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Addition theorems of probability - class-XI

Description: addition theorems of probability
Number of Questions: 41
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Tags: probability - i maths descriptive statistics and probability business maths statistics and probability
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A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then $\displaystyle P(A \cup B)$ is

  1. $\displaystyle \frac{2}{5}$

  2. $\displaystyle \frac{3}{5}$

  3. 0

  4. 1


Correct Option: D
Explanation:

Here, $\displaystyle A=\left{ 4,5,6\right} $
and  $\displaystyle B=\left{ 1,2,3,4,\right} $
We have $\displaystyle A\cup B=\left{1,2,3,4,5,6\right} =S$
Where S is the sample space of the experiment of throwing a die.
$\displaystyle P(S) =1$ since it is a sure event.Hence $\displaystyle P(A\cup B)=1$

Twenty bags or sugar, each marked 10 kg actually give the following data:

The weight of a bag (in kg) No. of bags
     $9.5-9.8$          1
     $9.8-9.9$          2
     $9.9-10.0$          5
     $10.0-10.1$         12

The lower limits of the classes are inclusive and the upper limits are exclusive.

What is the probability that the bag selected at random (without any reference) weighs 10 kg or more?

  1. $0.6$

  2. $0.8$

  3. $0.5$

  4. $0.3$


Correct Option: A
Explanation:

Total number of bags = $12 + 5 + 2 + 1 = 20$
Bags which weigh more than $10 kg = 12$
Thus Probability (bag weighs more than 10 kg) = $\dfrac{12}{20} = 0.6$

In an experiment, there are exactly three elementary events. The probability of two of them are $\displaystyle\frac{2}{7}$ and $\displaystyle\frac{1}{7}$. What is the probability of third event?

  1. $\displaystyle\frac{4}{7}$

  2. $\displaystyle\frac{3}{7}$

  3. $\displaystyle\frac{2}{7}$

  4. $\displaystyle\frac{1}{7}$


Correct Option: A
Explanation:

We know that the sum of all the elementary events = 1
$\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7} + P _3 = 1$ 

$\Rightarrow \quad P _3 = 1 - \left(\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7}\right) = \displaystyle\frac{4}{7}$ 

If $n(A)=6,n(B)=8$ and $n(A\cup B)=12$, then $n(A\cap B)=$

  1. $6$

  2. $2$

  3. $8$

  4. $12$


Correct Option: B
Explanation:
We know that $n(A)+n(B)=n(A\cup B)+n(A\cap B)$
$n(A)=6,$ $n(B)=8$ and $n(A\cup B)=12$
$n(A)+n(B)-n(A\cup B)=n(A\cap B)$
$n(A\cap B)=6+8-12=2$

Let A be a set of $4$ elements. From the set of all functions from A to A, the probability that it is an into function is?

  1. $\dfrac{3}{32}$

  2. $0$

  3. $\dfrac{29}{32}$

  4. $1$


Correct Option: A
Explanation:
Let set A be of n element and from A to A then no of onto element is n! and total function are $n^{2}$

$P(E)=\dfrac{n!}{n}=\dfrac{4!}{4\times 4\times 4\times 4}$

$=\dfrac{4\times 3\times 2}{4\times 4\times 4\times }=\dfrac{3}{32}$

From a well shuffled standard pack of $52$ playing cards, one card is drawn. What is the probability that it is either a King of hearts or a Queen of diamonds.

  1. $\dfrac {1}{2}$

  2. $\dfrac {1}{4}$

  3. $\dfrac {1}{8}$

  4. $\dfrac {1}{26}$


Correct Option: D
Explanation:

No. of Kings of heart $=$ 1, No. of Queen of diamonds $=$1
Probability of king of hearts or queen of diamonds $ = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$

If a coin is tossed twice, then the events 'occurrence of one head',  'occurrence of $2$ heads' and 'occurrence of no head' are -

  1. Independent

  2. Equally likely

  3. Not equally likely

  4. Both (A) and (B)


Correct Option: C
Explanation:
Possible outcomes when a coin is tossed $=$ H or T
Possible outcomes when a coin is tossed twice$=$ HH,TT,TH,HT
$=4$ outcomes
P(occurrence of $1H$) $=$ TH,HT
$=2/4=1/2$
P(occurrence of $2H$) $=$ HH
$=1/4$
P(occurrence of no H) $=$ TT
$=1/4$
Therefore, the probability of occurrence of no head and occurrence of no head and occurrence of two head is same but probability of occurrence of one Head is $1/2$. Thus the events are:-
Not equally likely.

