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NCERT Solutions

Class 12

Maths

Chapter 13 Probability Exercise 13.3

NCERT Solutions For Class 12 Maths Chapter 13 Probability Exercise 13.3 - 2025-26

Q: 1. How are NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3 helpful in understanding conditional probability?

These NCERT Solutions provide stepwise explanations on applying conditional probability and related theorems to problems as per the latest CBSE syllabus. By focusing on the method of defining events, clearly outlining what is 'given', and using formulas like P(A|B) = P(A ∩ B)/P(B), students develop a clear and exam-ready approach to tricky probability concepts.

Q: 2. What type of step-by-step methods should I follow when solving NCERT Exercise 13.3 problems?

Begin with defining all relevant events (e.g., A and B). Then, identify any ‘given that’ condition for clarity. Apply the multiplication law or conditional probability formula as needed, always calculating denominators first in conditional questions. Box key results and clearly mention each calculated probability step.”

Q: 3. Which are the most common mistakes students make while solving conditional probability questions in Exercise 13.3?

Mixing up formulas for independent and dependent eventsForgetting to calculate the correct denominator in conditional probabilityNot defining events properly before solvingSkipping calculation steps, leading to errorsAlways check for ‘given that’ in the question and ensure all events and steps are well-defined in your solution.

Q: 4. How do NCERT Solutions ensure the correct application of Bayes’ Theorem in board-style problems?

These solutions guide you to identify prior and conditional probabilities explicitly, then substitute correctly into Bayes’ Theorem as required by CBSE board exam standards. All calculations follow syllabus-approved logic, helping students avoid common pitfalls and ensuring full marks in proof-based answers.

Q: 5. What is the main difference between independent events and mutually exclusive events, and how is this shown in the solutions?

Independent events: The outcome of one event does not impact the probability of the other. Shown by using the multiplication law P(A ∩ B) = P(A) × P(B).Mutually exclusive events: Both cannot occur at the same time. Addressed by the addition law P(A ∪ B) = P(A) + P(B).NCERT stepwise solutions clarify these with clear event definitions so you apply the correct approach in each case.

Q: 6. How can I check if a question from Exercise 13.3 is asking for conditional probability?

Look for phrases like ‘given that’ or wording indicating an event has already occurred. If the problem requires you to find the probability of one event assuming another has happened, use P(A | B). Reading the question carefully for such keywords is essential to avoid applying the wrong formula.

Q: 7. Why are stepwise NCERT solutions preferred for Class 12 board exams in probability?

Stepwise solutions help you structure your answer logically, making it easy for examiners to award marks for each stage as per CBSE guidelines. You show clear event definitions, formulas used, substitution, and final boxed answers—this increases accuracy and partial marking opportunities.

Q: 8. What key probability formulas should I revise repeatedly before the board exam?

P(A ∩ B) = P(A) × P(B) (for independent events)P(A|B) = P(A ∩ B)/P(B) (conditional probability)P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (addition law)Bayes’ theorem for reversing conditional probabilityCreating a summary sheet with boxed formulas and practicing their usage in multiple questions from Exercise 13.3 reinforces exam readiness.

Q: 9. How do these solutions help in differentiating when to use Bayes’ Theorem versus direct conditional probability?

If you need to find the probability of a cause given an observed effect (i.e., reverse probability), Bayes’ Theorem is used. If you are calculating the probability of an event with another already occurring, use direct conditional probability. The solutions walk you through both situations in the syllabus context.

Q: 10. What strategies improve accuracy when answering probability word problems in the board exam?

Define all events and what is ‘given’ before you start calculationsDraw probability trees or diagrams for clarity if neededCheck if the question involves dependency or independenceDouble-check denominators in conditional calculationsPractice with a variety of solved examples from the NCERT Exercise 13.3

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Maths Class 12 Chapter 13 Questions and Answers - Free PDF Download

In NCERT Solutions Class 12 Maths Chapter 13 Exercise 13 3, you’ll dive deeper into Probability, especially conditional probability and event independence. These are important concepts for CBSE Class 12 students, and sometimes they can feel tricky at first. The step-by-step NCERT Solutions make things simpler, showing you exactly how to approach each type of problem so you feel confident—no matter how the question is asked in your board exams.

Table of Content

With Vedantu’s clear explanations and downloadable solutions, it’s easier to understand formulas, avoid common mistakes, and learn how to define events clearly for each probability question. Don’t forget to check the Class 12 Maths syllabus for the latest exam pattern and weightage for topics like Probability.

Practicing these NCERT solutions regularly will help you build speed and accuracy for your final exam. Plus, Probability usually carries 8 marks in your CBSE board, so mastering this chapter can really improve your total score!

