CBSE Class 12 Mathematics Chapter 13 Probability NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Mathematics Chapter 13, Probability, introduces students to the core principles of chance. This chapter meticulously explains conditional probability, the multiplication rule, and the nature of independent events. The NCERT Solutions for this chapter offer a clear, step-by-step approach to solving problems from Exercise 13.1. They guide you through calculating conditional probabilities, such as P(E|F) and P(F|E), and determining the probabilities of event intersections and unions. These solutions are crafted to solidify your grasp of probability concepts and enhance your ability to tackle complex problems, serving as an excellent tool for both exam preparation and ongoing revision. The focus is on building a strong foundational understanding through detailed problem-solving techniques.

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BoardCBSE
ClassClass 12
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 13

Chapter summary

Chapter 13, Probability, for Class 12 Mathematics focuses on conditional probability and its applications. The NCERT Solutions for Exercise 13.1 provide clear, step-by-step methods to solve problems involving the calculation of conditional probabilities, the intersection of events, and the union of events, using given probabilities and conditional probability formulas.

Learning outcomes

  • Understand the definition and calculation of conditional probability.
  • Apply the formula for conditional probability P(A|B) = P(A ∩ B) / P(B).
  • Calculate the probability of the intersection of two events.
  • Calculate the probability of the union of two events using the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
  • Solve problems involving given probabilities and conditional probabilities.

Topics covered

Paper topics

  • Conditional Probability
  • Multiplication Rule of Probability
  • Independent Events
  • Probability of Intersection of Events
  • Probability of Union of Events
  • Basic Probability Calculations

Important topics

  • Conditional Probability Formula
  • Calculating P(A ∩ B)
  • Calculating P(A ∪ B)
  • Relationship between P(A|B) and P(B|A)

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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 given 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|F) = \frac{P(E \cap F)}{P(F)}

Substituting the given values:

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

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

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

Substituting the given values:

P(F|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 probability of event B and the probability of the intersection of events A and B:

  • P(B) = 0.5
  • P(A \cap B) = 0.32

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

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

Substitute the given values into the formula:

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

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

P(A|B) = \frac{32}{50}

Reducing the fraction to its simplest form:

P(A|B) = \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):

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|B) = \frac{0.32}{0.5}

To simplify, multiply the numerator and denominator by 10:

P(A|B) = \frac{3.2}{5} = \frac{32}{50}

Reducing the fraction:

P(A|B) = \frac{16}{25} = 0.64

So, P(A|B) = 0.64.

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

We use the formula for the probability of the union of two events:

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

Substitute the given values and the value of P(A \cap B) from part (i):

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

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

So, 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 equation, we can find P(A) and P(B):

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

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

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

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}

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

Now 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 formula for conditional probability.
  • Confusing P(A|B) with P(B|A).
  • Errors in calculating the intersection or union of events.
  • Mistakes in arithmetic operations with fractions or decimals.

Revision tips

  • Review the formulas for conditional probability, intersection, and union of events.
  • Work through each solved example carefully, understanding each step.
  • Practice re-solving the problems without looking at the solution to test your understanding.
  • Focus on identifying the given information and what needs to be calculated in each problem.

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 events A and B, if P(A) = 0.8, P(B) = 0.5, and P(A ∩ B) = 0.32, what is P(A ∪ B)?

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 Chapter 13 Probability for Class 12?

Chapter 13 for Class 12 Mathematics primarily focuses on conditional probability, the multiplication rule, and independent events, building upon basic probability concepts.

How are conditional probabilities calculated in these NCERT Solutions?

Conditional probabilities like P(E|F) are calculated using the formula P(E|F) = P(E ∩ F) / P(F), where P(E ∩ F) is the probability of both events occurring and P(F) is the probability of the conditioning event.

What is the formula for the union of two events?

The formula for the union of two events A and B is P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Are these solutions helpful for exam revision?

Yes, these solutions provide clear, step-by-step explanations for each problem, helping students understand the concepts and methods required for exam revision.

What are the key concepts covered in Exercise 13.1?

Exercise 13.1 covers the calculation of conditional probabilities, the probability of the intersection of events, and the probability of the union of events, given certain probability values.

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