If events A and B are independent and P(A) $=$ 0.15, P(A $\cup $ B) $=$ 0.45, Then P(B) $=$ ..............

  1. $\displaystyle \frac{6}{13}$

  2. $\displaystyle \frac{6}{17}$

  3. $0.315$

  4. $0.352$


Correct Option: B,D
Explanation:
Events A and B are independent then $P\left( A\bigcap { B }  \right) =0$
$P\left( A\bigcup { B }  \right) =P(A)+P(B)-P(A).P(B)$
Given: $P(A)=0.15$
$P\left( A\bigcup { B }  \right) =0.45$
$\therefore P(B)=P\left( A\bigcup { B }  \right) -P(A)+P\left( A \right) P(B)$
$P(B)=0.45-0.15+1.15P(B)$
$P(B)=0.3+0.15P(B)$
$P(B)(1-0.15)=0.3$
$P(B)(0.85)=0.3$
$P(B)=\cfrac { 0.3 }{ 0.85 } =0.352=\cfrac { 6 }{ 17 } $

A card is randomly drawn from a well shuffled pack of $52$ playing cards. The probability that it is a club or numbered $5$ is 

  1. Not $(0.4 + 0.3)$

  2. $=$ $0.4 + 0.3$

  3. $=$$ 0.4 - 0.3$

  4. None of these


Correct Option: A
Explanation:

One card is both a 5 and a Club.
$\therefore$ Club or numbered 5 are not mutually exclusive.
P(club or numbered 5) $\neq$ 0.4 + 0.3

A die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $(A\cup B)$ is

  1. $\cfrac{2}{5}$

  2. $\cfrac{3}{5}$

  3. $0$

  4. $1$


Correct Option: D
Explanation:
$A \equiv \left\{4, 5, 6\right\}$
$B \equiv \left\{1, 2, 3, 4\right\}$
$\therefore \ A\cup B\equiv \left\{1, 2, 3, 4, 5, 6\right\}$
$\therefore \ P(A\cup B)=\dfrac {6}{6}=1$

Assume that the birth of a boy or girl to a couple to be equally likely,mutually exclusive exhaustive and independent of the other children in the family for a couple having $6$ children the probability that their 'three oldest are boy'is

  1. $\dfrac{{20}}{{64}}$

  2. $\dfrac{{1}}{{64}}$

  3. $\dfrac{2}{{64}}$

  4. $\dfrac{8}{{64}}$


Correct Option: C
Explanation:
Given that the birth of a boy or girl to a couple to be equally likely, 
$(1)$ mutually exclusive
$(2)$ exhaustive
$(3)$ independent
$\Rightarrow$ One event does not affect the other 
$P(E)=P(B).P(B).P(B), P(B\ or\ G). P(B\ or\ G)$
$P(B)=$ probability of boy is $\dfrac{1}{2}$
$P(G)=$ probability of girl is $\dfrac{1}{2}$
$P(E)=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times 2$ either boy or girls
$=\dfrac{2}{64}$

If  $A$ and $B$ are two independent events such that $P\left({A}^{\prime}\right)=0.7,  P\left({B}^{\prime}\right)=p$ and $P\left( A\cup B \right) =0.8,$ then the value of $p$ is 

  1. $1$

  2. $0.2$

  3. $0.7$

  4. $0.3$



Correct Option: D
Explanation:
$P(A')=0.7\Rightarrow P(A)=1-0.7=0.3$

$P(B)=1-p$ 

$P(A\cup B)=0.8$

$P(A\cap B)=P(A)\times P(B)=0.3(1-p)$ as they are independent events

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

$0.8=0.3+1-p-0.3(1-p)$

$0.8=0.3+1-p-0.3+0.3p$

$0.8=1-0.7p\Rightarrow 0.7p=0.2$

$p=\dfrac 27\approx0.3$

One card is drawn from a pack of $52$ cards. The probability that the card picked is either a spade or a king.

