NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

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Edited By Ravindra Pindel | Updated on Jul 5, 2022 - 3:02 p.m. IST

NCERT solutions for Class 12 Maths chapter 13 miscellaneous exercise consist of different types of questions covering all the concepts of probability Class 12 Maths NCERT syllabus. These questions can be practised, if you have solved all the previous NCERT book chapter 13 exercises. Miscellaneous exercise chapter 13 Class 13 questions are a bit difficult than all the previous exercises questions. You are advised to solve all exercise questions by yourself. If you find difficulties while solving them, you can go through NCERT solutions for Class 12 Maths chapter 13 miscellaneous exercise. There are very few questions asked in the board exams from miscellaneous exercises, it is not considered to be very important for board exams. You can solve these exercises to be on the safer side in the board exams. These exercises are very important for competitive exams, so you must solve them as well. You can also check for NCERT Solutions.

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Probability Class 12 Chapter 13-Miscellaneous Exercise

Question:1(i) A and B are two events such that $P(A)\neq 0.$ Find $P(B\mid A),$ if

$A$ is a subset of $B$

Answer:

A and B are two events such that $P(A)\neq 0.$

$A\subset B$

$\Rightarrow \, \, \, \, A\cap B=A$

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

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$P(B|A)=\frac{P( A)}{P(A)}$

$P(B|A)=1$

Question:1(ii) $A$ and $B$ are two events such that $P(A)\neq 0.$ Find $P(B\mid A),$ if

$A\cap B=\phi$

Answer:

A and B are two events such that $P(A)\neq 0.$

$P(A\cap B)=P(B\cap A)=0$

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$P(B|A)=\frac{0}{P(A)}$

$P(B|A)=0$

Question:2(i) A couple has two children,

Find the probability that both children are males, if it is known that at least one of the children is male.

Answer:

A couple has two children,

sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$

Let A be both children are males and B is at least one of the children is male.

$(A\cap B)=\left \{ (b,b) \right \}$

$P(A\cap B)=\frac{1}{4}$

$P(A)=\frac{1}{4}$

$P(B)=\frac{3}{4}$

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

$P(A|B)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Question:2(ii) A couple has two children,

Find the probability that both children are females, if it is known that the elder child is a female.

Answer:

A couple has two children,

sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$

Let A be both children are females and B be the elder child is a female.

$(A\cap B)=\left \{ (g,g) \right \}$

$P(A\cap B)=\frac{1}{4}$

$P(A)=\frac{1}{4}$

$P(B)=\frac{2}{4}$

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

$P(A|B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$

Question:3 Suppose that $5^{o}/_{o}$ of men and $0.25^{o}/_{o}$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Answer:

We have $5^{o}/_{o}$ of men and $0.25^{o}/_{o}$ of women have grey hair.

Percentage of people with grey hairs $=(5+0.25)\%=5.25\%$

The probability that the selected haired person is male :

$=\frac{5}{5.25}=\frac{20}{21}$

Question:4 Suppose that $90^{o}/_{o}$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Answer:

$90^{o}/_{o}$ of people are right-handed.

$P(right-handed)=\frac{9}{10}$

$P(left-handed)=q=1-\frac{9}{10}=\frac{1}{10}$

at most $6$ of a random sample of $10$ people are right-handed.

the probability that more than $6$ of a random sample of $10$ people are right-handed is given by,

$^{10}C_rP^{r}q^{10-r}$

$=\sum_{T}^{10}$ $^{10} C_r$ $\frac{9}{10}^r .(\frac{1}{10})^{10-r}$

the probability that at most $6$ of a random sample of $10$ people are right-handed is given by

$=1-\sum_{T}^{10}$ $^{10}C_r$ . $\frac{9}{10}^r .(\frac{1}{10})^{10-r}$

Question:5(i) An urn contains $25$ balls of which $10$ balls bear a mark $'X'$ and the remaining $15$ bear a mark $'Y'$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If $6$ balls are drawn in this way, find the probability that

all will bear $'X'$ mark.

Answer:

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

$P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}$

$P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}$

$6$ balls are drawn with replacement.

Let Z be a random variable that represents a number of balls with Y mark on them in the trial.

Z has a binomial distribution with n=6.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(Z=0)=^6C_0 (\frac{2}{5})^{6} \frac{3}{5}^0$

$P(Z=0)=^6C_0 (\frac{2}{5})^{6}$

Question:5(ii) An urn contains $25$ balls of which $10$ balls bear a mark $'X'$ and the remaining $15$ bear a mark $'Y'.$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If $6$ balls are drawn in this way, find the probability that

not more than $2$ will bear $'Y'$ mark.

Answer:

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

$P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}$

$P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}$

$6$ balls are drawn with replacement.

