Probability - Class 11 Commerce Maths - Extra Questions

A couple has two children . Find the probability that both the children are (i) males , if it is known that at least one of the children is male . (ii) females, if it is known that the elder child is a female .

$$i)$$If it is known that at least one of the children is male.
Then possible outcomes are:
i)Both Males ii)One male and one female.
Then probability that both the childrens are male is;$$\dfrac{1}{2}$$

$$ii)$$ If it is known that the elder child is a female.
Then possible outcomes are:
i)Both female   ii)One male and one female.
Then probability that both the childrens are female is:$$\dfrac{1}{2}$$




A card is drawn at random from well shuffled deck of playing cards. Find the probability that the card drawn is
i.     A king or a jack
ii.    A Non ace
iii.   A red ace
iv.   Neither a king nor a queen.


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Let $$A$$ and $$B$$ be two independent events. The probability that both $$A$$ and $$B$$ happen is $$\cfrac{1}{12}$$ and the probability that neither $$A$$ nor $$B$$ happen is $$1/2$$.If the sum of probabilities of occurence of $$A$$ and $$B$$ is $$\dfrac { k }{ 12 } $$, then the value of $$k$$ is

Total probability$$=1=P\left( A\cup B \right) +P\left( \overline { A\cup B }  \right) =P(A)+P(B)-P\left( A\cap B \right) +P\left( \overline { A\cup B }  \right) $$
It's given that$$P\left( \overline { A\cup B }  \right) =\dfrac { 1 }{ 2 } $$
$$\Rightarrow $$ $$1=P(A)+P(B)-\dfrac { 1 }{ 12 } +\dfrac { 1 }{ 2 } \Rightarrow P(A)+P(B)=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 12 } =\dfrac { 7 }{ 12 } $$
Hence,$$k=7$$


$$A$$ and $$B$$ are two candidates seeking admission in I.I.T. The probability that $$A$$ is selected is $$0.5$$ and the probability that both $$A$$ and $$B$$ are selected is atmost $$0.3$$. Is it possible that the probability of $$B$$ getting selected is $$0.9$$? If it is possible then enter 1, else enter 0.

If we have $$P(B)=0.9$$ ,then we will get $$P\left( A\cup B \right) =P(A)+P(B)-P\left( A\cap B \right) =0.5+0.9-0.3=1.1>1$$
Hence, this is not possible.


Verify the following identities where A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}
$$A \, \cup \, (B \, \cap \, C) \, = \, (A \, \cup \, B) \, \cap \, (A \, \cup \, C)$$

$$B\cap C = \{ 5, 6 \}$$
$$A \cup (B\cap C) = \{ 1, 2, 3, 4, 5, 6 \}$$    .................(1) 
$$A \cup B = \{ 1, 2, 3, 4, 5, 6 \}$$
$$A \cup C = \{ 1, 2, 3, 4, 5, 6, 7 \}$$
$$(A \cup B) \cap (A\cup C) = \{ 1, 2, 3, 4, 5, 6 \}$$  .................(2) 
From (1) and (2);
Hence proved.


In a survey of $$200$$ students of a higher secondary school, it was found that $$120$$ studied mathematics; $$90$$ studied physics; and $$70$$ studied chemistry; $$40$$ studied mathematics and physics; $$30$$ studied physics and chemistry; $$50$$ studied chemistry and mathematics, and $$20$$ studied none of these subjects. Find the number of students who studied all the three subjects.

let number of students studied all three subjects be 'x'

Maths only $$=120-(40+50-x)$$

Chemistry only $$=70-(30+50-x)$$

Physics only $$=90-(30+40-x)$$

No subject $$=20$$

$$200-20=120+70+90-40-50-30+x$$

$$180=160+x$$

$$x=20$$

$$20$$ students studied all the 3 subjects.



Rahim takes out all the hearts from the cards. What is the probability of
Picking out a diamonds.

As totally we have $$52$$ cards
Rahim taken out all the hearts.
So Remaining are $$39$$ cards 
So the probability of picking the diamond out of of $$13$$ is $$\dfrac{13}{39}=\dfrac{1}{3}$$.