  1. $ \displaystyle \frac{1}{26} $

  2. $ \displaystyle \frac{3}{26} $

  3. $ \displaystyle \frac{4}{13} $

  4. $ \displaystyle \frac{1}{9} $


Correct Option: C
Explanation:
Let $A:$ event of drawing a king
$\displaystyle \Rightarrow P(A)=\dfrac{4}{52}=\dfrac{1}{13}$
$B:$ event of drawing a spade $\displaystyle =\dfrac{13}{52}=\dfrac{1}{4}$
$\Rightarrow \displaystyle P\left( A\cap B \right) =\dfrac { 1 }{ 52 } $
$\displaystyle \therefore P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) =\dfrac { 1 }{ 13 } +\dfrac { 1 }{ 4 } -\dfrac { 1 }{ 52 } =\dfrac { 4 }{ 13 } $

In a survey conducted among 400 students of X standard in Pune district, 187 offered to join Science faculty after X std. and 125 students offered to join Commerce faculty after X, If a student is selected at random from this group. Find the probability that student prefers Science or Commerce faculty.

  1. $\displaystyle \frac{39}{50}$

  2. $\displaystyle \frac{4}{5}$

  3. $\displaystyle \frac{41}{50}$

  4. $\displaystyle \frac{43}{50}$


Correct Option: A
Explanation:

Total number of students $= 400$
No. of science students $= 187$
No. of commerce students $ = 125$
Thus, Probabilty of the student being either science or commerce = $\dfrac{187}{400} + \dfrac{125}{400}$
= $\dfrac{312}{400}$
= $\dfrac{39}{50}$

An integer is chosen at random from the first two hundred digits.What is the probability that the integer chosen is divisible by 6 or 8 ?

  1. $\displaystyle\frac{29}{100}.$

  2. $\displaystyle\frac{1}{4}.$

  3. $\displaystyle\frac{1}{8}.$

  4. $\displaystyle\frac{21}{100}.$


Correct Option: B
Explanation:

Let A=the integer is divisible by 6 
$A={6,12,18,....198}$
$198=6+(n-1)6$
$\Rightarrow n =33$
So, $P(A)=\dfrac{33}{200}$

B=the integer is divisible by 8
$B={8,16,24,....200}$
$200=8+(n-1)8$
$\Rightarrow n=25$
So, $P(B)=\dfrac{25}{200}$

Both are divisible by 6 and 8 both 
$A\cap B=24,48,.....192$
$192=24+(n-1)24$
$\Rightarrow n=8$
So, $P(A\cap B)=\dfrac{8}{200}$

$\displaystyle \therefore P\left ( A\cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( AB \right )$
$\displaystyle=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{1}{4}.$

A die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $P(A \cup B)$

  1. $3/5$

  2. $0$

  3. $1$

  4. $2/5$


Correct Option: C
Explanation:

Simply speaking, any number on the dice is either less than $5$ or more than $3$ (since the range covers all the numbers from $1$ to $6$). Hence, required probability = $1$.

Alternate method:
$P(A\cup  B)=P(A)+P(B)-P(A\cap B)=\dfrac { 3 }{ 6 } +\dfrac { 4 }{ 6 } -\dfrac { 1 }{ 6 } =1$

A die is thrown. Let A be the event that the number obtained is greater than 3. Let$B$ be the event that the number obtained is less than 5. Then $P(A\cup B)$ is 

  1. 1

  2. $\displaystyle \frac{2}{5}$

  3. $\displaystyle \frac{3}{5}$

  4. $0$


Correct Option: A
Explanation:

$n(A\cup B)={1,2,3,4,5,6}\therefore  P(A\cup B)=1$

A card is drawn is from pack of 52 cards. Find the probability of drawing '5' of spade or '8' of hearts

  1. $\dfrac{1}{52}$

  2. $\dfrac{1}{26}$

  3. $\dfrac{3}{52}$

  4. $\dfrac{1}{13}$


Correct Option: B
Explanation:

$\displaystyle P(A)=\frac { 1 }{ 52 } +\frac { 1 }{ 52 } =\frac { 2 }{ 52 } =\frac { 1 }{ 26 } $

A box contains 5 red balls, 8 green balls and 10 pink balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?