Access NCERT Solutions for Maths Class 12 Chapter 13 - Probability

```html 1. An urn contains 5 red and 5 black balls. A ball is drawn at random; its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Summary: The question asks for the probability that the second ball drawn is red, after updating the urn based on the first draw and then drawing again.

If a red ball is drawn first (probability = 1/2), then the urn has 7 red, 5 black balls. Probability second is red: 7/12.
If a black ball is drawn first (probability = 1/2), then the urn has 5 red, 7 black balls. Probability second is red: 5/12.
Total probability = (1/2 × 7/12) + (1/2 × 5/12) = 1/2.

2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
The probability that the red ball was drawn from the first bag is:

Let E₁: first bag chosen, E₂: second bag chosen. P(E₁) = P(E₂) = 1/2.
P(A|E₁) = 4/8 = 1/2 (red from first bag), P(A|E₂) = 2/8 = 1/4 (red from second bag).
By Bayes’ theorem:
P(E₁|A) = [1/2 × 1/2] / [1/2 × 1/2 + 1/2 × 1/4]
= (1/4) / (1/4 + 1/8) = 2/3

3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?
Summary: To find the probability a student who got an A grade is a hosteller.

P(E₁) = 0.6 (hosteller), P(E₂) = 0.4 (day scholar).
P(A|E₁) = 0.3, P(A|E₂) = 0.2.
By Bayes’ theorem:
P(E₁|A) = [0.6 × 0.3] / [0.6 × 0.3 + 0.4 × 0.2] = 0.18/0.26 = 9/13

4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?
The probability the student actually knew the answer (given the correct answer) is 12/13.

5. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Summary: This is a Bayes’ theorem question; it asks for the probability a person who tested positive really has the disease.

P(E₁) = 0.001 (disease), P(E₂) = 0.999 (no disease).
P(A|E₁) = 0.99 (true positive), P(A|E₂) = 0.005 (false positive).
P(E₁|A) = [0.001 × 0.99] / [0.001 × 0.99 + 0.999 × 0.005] = 0.00099/0.005985 = 22/133

6. There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Given that heads appears, the probability that the coin was the two-headed one is 4/9.

7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Summary: We need the probability that a person who had an accident was a scooter driver, using Bayes’ theorem.

Proportion of drivers: scooter (1/6), car (1/3), truck (1/2).
Probability of accident: scooter (1/100), car (3/100), truck (15/100).
Apply Bayes’ theorem:
P(scooter|accident) = [1/6 × 1/100] / [1/6 × 1/100 + 1/3 × 3/100 + 1/2 × 15/100]
= 1/6 ÷ 104/12 = 1/52

8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Future, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen random from this and is found to be defective. What is the probability that was produced by machine B?
The probability that a defective item was made by machine B is 1/4.

9. Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Summary: Using Bayes’ theorem, we find the probability the second group introduced the product.

First group wins (0.6), new product given first group (0.7).
Second group wins (0.4), new product given second group (0.3).
P(second group|product) = [0.4 × 0.3] / [0.6 × 0.7 + 0.4 × 0.3] = 2/9

10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
The probability the girl rolled a 1, 2, 3, or 4, given she got exactly one head, is 8/11.

11. A manufacture has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?
Summary: We want the probability that a defective item was produced by operator A, using their working times and defect rates.

Operator times: A (1/2), B (3/10), C (1/5).
Defect rates: A (1/100), B (5/100), C (7/100).
P(A|defective) = [1/2 × 1/100]/[1/2 × 1/100 + 3/10 × 5/100 + 1/5 × 7/100] = 5/34

12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
The probability that the lost card is a diamond, given two drawn are diamonds, is 11/50.

13. Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
A) 4/5
B) 1/2
C) 1/5
D) 2/5

The probability that there was actually a head is 4/5.

14. If A and B are two events such that A ⊆ B and P(B) ≠ 0, then which of the following is correct?
A) P(A|B)=P(B)/P(A)
B) P(A|B)<P(A)
C) P(A|B) ≥ P(A)
D) None of these

The correct option is: P(A|B) ≥ P(A)

Main Points from Probability Class 12 Exercise 13.3

Conditional probability uses the formula P(A|B) = P(A ∩ B)/P(B).
Bayes’ theorem helps to reverse conditional probabilities for real-life questions.
Identifying independent and mutually exclusive events is important in 13.3 class 12 maths.
Define your events and ‘given that’ conditions clearly before solving.
Many questions involve breaking down problems and step-by-step calculation.

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FAQs on NCERT Solutions For Class 12 Maths Chapter 13 Probability Exercise 13.3 - 2025-26

1. How are NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3 helpful in understanding conditional probability?

These NCERT Solutions provide stepwise explanations on applying conditional probability and related theorems to problems as per the latest CBSE syllabus. By focusing on the method of defining events, clearly outlining what is 'given', and using formulas like P(A|B) = P(A ∩ B)/P(B), students develop a clear and exam-ready approach to tricky probability concepts.

2. What type of step-by-step methods should I follow when solving NCERT Exercise 13.3 problems?