Three unbiased coins are tossed together . Find the probability of getting 
$$1.$$ Two heads. 
$$2.$$ At least two heads. 
$$3.$$ No heads

When $$3$$ unbiased coins are tossed, the sample space is $$\left\{HHH, HTH, HHT, THH, THT, TTH, HTT, TTT\right\}= 8$$
 $$1.$$Event of getting two heads $$=\left\{HTH, HHT, THH\right\} = 3$$
 $$P\left(Two \,\,Heads\right) =\dfrac{3}{8}$$
 $$2.$$Event of getting At least two heads (two or more than $$2$$ heads) $$=\left\{HHH, HTH, HHT, THH\right\} = 4$$
 $$P\left(At \,\, least \,\, Two \,\, Heads\right) = \dfrac{4}{8}=\dfrac{1}{2}$$
$$3.$$ Event of getting No heads $$=\left\{TTT\right\} = 1$$
 $$P\left(No\,\, Heads\right) =\dfrac{1}{8}$$


Meeta and Reeta are two sisters one is in class VI another is in Class VII, The numbers of student in class VI is 25 and in class VII isFind the probability that both of them get first position in their respective classes. 

$$A\rightarrow$$ Meeta got 1st in a class of $$25$$ strudents

$$P(A)=\dfrac {1}{25}$$

$$B\rightarrow$$ Reeta got 1st in a class of $$30$$ strudents

$$P(B)=\dfrac {1}{30}$$

$$P(A\cap B)=P(A)P(B)\\ \quad=\cfrac{1}{25}\times\cfrac{1}{30}=\cfrac{1}{750}$$


In a hostel 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A students is selected at random.
Find the probability that the reads neither Hindi nor English news papers. 

Let the no. of students in Hostel$$=P$$

No. of students who read Hindi$$=\cfrac { 60 }{ 100 } \times P=\cfrac { 3P }{ 5 } $$

No. of students who read English$$=\cfrac { 40 }{ 100 } \times P=\cfrac { 2P }{ 5 } $$

No. of students who read both$$=\cfrac { 20 }{ 100 } \times P=\cfrac { P }{ 5 } $$

P(Hindi)$$=\cfrac { n(Hindi) }{ total } =\cfrac { 3P }{ 5\times P } =\cfrac { 3 }{ 5 } $$

P(English)$$=\cfrac { n(English) }{ total } =\cfrac { 2P }{ 5\times P } =\cfrac { 2 }{ 5 } $$

P(Both)$$=\cfrac { n(both) }{ total } =\cfrac { P }{ 5P } =\cfrac { 1 }{ 5 } $$

$$P\left( \overline { Hindi\cup { English }  }  \right) =1-P\left( English\cup { Hindi }  \right) $$

$$=1-\left( P\left( English \right) +P(Hindi)-P(both) \right) $$

$$=1-\left( \cfrac { 2 }{ 5 } +\cfrac { 3 }{ 5 } -\cfrac { 1 }{ 5 }  \right) $$

$$=1-1+\cfrac { 1 }{ 5 } =\cfrac { 1 }{ 5 } $$

$$=0.2$$


Suppose A and B are independent events with $$P(A)=0.6,P(B)=0.7$$.Then compute:
a)$$P(A \bigcap B)$$
b) $$P(A \bigcup B)$$
c)$$P (B/A)$$
d) $$P(A^c \bigcap B^c)$$

Given,

$$P(A)=0.6,P(B)=0.7$$

Since these are independent event,

So, 
a). $$P(A\cap B)=P(A).P(B)=0.6\times 0.7=0.42$$

b). $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.7+0.6-0.42=0.88$$

c).  $$P(B/A)=\dfrac{P(A \cap B)}{P(A)}=\dfrac{0.42}{0.6}=0.7$$

d). $$P(A^c \cap B^c)=1-P(A \cup B)=1-0.88=0.12$$


$$P(A \cup B)=P(A \cap B)$$ if and only if the relation between P(A ) and P(B) is..................