  1. $\dfrac{13}{23}$

  2. $\dfrac{10}{23}$

  3. $\dfrac{11}{23}$

  4. $\dfrac{13}{529}$


Correct Option: A
Explanation:

Total Number of Balls $= 5+8+10 = 23$ balls

Probability of getting a red ball  $= \dfrac{5}{23} $

Probability of getting a green ball  $= \dfrac{8}{23} $

Probability of getting a pink ball  $= \dfrac{10}{23} $

Then, probability of getting a red  or a green ball  $= \dfrac{5+8}{23} = \dfrac{13}{23}$

An institute organised a fete and ${1}/{5}$ of the girls and ${1}/{8}$ of the boys participated in the same. What fraction of the total number of students took part in the fete?

  1. ${2}/{13}$

  2. ${13}/{40}$

  3. Data inadequate

  4. None of these


Correct Option: C
Explanation:
Ans is "Data inadequate".

Suppose ther are 18 students (10 girls, 8 boys).

1/5 of 10=2 (girls).

1/8 of 8=1 (boy).

Total participation=2+1=3.

Total students=10+8=18.

Answer = 3/18 = 1/6.

Answer cannot be determined until we have the "Ratio of girls to boys".

For each different number of boys and girls, we get different answers. So we cannot obtain the required answers without knowing the ratio of boys and girls.

A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband`s selection is $\dfrac {1}{7}$ and that of wife's selection is $\dfrac {1}{5}$. What is the probability that only one of them will be selected?

  1. $\dfrac {1}{7}$

  2. $\dfrac {2}{7}$

  3. $\dfrac {3}{7}$

  4. $\dfrac {4}{7}$


Correct Option: B
Explanation:

There would be three cases.

1. Husband is selected while wife doesn't get selected.
2. Wife is selected while husband doesn't get selected.
3. Both of them get selected.

Probability of husband not to get selected $=\dfrac{6}{7}$
Probability of wife not to get selected $=\dfrac{4}{5}$
Probability only one to get selected $=\dfrac{6}{7} \times \dfrac{1}{5} +\dfrac{1}{7} \times \dfrac{4}{5} $
$=\dfrac{10}{35} =\dfrac{2}{7}$

A number $x$ is selected from first $100$ natural numbers. Find the probability that $x$ satisfies the condition $x+ \dfrac{100}{x} >50$

  1. $\dfrac{55}{100}$

  2. $\dfrac{45}{100}$

  3. $1$

  4. $\dfrac{50}{100}$


Correct Option: A
Explanation:

$x+\dfrac {100}x>50\\implies x^2 +100>50x\\implies x^2-50x+100+(25)^2-(25)^2>0\\implies x^2+(25)^2-50x-525>0\\implies (x-25)^2>525\\implies x-25>\pm\sqrt{525}\\implies x>25+22.91, x<25-22.91\\implies x>47.91, x<2.09\\implies x\ge 48, x\le 2$

As we have to select from 1 to 100 and x is greater than or equal to 48. Hence we have to select from 49 to 100 which is 53 numbers and favorable cases are 1, 2.
Total favorable case = 53+2=55.
Total number of cases = 100
Hence the required probability, $=\dfrac {55}{100}$

$A$ speaks the truth in $60\%$ cases and $B$ in $70\%$ cases. The probability that they will say the same thing while describing a single event is:

  1. $0.56$

  2. $0.54$

  3. $0.38$

  4. $0.94$


Correct Option: B
Explanation:

Both will speak the same thing means both can say truth or both can say false

The probability of truth for $A$ is $0.6$ and for $B$ is $0.7$
The probability of false for $A$ is $0.4$ and for $B$ is $0.3$
The probability that both say same thing is $0.6 \times 0.7 + 0.4 \times 0.3 = 0.42+0.12=0.54$
Therefore option $B$ is correct

In a single throw of two dice, the probability of obtaining a total of $7$ or $9,$ is:

  1. $\dfrac{4}{9}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{7}{18}$

  4. None of these


Correct Option: D
Explanation:

Let the outcomes of two dice be $x,y$.

