Class 11 Maths Chapter 14 Probability – Exercise 14.3 NCERT Solutions

Introduction

Exercise 14.3 focuses on the multiplication theorem of probability. You will learn how to calculate probabilities of intersection of events, apply formulas for independent events, and solve problems involving combined probabilities. This exercise builds the foundation for conditional probability and Bayes’ theorem in Class 12.

Key Concepts

1. Multiplication Theorem: For events $A$ and $B$:

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

2. Independent Events: If $A$ and $B$ are independent:

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

3. Conditional Probability:

$$P(A\mathrm{\mid }B)=\frac{P(A\cap B)}{P(B)},P(B)\mathrm{\ne }0$$

NCERT Questions with Solutions (10)

Q1. A coin is tossed twice. Find probability of getting head on both tosses.

$$P(H_1\cap H_2)=P(H_1)\cdot P(H_2)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$$

Q2. A die is thrown twice. Find probability of getting 4 on both throws.

$$P(4_1\cap 4_2)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$$

Q3. A card is drawn twice with replacement. Find probability of getting ace both times.

$$P(A_1\cap A_2)=\frac{4}{52}\cdot \frac{4}{52}=\frac{1}{169}$$

Q4. A card is drawn twice without replacement. Find probability of getting ace both times.

$$P(A_1\cap A_2)=\frac{4}{52}\cdot \frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$$

Q5. A coin is tossed thrice. Find probability of getting all heads.

$$P(HHH)=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$$

Q6. A die is thrown twice. Find probability of getting even number on first and prime on second. Even = $\frac{3}{6}=\frac{1}{2}$, Prime = $\frac{3}{6}=\frac{1}{2}$.

$$P(E\cap P)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$$

Q7. A card is drawn from deck. Find probability of getting red card and ace. Red cards = 26, Aces = 4, overlap = 2 red aces.

$$P(R\cap A)=\frac{2}{52}=\frac{1}{26}$$

Q8. A die is thrown twice. Find probability of getting 6 on first and odd number on second.

$$P(6)=\frac{1}{6},P(Odd)=\frac{3}{6}=\frac{1}{2}$$

$$P=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$$

Q9. A coin is tossed twice. Find probability of getting head on first and tail on second.

$$P(H\cap T)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$$

Q10. A card is drawn twice with replacement. Find probability of getting king first and queen second.

$$P(K\cap Q)=\frac{4}{52}\cdot \frac{4}{52}=\frac{1}{169}$$

FAQs (10)

1. What is multiplication theorem of probability? $P(A\cap B)=P(A)\cdot P(B\mathrm{\mid }A)$.

2. What is conditional probability? Probability of event given another event has occurred.

3. What is formula for conditional probability? $P(A\mathrm{\mid }B)=\frac{P(A\cap B)}{P(B)}$.

4. What are independent events? Events where occurrence of one does not affect the other.

5. What is probability of intersection of independent events? Product of their probabilities.

6. Can probability of intersection exceed probability of individual events? No, always less or equal.

7. What is probability of two independent coin tosses both heads? $\frac{1}{4}$.

8. What is probability of ace in two draws with replacement? $\frac{1}{169}$.

9. What is probability of ace in two draws without replacement? $\frac{1}{221}$.

10. Why is this exercise important? It builds foundation for conditional probability and Bayes’ theorem.

Conclusion

Exercise 14.3 has 10 solved questions and 10 FAQs that strengthen your understanding of multiplication theorem and intersection of events in probability. This builds the foundation for conditional probability in Class 11 Maths.

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New Syllabus -Class 11 Maths Chapter 14 Probability – Exercise 14.3 NCERT