NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Komal MiglaniUpdated on 28 Apr 2025, 12:44 AM IST

You need two essential factors to plan your road trip since the driving conditions must have a working vehicle plus safe weather conditions. The likelihood of your planned journey becomes minimal once you know your car works properly, since rainfall conditions could prevent it. The NCERT solutions of Miscellaneous Exercise in Class 12 Maths Chapter 13 explores events which occur simultaneously or sequentially, thus making this a core decision-making technique.

This Story also Contains

  1. NCERT Solutions Class 12 Maths Chapter 13: Miscellaneous Exercise
  2. Topics Covered in Chapter 13 Probability: Miscellaneous Exercise
  3. NCERT Solutions of Class 12 Subject Wise
  4. NCERT Solutions for Class 12 Maths
  5. Subject Wise NCERT Exampler Solutions

This NCERT exercise contains problems about basic probability and conditional probability, and includes independent and dependent events alongside the multiplication and addition theorems. This exercise combines all concepts to check comprehension while aiding students during board examinations. Here are the detailed solutions.

NCERT Solutions Class 12 Maths 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:

$90^{\circ} \%$ of people are right-handed.

$
\begin{aligned}
& P(\text { right }- \text { handed })=\frac{9}{10} \\
& P(\text { left }- \text { handed })=q=1-\frac{9}{10}=\frac{1}{10}
\end{aligned}
$

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,

$
\begin{aligned}
& { }^{10} C_r P^r q^{10-r} \\
& =\sum_T^{1010} C_r \frac{9}{10}^r \cdot\left(\frac{1}{10}\right)^{10-r}
\end{aligned}
$

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

$
=1-\sum_T^{1010} C_r \cdot \frac{9}{10}^r \cdot\left(\frac{1}{10}\right)^{10-r}
$

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 6 (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 6 (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 6 (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 7: 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 8: 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 9 (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 9 (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 10: 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 11: 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) $B\subset A$

(C) $B=\phi$

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

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

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

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

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

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

(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.

Topics Covered in Chapter 13 Probability: Miscellaneous Exercise

1. Conditional Probability
Formula:

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

2.
Multiplication Theorem
$P(A \cap B)=P(B) \cdot P(A \mid B)$
3. Independent Events
Events A and B are independent if:

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

4. Bayes' Theorem
Used to revise probability with new information:

$
P\left(E_i \mid A\right)=\frac{P\left(E_i\right) \cdot P\left(A \mid E_i\right)}{\sum P\left(E_j\right) \cdot P\left(A \mid E_j\right)}
$

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Frequently Asked Questions (FAQs)

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

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

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

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

Q: What is the probability of an impossible event ?
A:

The probability of an impossible event is 0.

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

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

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

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

Q: Give an example of certain event .
A:

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

Q: Give an example of impossible event ?
A:

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

Q: Where can I get NCERT solutions ?
A:

Here you can get NCERT Solutions.

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The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

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Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

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Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

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Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



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For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.