For the sum to be $7$ , $x+y=7$. There are $6$ possibilities.
For the sum to be $9$ , $x+y=9$. There are $4$ possibilities.
Total number of possibilities for $7$ or $9$ is $10$.
The probability of obtaining total $7$ or $9$ is $\cfrac{10}{36}=\cfrac{5}{18}$.
Therefore option $D$ is correct.

The chance of throwing a total of $3$ or $5$ or $11$ with two dice is:

  1. $\dfrac{5}{36}$

  2. $\dfrac{1}{9}$

  3. $\dfrac{2}{9}$

  4. $\dfrac{19}{36}$


Correct Option: C
Explanation:

The number of possible combinations with two dice is $6 \times 6=36$.

The number of ways of getting $3$ as sum is $2$.
The number of ways of getting $5$ as sum is $4$.
The number of ways of getting $11$ as sum is $2$.
The number of ways of getting $3$ or $5$ or $11$ as sum is $8$.
The probability is $\cfrac{8}{36}=\cfrac{2}{9}$.
Therefore option $C$ is correct.

If A and B are mutually exclusive events such that $P(A)=\frac{3}{5}$ and $ P(B)=\frac{1}{5}$, then find $P(A \cup B)$. 

  1. $\dfrac{2}{5}$

  2. $\dfrac{3}{5}$

  3. $\dfrac{4}{5}$

  4. None of these


Correct Option: C
Explanation:

$A\quad and\quad B\quad are\quad mutually\quad exclusive\quad events\quad than\quad P(A\cap B)=0\ P(A)=\frac { 3 }{ 5 } \quad and\quad P(B)=\frac { 1 }{ 5 } \ P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \qquad \qquad =\frac { 3 }{ 5 } +\frac { 1 }{ 5 } -0=\frac { 4 }{ 5 } $

Two dice each numbered from $1$ to $6$ are thrown together. Let $A$ and $B$ be two events given by
$A:$ even number on the first die
$B:$  number on the second die is greater than $4$

What is $P(A\cup B)$ equal to?

  1. $1/2$

  2. $1/4$

  3. $2/3$

  4. $1/6$


Correct Option: C
Explanation:

Given:

Two dice are thrown, hence the total number of all possible ways, n(S) = $6\times 6=36$

A:  even number on the first die
B: the number on the second die is greater than 4

To find:
$P(A\cup B) $
favourable ways of event A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Hence n(A)=18
$\therefore, P(A)=\dfrac {n(A)}{n(S)}=\dfrac {18}{36}=\dfrac 12$
favourable ways of event B = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)}
Hence n(B) = 12
$\therefore, P(B)=\dfrac {n(B)}{n(S)}=\dfrac {12}{36}=\dfrac 13$
Hence, $P(\cup B)=P(A)+P(B)-P(A)(P(B)=\dfrac 12+\dfrac 13-\dfrac 12\times \dfrac 13=\dfrac {3+2-1}6=\dfrac 46=\dfrac 23$

If $A$ and $B$ are two events such that $P(A\cup B)=\cfrac { 3 }{ 4 } ,P(A\cap B)=\cfrac { 1 }{ 4 } ,P(\bar { A } )=\cfrac { 2 }{ 3 } $ where $\bar { A } $ is the complement of $A$, then what is $P(B)$ equal to?

  1. $\dfrac{1}{3}$

  2. $\dfrac{2}{3}$

  3. $\dfrac{1}{9}$

  4. $\dfrac{2}{9}$


Correct Option: B
Explanation:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\ \Rightarrow P(A\cup B)=1-P(\overset { \ _  }{ A } )+P(B)-P(A\cap B)\ \Rightarrow \dfrac { 3 }{ 4 } =1-\dfrac { 2 }{ 3 } +P(B)-\dfrac { 1 }{ 4 } \ \Rightarrow P(B)=\dfrac { 2 }{ 3 } $
Hence, option B is correct.