Begin with defining all relevant events (e.g., A and B). Then, identify any ‘given that’ condition for clarity. Apply the multiplication law or conditional probability formula as needed, always calculating denominators first in conditional questions. Box key results and clearly mention each calculated probability step.”

3. Which are the most common mistakes students make while solving conditional probability questions in Exercise 13.3?

Mixing up formulas for independent and dependent events
Forgetting to calculate the correct denominator in conditional probability
Not defining events properly before solving
Skipping calculation steps, leading to errors

Always check for ‘given that’ in the question and ensure all events and steps are well-defined in your solution.

4. How do NCERT Solutions ensure the correct application of Bayes’ Theorem in board-style problems?

These solutions guide you to identify prior and conditional probabilities explicitly, then substitute correctly into Bayes’ Theorem as required by CBSE board exam standards. All calculations follow syllabus-approved logic, helping students avoid common pitfalls and ensuring full marks in proof-based answers.

5. What is the main difference between independent events and mutually exclusive events, and how is this shown in the solutions?

Independent events: The outcome of one event does not impact the probability of the other. Shown by using the multiplication law P(A ∩ B) = P(A) × P(B).
Mutually exclusive events: Both cannot occur at the same time. Addressed by the addition law P(A ∪ B) = P(A) + P(B).

NCERT stepwise solutions clarify these with clear event definitions so you apply the correct approach in each case.

6. How can I check if a question from Exercise 13.3 is asking for conditional probability?

Look for phrases like ‘given that’ or wording indicating an event has already occurred. If the problem requires you to find the probability of one event assuming another has happened, use P(A | B). Reading the question carefully for such keywords is essential to avoid applying the wrong formula.

7. Why are stepwise NCERT solutions preferred for Class 12 board exams in probability?

Stepwise solutions help you structure your answer logically, making it easy for examiners to award marks for each stage as per CBSE guidelines. You show clear event definitions, formulas used, substitution, and final boxed answers—this increases accuracy and partial marking opportunities.

8. What key probability formulas should I revise repeatedly before the board exam?

P(A ∩ B) = P(A) × P(B) (for independent events)
P(A|B) = P(A ∩ B)/P(B) (conditional probability)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (addition law)
Bayes’ theorem for reversing conditional probability

Creating a summary sheet with boxed formulas and practicing their usage in multiple questions from Exercise 13.3 reinforces exam readiness.

9. How do these solutions help in differentiating when to use Bayes’ Theorem versus direct conditional probability?

If you need to find the probability of a cause given an observed effect (i.e., reverse probability), Bayes’ Theorem is used. If you are calculating the probability of an event with another already occurring, use direct conditional probability. The solutions walk you through both situations in the syllabus context.

10. What strategies improve accuracy when answering probability word problems in the board exam?

Define all events and what is ‘given’ before you start calculations
Draw probability trees or diagrams for clarity if needed
Check if the question involves dependency or independence
Double-check denominators in conditional calculations
Practice with a variety of solved examples from the NCERT Exercise 13.3

11. In what ways are conditional probability and independence of events tested in Exercise 13.3?

Exercise 13.3 includes questions where you assess if two events are independent (their joint probability equals the product of their probabilities) and apply conditional probability when the occurrence of one event affects the probability of another. Some problems require you to use both concepts together to test conceptual understanding.

12. How do NCERT Solutions address the hardest board-style questions on probability?

The hardest questions usually combine multiple concepts, such as Bayes’ Theorem, conditional probability, and event independence. NCERT Solutions demonstrate how to break down such problems into smaller definable events, guide through all steps, and highlight where most students make mistakes, providing strategies for systematic, accurate answers.

13. What should I do if I get confused between mutually exclusive and independent events in a problem?

Remember that mutually exclusive events never occur together (P(A ∩ B) = 0), while independent events can occur together but do not influence each other’s probabilities. Re-read definitions in the NCERT textbook and review example problems that clarify these distinctions.

14. How are real-life scenarios like medical tests or coin tosses represented in probability solutions?

Real-life scenarios are translated into defined events (such as test outcomes as A/B, or heads/tails). The NCERT Solutions teach students to model these with formal probability language, apply the appropriate formulas, and interpret results as per the Class 12 Maths syllabus for probability.