Let Z be random variable that represents number of balls with Y mark on them in trial.

Z has binomail distribution with n=6.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(not\, more\, than\, 2\, bear\, Y) =P(Z\leq 2)$

$=P(Z=0)+P(Z=1)+P(Z=2)$

$=^6C_0 (\frac{2}{5})^{6} (\frac{3}{5})^0+^6C_1 (\frac{2}{5})^{5} (\frac{3}{5})^1+^6C_2 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{6} +6 (\frac{2}{5})^{5} (\frac{3}{5})^1+15 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{4} [(\frac{2}{5})^{2}+6 (\frac{2}{5}) (\frac{3}{5})+15(\frac{3}{5})^2]$

$= (\frac{2}{5})^{4} [\frac{4}{25}+\frac{36}{25}+\frac{135}{25}]$

$= (\frac{2}{5})^{4} [\frac{175}{25}]$

$= (\frac{2}{5})^{4} [7]$

Question:5(iii) An urn contains 25 balls of which 10 balls bear a mark $'X'$ and the remaining 15 bear a mark $'Y'.$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

at least one ball will bear $'Y'$ mark.

Answer:

$P(atleast\, one\, bear\, \, Y\, mark) =P(Z\geq 1)=1-P(Z=0)$

$=1-(\frac{2}{5})^6$

Question:5(iv) An urn contains 25 balls of which 10 balls bear a mark $'X'$ and the remaining 15 bear a mark $'Y'.$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

the number of balls with $'X'$ mark and $'Y'$ mark will be equal.

Answer:

$P(equal\, to\, the \, number\, of\, balls \, with \, X \, mark \, and\, Y \, mark )=P(Z=3)$

$P(Z=3)=^6C_3.(\frac{2}{5})^2.(\frac{3}{5})^3$

$P(Z=3)=\frac{20\times 8\times 27}{15625}$

$P(Z=3)=\frac{864}{3125}$

Question:6 In a hurdle race, a player has to cross $10$ hurdles. The probability that he will clear each hurdle is $\frac{5}{6}$ . What is the probability that he will knock down fewer than $2$ hurdles?

Answer:

Let p and q respectively be probability that the player will clear and knock down the hurdle.

$p=\frac{5}{6}$

$q=1-p=1-\frac{5}{6}=\frac{1}{6}$

Let X represent random variable that represent number of times the player will knock down the hurdle.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(Z< 2)=P(Z=0)+P(Z=1)$

$=^{10}C_0 .( \frac{5}{6})^{10}.( \frac{1}{6})^{0}+^{10}C_1 .( \frac{5}{6})^{9}.( \frac{1}{6})^{1}$

$=( \frac{5}{6})^{10}+10.( \frac{5}{6})^{9}.( \frac{1}{6})$

$=( \frac{5}{6})^{9}( \frac{5}{6}+10\times . \frac{1}{6})$

$=( \frac{5}{6})^{9} \times \frac{5}{2}$

$=\frac{5^1^0}{2\times 6^9}$

Question:7 A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Answer:

Probability of 6 in a throw of die =P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

Probability that 2 sixes come in first five throw of die :

$=^5C_2 (\frac{5}{6})^3 (\frac{1}{6})^2$

$=\frac{10\times 5^3}{6^5}$

Probability that third six comes in sixth throw :

$=\frac{10\times 5^3}{6^5}\times \frac{1}{6}$

$=\frac{10\times 125}{6^6}$

$=\frac{1250}{23328}$

Question:8 If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Answer:

In a leap year, there are 366 days.

In 52 weeks, there are 52 Tuesdays.

The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.

The remaining 2 days can be :

1. Monday and Tuesday

2. Tuesday and Wednesday

3. Wednesday and Thursday

4. Thursday and Friday

5.friday and Saturday

6.saturday and Sunday

7.sunday and Monday

Total cases = 7.

Favorable cases = 2

Probability of having 53 Tuesday in a leap year = P.

$P=\frac{2}{7}$

Question:9 An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Answer:

Probability of success is twice the probability of failure.

Let probability of failure be X

then Probability of success = 2X

Sum of probabilities is 1.

$\therefore \, \, \, X+2X=1$

$\Rightarrow \, \, \, 3X=1$

$\Rightarrow \, \, \, X=\frac{1}{3}$

Let $P=\frac{1}{3}$ and $q=\frac{2}{3}$

Let X be random variable that represent the number of success in six trials.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X\geq 4)=P(X=4)+P(X=5)+P(X=6)$

$=^6C_4 \left [ \frac{2}{3} \right ]^4\left [ \frac{1}{3} \right ]^2+^6C_5 \left [ \frac{2}{3} \right ]^5\left [ \frac{1}{3} \right ]^1+^6C_6 \left [ \frac{2}{3} \right ]^6\left [ \frac{1}{3} \right ]^0$

$=\frac{15\times 2^4}{3^6}+\frac{6\times 2^5}{3^6}+\frac{2^6}{3^6}$

$=\frac{ 2^6}{3^6}(15+12+4)$

$=\frac{ 2^6}{3^6}(31)$

$=\frac{31}{9}\left [ \frac{2}{3} \right ]^4$

Question:10 How many times must a man toss a fair coin so that the probability of having at least one head is more than $90^{o}/_{o}$ ?