A is interviewed for $3$ posts. There are $3$ candidates for post $1,4$ for second post and $2$ for post No. three. The probability of A's being selected for at least one post is:

  1. $\dfrac{3}{4}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{5}$


Correct Option: A
Explanation:
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$    $P(B)=\dfrac{1}{4}$     $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{4} \right)-\left( \dfrac { 1 }{ 3 }\times \dfrac{1}{2} \right)-\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2} \right)+\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4}\times\dfrac{1}{2}\right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{8+6+12-2-4-3+1}{24}$
                                 $=\dfrac{3}{4}.$
Hence, the answer is $\dfrac{3}{4}.$

Find the probability of getting a total of $7$ or $11$ when a pair of dice is tossed.

  1. $\dfrac{1}{9}$

  2. $\dfrac{7}{9}$

  3. $\dfrac{2}{9}$

  4. $\dfrac{5}{9}$


Correct Option: C
Explanation:
Total number of all possible outcomes, $n(S)=6\times 6=36$
Let A: be the event that the sum is $7$.
Hence the favourable outcomes are ${(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$, i.e., $n(A)=6$
Hence the probability of getting a total of $7$. $P(E)=\dfrac {n(A)}{n(S)}=\dfrac {6}{36}$
Similarly, let B: be the event that the sum is $11$
Hence the favourable outcomes are {(5, 6), (6, 5)}, i.e., $n(E)=2$
Hence the probability of getting a total of 11 $P(B)=\dfrac {n(B)}{n(S)}=\dfrac {2}{36}$
Now, the probability of getting a total of $7$ or $11$ is
$P(A)+P(B)\\\implies =\dfrac 6{36}+\dfrac 2{36}=\dfrac 8{36}=\dfrac 29$

A is interviewed for $3$ posts. There are $3$ candidates for post $1, 4$ for second post and $2$ for post No. three. The probability of A's being selected for none of the posts is:

  1. $\dfrac{3}{4}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{4}$

  4. $\dfrac{1}{5}$


Correct Option: C
Explanation:
Let $P(A)=$ Probability of getting first company by the candidates Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(A\cap C)-P(B\cap C)+P(A\cap B \cap C)$
$\Rightarrow P(A)=\dfrac{1}{3}$   $P(B)$  $P(C)=\dfrac{1}{2}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\left( \dfrac {1  }{  3} \times\dfrac{1}{4}\right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{2}  \right) -\left( \dfrac { 1 }{ 4 }\times\dfrac{1}{2}  \right)=\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{4} \dfrac{1}{2} \right)$
                                 $=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{24}$
                                 $=\dfrac{3}{4}$
$\Rightarrow P$ ( not getting selected for any of the post ) $=1-\dfrac{3}{4}$
                                                                                $=\dfrac{1}{4}$
Hence, the answer is $\dfrac{1}{4}.$

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either $3$ or $13$?

  1. $\dfrac{2}{5}$

  2. $\dfrac{1}{9}$

  3. $\dfrac{11}{27}$

  4. $\dfrac{9}{27}$


Correct Option: A
Explanation:

Given: first $30$ natural numbers

To find: the probability of getting a multiple of either $3$ or $13$
According to the question, 
$n(S)=30$
Multiple of 3 in first 30 natural numbers are ${3, 6, 9, 12, 15, 18, 21, 24, 27, 30} = 10$
Multiple of $13$ in first $30$ natural numbers are ${13, 26}$
Hence the probability of getting a multiple of either $3$ or $13$ = probability of getting a multiple of 3 or probability of getting a multiple of 13
$\implies \dfrac {10}{30}+\dfrac 2{30}=\dfrac {12}{30}=\dfrac 25$
is the required probability.

If $P(A)=\dfrac {1}{8}$ and $P(B)=\dfrac {5}{8}$. Which of the following statement is/are not correct?

