CBSE Class 12 Mathematics Probability NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

This comprehensive set of NCERT Solutions for CBSE Class 12 Mathematics, Chapter 13 on Probability, provides detailed explanations and step-by-step solutions for Exercise 13.1. The solutions cover fundamental concepts of conditional probability, including calculating P(E|F) and P(F|E), and applying the formulas for the intersection and union of events. Students will find clear derivations for problems involving given probabilities of individual events and their intersections. These solutions are designed to reinforce understanding of key probability theorems and formulas, aiding students in mastering the subject matter. They serve as an excellent resource for exam preparation, offering clarity on complex calculations and problem-solving techniques, ensuring students are well-prepared for their board examinations.

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Quick info

BoardCBSE
ClassClass 12
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterProbability

Chapter summary

This chapter focuses on the core concepts of conditional probability and the relationships between events. The NCERT Solutions for Probability (Chapter 13) provide a clear walkthrough of calculating conditional probabilities like P(E|F) and P(F|E), and applying the formulas for the union and intersection of events. Exercise 13.1 specifically tests the understanding and application of these fundamental formulas through various numerical problems.

Learning outcomes

  • Understand the concept of conditional probability.
  • Calculate conditional probabilities P(E|F) and P(F|E).
  • Apply the formula for the intersection of two events.
  • Apply the formula for the union of two events.
  • Solve problems involving given probabilities of events and their intersections.

Topics covered

Paper topics

  • Conditional Probability
  • Intersection of Events
  • Union of Events
  • Probability Formulas
  • Calculating P(E|F)
  • Calculating P(F|E)
  • Calculating P(A ∩ B)
  • Calculating P(A ∪ B)

Important topics

  • Conditional Probability Formula
  • Intersection of Events
  • Union of Events Formula
  • Application of Probability Formulas

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Questions and Solutions

Question 1

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3, and P(E \cap F) = 0.2, find P(E|F) and P(F|E).
Solution:

We are provided with the probabilities of two events E and F, and their intersection:

  • P(E) = 0.6
  • P(F) = 0.3
  • P(E \cap F) = 0.2

To find the conditional probability P(E|F), we use the formula:

P(E \mid F) = \frac{P(E \cap F)}{P(F)}

Substituting the given values:

P(E \mid F) = \frac{0.2}{0.3} = \frac{2}{3}

Similarly, to find the conditional probability P(F|E), we use the formula:

P(F \mid E) = \frac{P(E \cap F)}{P(E)}

Substituting the given values:

P(F \mid E) = \frac{0.2}{0.6} = \frac{1}{3}

Thus, P(E|F) = \frac{2}{3} and P(F|E) = \frac{1}{3}.

Question 2

Compute P(A|B), if P(B) = 0.5 and P(A \cap B) = 0.32.
Solution:

We are given the following probabilities:

  • P(B) = 0.5
  • P(A \cap B) = 0.32

The formula for conditional probability P(A|B) is:

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

Substitute the given values into the formula:

P(A \mid B) = \frac{0.32}{0.5}

To simplify the fraction, we can multiply the numerator and denominator by 100:

P(A \mid B) = \frac{32}{50} = \frac{16}{25}

Therefore, P(A|B) = \frac{16}{25}.

Question 3

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A \cap B) (ii) P(A|B) (iii) P(A \cup B).
Solution:

We are given the following probabilities:

  • P(A) = 0.8
  • P(B) = 0.5
  • P(B|A) = 0.4

(i) Find P(A \cap B):

We use the definition of conditional probability P(B|A) = \frac{P(A \cap B)}{P(A)}. Rearranging this formula to solve for P(A \cap B), we get:

P(A \cap B) = P(B|A) \times P(A)

Substitute the given values:

P(A \cap B) = 0.4 \times 0.8 = 0.32

So, P(A \cap B) = 0.32.

(ii) Find P(A|B):

We use the formula for conditional probability P(A|B) = \frac{P(A \cap B)}{P(B)}. Using the value of P(A \cap B) found in part (i):

P(A \mid B) = \frac{0.32}{0.5}

To simplify, multiply the numerator and denominator by 100:

P(A \mid B) = \frac{32}{50} = \frac{16}{25} = 0.64

So, P(A|B) = 0.64.

(iii) Find P(A \cup B):

We use the formula for the union of two events: P(A \cup B) = P(A) + P(B) - P(A \cap B). Using the given values and the result from part (i):

P(A \cup B) = 0.8 + 0.5 - 0.32

P(A \cup B) = 1.3 - 0.32 = 0.98

Therefore, P(A \cup B) = 0.98.

Question 4

Evaluate P (A \cup B), if 2P (A) = P (B) = \frac{5}{13} and P(A|B) = \frac{2}{5}.
Solution:

We are given the following information:

  • 2P(A) = P(B) = \frac{5}{13}
  • P(A|B) = \frac{2}{5}

From the first condition, we can find P(A) and P(B):

2P(A) = \frac{5}{13} \Rightarrow P(A) = \frac{5}{2 \times 13} = \frac{5}{26}

P(B) = \frac{5}{13}

Now, we use the formula for conditional probability P(A|B) = \frac{P(A \cap B)}{P(B)}. We can rearrange this to find P(A \cap B):

P(A \cap B) = P(A|B) \times P(B)

Substitute the given values:

P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2 \times 5}{5 \times 13} = \frac{2}{13}

Finally, we need to evaluate P(A \cup B). We use the formula for the union of two events:

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

Substitute the values we have found:

P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}

To add and subtract these fractions, we find a common denominator, which is 26:

P(A \cup B) = \frac{5}{26} + \frac{5 \times 2}{13 \times 2} - \frac{2 \times 2}{13 \times 2}

P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}

P(A \cup B) = \frac{5 + 10 - 4}{26} = \frac{11}{26}

Therefore, P(A \cup B) = \frac{11}{26}.

Common mistakes

  • Incorrectly applying the conditional probability formula.
  • Confusing the intersection (AND) and union (OR) of events.
  • Errors in algebraic manipulation when solving for unknown probabilities.
  • Misinterpreting the given information in probability problems.

Revision tips

  • Review the formulas for conditional probability, intersection, and union of events.
  • Work through each solved example carefully to understand the step-by-step process.
  • Practice solving similar problems to build confidence and speed.
  • Pay attention to the notation used for events and probabilities.

Practice MCQs

Q1. If P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2, what is P(E|F)?

Q2. Given P(B) = 0.5 and P(A ∩ B) = 0.32, what is P(A|B)?

Q3. If P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4, what is P(A ∩ B)?

Q4. For any two events A and B, which formula represents the union of events?

Q5. If 2P(A) = P(B) = 5/13 and P(A|B) = 2/5, what is P(A ∩ B)?

Frequently asked questions

What is the main focus of the NCERT Solutions for Class 12 Probability, Chapter 13?

The solutions focus on conditional probability, including calculating P(E|F) and P(F|E), and applying formulas for the intersection and union of events, as presented in Exercise 13.1.

How do these solutions help in understanding conditional probability?

They provide step-by-step calculations and clear explanations for problems involving conditional probabilities, helping students grasp the concept and its application.

Are the formulas for intersection and union of events covered?

Yes, the solutions demonstrate how to use and apply the formulas for P(A ∩ B) and P(A ∪ B) in various problem-solving scenarios.

What is the value of P(E|F) if P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2?

The value of P(E|F) is calculated as P(E ∩ F) / P(F) = 0.2 / 0.3 = 2/3.

How can these NCERT Solutions be used for exam revision?

Students can use these solutions to review key concepts, understand problem-solving methods, and practice applying probability formulas, which is crucial for exam preparation.

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