Answer:

Let the man toss coin n times.

Probability of getting head in first toss = P

$P=\frac{1}{2}$

$q=\frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^nC_x.\left ( \frac{1}{2} \right )^{n-x}.\left ( \frac{1}{2} \right )^x$

$P(getting\, atleast\, one\, head)> \frac{90}{100}$

$P(X\geq 1)> 0.9$

$1-P(X=0)> 0.9$

$1-^nC_0 \frac{1}{2^n}> 0.9$

$^nC_0 \frac{1}{2^n}< 0.1$

$\frac{1}{2^n}< 0.1$

$\frac{1}{0.1}< 2^n$

$10< 2^n$

The minimum value to satisfy the equation is 4.

The man should toss a coin 4 or more times.

Question:11 In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.

Answer:

In a throw of die,

probability of getting six = P

$P=\frac{1}{6}$

probability of not getting six = q

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

There are three cases :

1. Gets six in the first throw, required probability is $\frac{1}{6}$

The amount he will receive is Re. 1

2.. Does not gets six in the first throw and gets six in the second throw, then the probability

$=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$

The amount he will receive is - Re.1+ Re.1=0

3. Does not gets six in first 2 throws and gets six in the third throw, then the probability

$=\frac{5}{6}\times\frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$

Amount he will receive is -Re.1 - Re.1+ Re.1= -1

Expected value he can win :

$=1\times \frac{1}{6}+0\times \frac{5}{36}+(-1)\times \frac{25}{216}$

$= \frac{1}{6}-\frac{25}{216}$

$= \frac{11}{216}$

Question:12(i) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A ?

Answer:

Let R be the event of drawing red marble.

Let $E_A,E_B,E_C$ respectively denote the event of selecting box A, B, C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box A is $P(E_A|R)$

$P(E_A|R)=\frac{P(E_A\cap R)}{P(R)}$

$=\frac{\frac{1}{40}}{\frac{3}{8}}$

$=\frac{1}{15}$

Question:12(ii) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box B?

Answer:

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box B is $P(E_B|R)$

$P(E_B|R)=\frac{P(E_B\cap R)}{P(R)}$

$=\frac{\frac{6}{40}}{\frac{3}{8}}$

$=\frac{2}{5}$

Question:12(iii) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box C?

Answer:

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box C is $P(E_C|R)$

$P(E_C|R)=\frac{P(E_C\cap R)}{P(R)}$

$=\frac{\frac{8}{40}}{\frac{3}{8}}$

$=\frac{8}{15}$

Question:13 Assume that the chances of a patient having a heart attack is $40^{o}/_{o}.$ It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30^{o}/_{o}$ and prescription of certain drug reduces its chances by $25^{o}/_{o}.$ At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer:

Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted

the drug prescription.

$\therefore \, \, \, \, P(A)=0.40$

$\therefore \, \, \, \, P(E1)=P(E2)=\frac{1}{2}$

$P(A|E1)=0.40\times 0.70=0.28$

$P(A|E2)=0.40\times 0.75=0.30$

the probability that the patient followed a course of meditation and yoga is $P(E1,A)$

$P(E1,A)=\frac{P(E1).P(E1|A)}{P(E1).P(E1|A)+P(E2).P(E2|A)}$

$P(E1,A)=\frac{0.5\times 0.28}{0.5\times 0.28 + 0.5\times 0.30}$

$=\frac{14}{29}$

Question:14 If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ).

Answer:

Total number of determinant of second order with each element being 0 or 1 is $2^4=16$

The values of determinant is positive in the following cases $\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 0\\ 1 & 1\end{bmatrix}$

Probability is

$=\frac{3}{16}$

Question:15(i) An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = $0.2$

P(B fails alone) = $0.15$

P(A and B fail) = $0.15$

Evaluate the following probabilities

$P(A \; fails\mid B\; has\; failed)$

Answer:

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_A\, and \, E_B)=0.15$

$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$

$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$

$\Rightarrow \, \, \, P(E_B)=0.3$

$P(E_A|E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}$

$=\frac{0.15}{0.3}=0.5$

Question:15(ii) An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = $0.2$

P(B fails alone) = $0.15$

P(A and B fail) = $0.15$

Evaluate the following probabilities

$P(A\; fails \; alone)$

Answer:

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_A\, and \, E_B)=0.15$

$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$

$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$

$\Rightarrow \, \, \, P(E_B)=0.3$

$P(A\, fails\, \, alone)=P(E_A)-P(E_A\, and\, E_B)$

$=0.2-0.15=0.05$

Question:16 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer:

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.

