 7, 9\right\}$
0%
$None$

Explanation

$A=\{ 2,4,6,8,10\} \\ B=\{ 1,3,5,7,9\} \\ A-B=\{ 2,4,6,8,10\} -\{ 1,3,5,7,9\} =\{ 2,4,6,8,10\}$

If A and B are two sets such that

$n(A)=17, n(B)=23, n(A \cup B)=38$ , find

$n(A \cap B)$ .

0%
1
0%
2
0%
3
0%
4

Explanation

We know,

$n(A \cap B)=n(A)+n(B)-n(A \cup B)$

$n(A \cap B)=17+23-38=2$

Which of the following is set ?

0%
The collection of months having names starting with J.
0%
The collection of smart boys in your class.
0%
The collection of most talented persons.
0%
The collection of sand grains in a Earth.

If X and Y are two sets such that

$n(X)=45, n(X \cup Y)=76, n(X \cap Y)=12,$ find

$n(Y)$ .

0%
$41$
0%
$43$
0%
$49$
0%
$47$

Explanation

$n(X \cup Y)=n(X) +n(Y)-n(X \cap Y)$

$76=45+n(Y)-12$

$n(Y)=43$

State whether the following statement is true or false. Give reason to support your answer.
A set can have infinitely many subsets.

0%
True
0%
False

Explanation

It is not true for every set.

Suppose

$A$ is a set :

$A=\{1,2\}$

Now,

Clearly it will have finite number of subsets,such that

$B=\{ 1 \} , C=\{2\} ,D=\{1,2\} ,E=\phi$ will be only subsets of A.

The number of subsets of the set

$A=\{ { a }_{ 1 },{ a }_{ 2 },.........{ a }_{ n }\}$ which contain even number of elements is

0%
${ 2 }^{ n-1 }$
0%
${ 2 }^{ n }-1$
0%
${ 2 }^{ n }-2$
0%
${ 2 }^{ n }$

Explanation

The total no of subsets of

$A$ is the cardinality of the power set of

$A$ .

So, if

$|A|=n$ then

$|P(A)|=2^n$ .

Therefore total no of subsets of

$A$ is

$2^n$ .

Similarly,

The even number of events is given by,

$2^{n-1}$

Let

$S=\{2,4,6,8,......20\}$ . What is the maximum number of subsets does

$S$ have ?

0%
$10$
0%
$20$
0%
$512$
0%
$1024$

Explanation

Given,

$S={2,4,6,8........,20}$

There are a total of

$10$ elements.

Therefore we have

$2^{10}=1024$ subsets.

State whether the following statements are true(T) or false(F).Justify your answer.
A collection of stamps is a set.

0%
True
0%
False

State whether the following statements are true(T) or false(F).Justify your answer.
A group of boys playing cricket is a set.

0%
True
0%
False

State whether the following statements are true(T) or false(F).Justify your answer.
A collection of some fruits is a set.

0%
True
0%
False

Which of the following collections is a set?

0%
Collection of all tasty fruits
0%
Collection of all good football players of your school
0%
Collection of all months of a year
0%
Collection of $5$ most intelligent students of your class.

Explanation

Collection of all months of a year is a set If we denote the given set, then

$A =\{\text{January, February, March, April, ........December}\}$

Examine whether the following statements are true or false:

$(1,2,3)\subset (1,3,5)$

0%
True
0%
False

Explanation

False.

$2\epsilon (1,2,3)$ ; however,

$2\not{\subset}(1,3,5)$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 3 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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