  1. $P(A\cup B)\leq \dfrac {3}{4}$

  2. $P(A\cap B)\leq \dfrac {1}{8}$

  3. $P(\overline A\cap B)\leq \dfrac {5}{8}$

  4. None of these


Correct Option: A,B,C
Explanation:

Given that $P\left( A \right) =\cfrac { 1 }{ 8 } ,\quad P\left( B \right) =\cfrac { 5 }{ 8 } $

we know that $P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $
a. $P\left( A\cup B \right) =\cfrac { 3 }{ 4 } -P\left( A\cup B \right)$
 $ \ P\left( A\cup B \right) $ would be max when $P\left( A\cap B \right) =0$
so,  
$P\left( A\cup B \right) \quad \le \quad \cfrac { 3 }{ 4 } $
So option A is correct.

b. 
$P\left( A\cup B \right) \quad \le \quad min\left( P\left( A \right) ,P\left( B \right)  \right) \ $
In this case, $min\left( P\left( A \right) ,P\left( B \right)  \right) \quad =\quad \frac { 1 }{ 8 } $
So, $P\left( A\cup B \right) \quad \le \quad \cfrac { 1 }{ 8 } $
Option B is correct.

c
$P\left( \bar { A }  \right) =1-P\left( A \right) \ \qquad \quad \quad =1-\cfrac { 1 }{ 8 } \ \quad \quad \quad \quad \quad =\cfrac { 7 }{ 8 } \ P\left( A\cup B \right) \quad \le \quad min\left( P\left( A \right) ,P\left( B \right)  \right) $
So, 
$P\left( \bar { A } \cap B \right) \quad \le \quad \cfrac { 5 }{ 8 } $
Option C is correct.

Addition Theorem of Probability states that for any two events $A$ and $B$,

  1. $P(A$ $\cup$ $B) = P(A) + P(B)$

  2. $P(A$ $\cup$ $B) = P(A) + P(B) + P(A$ $\cap$ $B)$

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

  4. $P(A$ $\cup$ $B) = P(A) \times P(B)$


Correct Option: C
Explanation:
For Addition theorem,
If $A$ and $B$ are any two events associated with a random experiment, then
$\Rightarrow P(A\cup B) = P(A) + P(B)- P(A\cap B)$
Hence, the answer is $P(A\cup B) = P(A) + P(B)- P(A\cap B).$

A, B and C in order toss a coin. the first one to throw a head wins. If A starts to toss, then 

  1. $P(A)=\dfrac{4}{7}$

  2. $P(B)=\dfrac{2}{7}$

  3. $P(C)=\dfrac{1}{7}$

  4. $P(C)=\dfrac{2}{7}$


Correct Option: B
Explanation:
P(A)=H+TTTH+TTTTTTH+.....
        =$1/2+{1 (1/2 })^{ 4 }+{ (1/2 })^{ 7 }+....$
        $=\frac { 1/2 }{ 1-1/8 } =4/7$
P(B)=TH+TTTTH+TTTTTTTH+.....
        =${ (1/2 })^{ 2 }+{ (1/2 })^{ 5 }+....$
        $=\frac { 1/4 }{ 1-1/8 } =2/7$
P(C)=TTH+TTTTTH+TTTTTTTTH+.....
        =${ (1/2 })^{ 3 }+{ (1/2 })^{ 6 }+....$
        $=\frac { 1/8 }{ 1-1/8 } =1/7$

A bag contain $5$ balls of unknown colors. A ball is drawn at random from it and is found to be white. The probability that bag contains only white ball is

  1. $\dfrac{3}{5}$

  2. $\dfrac{1}{5}$

  3. $\dfrac{2}{3}$

  4. $\dfrac{1}{3}$


Correct Option: D
Explanation:
Let $Q$ be the event that the drawn balls are white

$A$ be the event that the bag contains $1$ white ball
$B$ be the event that the bag contains $2$ white balls
$C$ be the event that the bag contains $3$ white balls
$D$ be the event that the bag contains $4$ white balls
$E$ be the event that the bag contains $5$ white balls