Given that
$$P(A \cup B)=P(A \cap B)$$
$$\Rightarrow P(A)+P(B)-P(A \cap B)=P(A \cap B)$$ 
$$\Rightarrow [P(A)-P(A \cap B)]+[P(B)-P(A\cap B)]=0$$
But
$$P(A)-P(A \cap B)\geq 0$$ and $$P(B)-P(A \cap B)\geq 0$$  [$$\because P(A \cap B)\leq P(A),P(B)$$]
$$\Rightarrow P(A)-P(A \cap B)=0$$ and $$P(B)-P(A \cap B)=0$$ [since sum of two non-negative numbers can be zero only when these numbers are zeros.]
$$\Rightarrow P(A)=P(B)=P(A\cap B)$$
Which is the required relationship.


If $$(1+3p)/3,(1-p)/4$$ and $$(1-2p)/2$$ are the probabilities of three mutually exclusive events, then the set of all values of p is.............

Let,
$$P(A)=\dfrac{1+3p}{3},P(B)=\dfrac{1-p}{4},P(C)=\dfrac{1-2p}{2}$$
A,B and C are three mutually exclusive events.
$$\therefore P(A)+P(B)+P(C)\leq 1$$
$$\Rightarrow \dfrac{1+3p}{3}+\dfrac{1-p}{4}+\dfrac{1-2p}{2}\leq 1 \leq 1$$
$$\Rightarrow 4+12p+3-3p+6-12p \leq 12$$
$$\Rightarrow p \leq 1/3$$................(1)
Also,
$$0 \leq P(A) \leq 1 \Rightarrow 0 \leq \dfrac{1+3p}{3}\leq 1$$
$$0 \leq 1+3p \leq 3$$
$$\Rightarrow -\dfrac{1}{3}\leq p \leq \dfrac{2}{3}$$............(2)
$$0\leq P(B)\leq 1\Rightarrow 0 \leq \dfrac{1-p}{4}\leq 1$$
$$\Rightarrow 0 \leq 1-p\leq 4$$
$$\Rightarrow -3 \leq p \leq 1$$.................(3)
$$0 \leq P(C)\leq 1\Rightarrow 0 \leq \dfrac{1-2p}{2}\leq 1$$
$$\Rightarrow -\dfrac{1}{2}\leq p \leq \dfrac{1}{2}$$.................(4)
Combining eq.(1),(2),(3) and (4) ,we get
$$\dfrac{1}{3}\leq p \leq \dfrac{1}{2}$$



If $$ P(A) = \dfrac{1}{2} , P(B) = 0 $$ , then find $$ P ( A / B) $$ .

We have , 
            $$ P(A) = \dfrac{1}{2} , P(B) = 0 $$ 

$$ \therefore \, $$    $$ P(A / B) = \dfrac{P(A \bigcap B)}{A(B)} $$ 

Since , $$ P(B) = 0 $$ so $$ P(A / B) $$ is not defined .




A  and B are two independent events.C is an event in which exactly one of A or B occurs.Prove that $$P(C)\geq P(A \cup B) P(A \cap B)$$.

Given that A and B are two independent events.C is an event in which exactly one of A and B occurs.Let P(A)=x,P(B)=y.
Then,
$$P(C)=P(A \cap \overline{B})+P(\overline{A}\cap B)$$
$$=P(A)P(\overline{B})+P(\overline{A})P(B)$$ [$$\because $$ if A and B are independent so are 'A and $$\overline{B}$$ and $$\overline{A}$$ and B]
$$\Rightarrow P(C)=x(1-y)+y(1-x)$$..................(1)
Now consider,
$$P(A \cap B)P(\overline{A} \cap \overline{B})=[P(A)+P(B)-P(A)P(B)][P(\overline{A})P(\overline{B})]$$
$$=(x+y-xy)(1-x)(1-y)$$
$$=(x+y)(1-x)(1-y)-xy(1-x)(1-y)......x(x+y)(1-x)(1-y)$$ $$ [\because x,y \varepsilon(0,1)]$$
$$\leq x(1-x)(1-y)+y(1-x)(1-y)$$
$$\leq x(1-y)+y(1-x)-x^2(1-y)-y^2(1-x)+.....+x(1-y)+y(1-x)$$
$$\leq P(C)$$ [Using eq.1]
Thus $$P(C) \geq P(A \cup B)P(\overline{A} \cap \overline{B})$$  is proved.


Class 11 Commerce Maths Extra Questions