$P(E1)=\frac{3}{7}$ and $P(E2)=\frac{4}{7}$

Let A be the event that ball drawn is red.

When a red ball is transfered from bag 1 to bag 2.

$P(A|E1)=\frac{5}{10}=\frac{1}{2}$

When a black ball is transfered from bag 1 to bag 2.

$P(A|E2)=\frac{4}{10}=\frac{2}{5}$

$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E2).P(A|E2)+P(E1).P(A|E1)}$

$=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{4}{7}\times \frac{2}{5}+\frac{3}{7}\times \frac{1}{2}}$

$=\frac{16}{31}$

Question:17 If A and B are two events such that $P(A\neq 0)$ and $P(B\mid A)=1,$ then

Choose the correct answer of the following:

(A) $A\subset B$

(B) $B\subset A$

(D) $A=\phi$

Answer:

A and B are two events such that $P(A\neq 0)$ and $P(B\mid A)=1,$

$P(B|A)=\frac{P(B\cap A)}{P(A)}$

$1=\frac{P(B\cap A)}{P(A)}$

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

$\Rightarrow \, \, \, A\subset B$

Option A is correct.

Question:18 If $P(A\mid B)> P(A)$ , then which of the following is correct :

(A) $P(B\mid A)< P(B)$

(B) $P(A\cap B)< P(A).P(B)$

(D) $P(B\mid A)= P(B)$

Answer:

$P(A\mid B)> P(A)$

$\Rightarrow \, \, \frac{P(A\cap B)}{P(B)}> P(A)$

$\Rightarrow \, \, P(A\cap B)> P(A).P(B)$

$\Rightarrow \, \, \frac{P(A\cap B)}{P(A)}> P(B)$

$\Rightarrow \, \, P(B|A)> P(B)$

Option C is correct.

Question:19 If A and B are any two events such that $P(A)+P(B)-P(A \; and\; B)=P(A),$ then

(A) $P(B\mid A)=1$

(B) $P(A\mid B)=1$

(D) $P(A\mid B)=0$

Answer:

$P(A)+P(B)-P(A \; and\; B)=P(A),$

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

$\Rightarrow \, \, \, P(B)-P(A\cap B)=0$

$\Rightarrow \, \, \, P(B)=P(A\cap B)$

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$

Option B is correct.

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Miscellaneous Exercise:-

NCERT solutions for Class 12 Maths chapter 13 miscellaneous exercise are explained in a very detailed manner, so you can understand them very easily.
There are 19 questions and 5 examples given in the miscellaneous exercise which you can solve to get a command on the probability theory.
NCERT solutions for Class 12 Maths chapter 13 miscellaneous exercise are very helpful for the students who are preparing for the competitive exams.
Miscellaneous exercise chapter 13 Class 12 is a mixture of the questions from probability theory which will check your understanding of the concept.
NCERT solutions for Class 12 Maths chapter 13 miscellaneous Exercise can be used for reference.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

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Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Question: Do questions from the miscellaneous exercises comes in 12th board exams?

Answer:

More than 95% of questions in the board's exam are not asked from miscellaneous exercises but it is very important for competitive exams.

Question: What are the topics in miscellaneous exercise probability Class 12 ?

Answer:

Miscellaneous exercise is consists of different questions from all the topics of probability Class 12 Maths.

Question: What is the probability of an impossible event ?

Answer:

The probability of an impossible event is 0.

Question: What is the probability of getting 0 when we roll a die ?

Answer:

The probability of getting 0 when we roll a die is zero.

Question: What is the probability of getting number less than 7 when we roll a die ?

Answer:

The probability of a number less than 7 when we roll a die is one.

Question: Give an example of certain event .

Answer:

The event of getting a number less than 7 when we roll a die is an example of a certain event.

Question: Give an example of impossible event ?

Answer:

The event of getting number 0 we roll a die is an example of an impossible event.

Question: Where can I get NCERT solutions ?

Answer:

Here you can get NCERT Solutions.

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NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Probability Class 12 Chapter 13-Miscellaneous Exercise

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Frequently Asked Question (FAQs) - NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Question: Do questions from the miscellaneous exercises comes in 12th board exams?

Question: What are the topics in miscellaneous exercise probability Class 12 ?

Question: What is the probability of an impossible event ?

Question: What is the probability of getting 0 when we roll a die ?

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