$P(A)=1/5,\ P(B)=1/5,\ P(C)=1/5,\ P(D)=1/5,\ P(E)=1/5$
$P(Q/A)=\dfrac {^1C _1}{^5C _1}=\dfrac {1}{5}$
$P(Q/B)=\dfrac {^2C _1}{^5C _1}=\dfrac {2}{5}$
$P(Q/C)=\dfrac {^3C _1}{^5C _1}=\dfrac {3}{5}$
$P(Q/D)=\dfrac {^4C _1}{^5C _1}=\dfrac {4}{5}$
$P(Q/E)=\dfrac {^5C _1}{^5C _1}=1$

using Baye's thorem,

$P(A/Q)=\dfrac {P(E)\ P(Q/E)}{P(A)\ P(Q/A)+P(B)\ P(Q/B)+P(C)\ P(Q/C)+P(D)\ P(Q/D)+P(E)\ P(Q/E)}$

$=\dfrac {1/5. 1}{1/5. 1/5+1/5.2/5+1/5.3/5 +1/5.4/5+1/5.1.}$

$=\dfrac {1/5}{1/5 (1/5)+2/5+3/5+4/5+1}$

$=\dfrac {1}{\dfrac {1+2+3+4+5}{5}}=\dfrac {5}{1+2+3+4+5}=\dfrac {5}{15}=\dfrac {1}{3}$

n men and n women are seated at round table in random order. The probability that they can be divided into n non-interrecting pairs so that each pair consists of a man and a women is

  1. 1/2n

  2. $2(2^n-1)/^{2n}C _n$

  3. $2n/^{2n}C _n$

  4. $1/(^nC _n)^2$


Correct Option: B
Explanation:

The women can be seated in $^{2n}C _n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left {(M, W), (M, W), ...(M, W)\right }$ of A(B) and $\left {(W, M), (W, M), ..., (W, M)\right }$ of B(A).
Therefore, $P(A\cup B)=P(A)+P(B)-P(A\cup B)$
$\displaystyle =\frac {2^n+2^n-2}{^{2n}C _n}=\frac {2^{n+1}-2}{^{2n}C _n}$.

A is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. The probability of A's being selected for at least one post is:

  1. $\dfrac{3}{4}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{3}{5}$


Correct Option: D
Explanation:
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A)=\dfrac{1}{4}$    $P(B)=\dfrac{1}{3}$     $P(C)=\dfrac{1}{5}$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$

                                 $=\dfrac{15+20+12-5-3-4+1}{60}$
                                 $=\dfrac{3}{5}.$
Hence, the answer is $\dfrac{3}{5}.$

$A$ is interviewed for $3$ posts. There are $4$ candidates for post $1, 3$ for second post and $5$ for post No. three. the probability of $A$'s being selected for none of the posts is:

  1. $\dfrac{3}{4}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{1}{4}$

  4. $\dfrac{1}{5}$


Correct Option: B
Explanation:
Let $P(A)=$ Probability of getting first company by the candidates.
Similarly, $P(B)$ $\xi$ $P(C)$ be the probability of getting into second and third company respectively.
$\Rightarrow P(A\cup B\cup C)= P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
$\Rightarrow P(A\cup B\cup C)=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{3} \right)-\left( \dfrac { 1 }{ 4 }\times \dfrac{1}{5} \right)-\left( \dfrac { 1 }{ 3 }\times\dfrac{1}{5} \right)+\left( \dfrac { 1 }{ 34}\times\dfrac{1}{3}\times\dfrac{1}{5}\right)$
                                 $=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{15}+\dfrac{1}{60}$
                                 $=\dfrac{3}{5}.$
$\Rightarrow P$ ( not getting selected for none of the post ) $=1-\dfrac{3}{5}$
                                                                                   $=\dfrac{2}{5}$
Hence, the answer is $\dfrac{2}{5}.$

Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:

Sum  Frequency
2 14
3 30
4 43
5 55
6 72
7 75
8 70
9 53
10 46
11 28
12 15

If the dice are thrown once more, what is the probability of getting a sum between 8 and 12?

  1. $0.154$

  2. $0.20$

  3. $0.254$

  4. $0.30$


Correct Option: C
Explanation:

If we rolled dice around 500 times then it can be considered as its expected outcome.
Hence,
Probability of obtaining between 8 and 12 as sum = 53 + 46 +28/500= 0.254

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