Class 12 Maths Chapter 13 Probability – Exercise 13.1 NCERT Solutions

Introduction

Exercise 13.1 introduces the basic concepts of probability. Students learn about random experiments, sample space, events, and probability calculation. This exercise builds the foundation for advanced probability topics such as conditional probability, Bayes’ theorem, and random variables.

Key Concepts

1. Random Experiment: An experiment with uncertain outcome.

2. Sample Space (S): Set of all possible outcomes.

3. Event (E): Subset of sample space.

4. Probability of Event: If $n(S)$ = total outcomes, $n(E)$ = favorable outcomes, then

$$P(E)=\frac{n(E)}{n(S)}$$

Students Frequently Make Mistakes

• Forgetting to count sample space correctly.

• Confusing favorable outcomes with total outcomes.

• Errors in probability fraction simplification.

• Skipping complement rule $P(E)+P(E^{\mathrm{\prime }})=1$.

• Misinterpreting mutually exclusive events.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. A coin is tossed once. Find probability of getting head.

$$S=\{H,T\},n(S)=2,n(E)=1$$

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

Q2. A die is thrown once. Find probability of getting even number.

$$S=\{1,2,3,4,5,6\},E=\{2,4,6\},n(E)=3$$

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

Q3. A card is drawn from 52 cards. Find probability of getting ace.

$$n(S)=52,n(E)=4$$

$$P(E)=\frac{4}{52}=\frac{1}{13}$$

Q4. A coin is tossed twice. Find probability of getting two heads.

$$S=\{HH,HT,TH,TT\},n(S)=4,n(E)=1$$

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

Q5. A die is thrown once. Find probability of getting prime number. Primes: $\{2,3,5\}$.

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

Q6. A card is drawn. Find probability of getting red card. Red cards = 26.

$$P(E)=\frac{26}{52}=\frac{1}{2}$$

Q7. A coin is tossed thrice. Find probability of getting exactly two heads.

$$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\},n(S)=8$$

\text{Favorable: } \{HHT, HTH, THH\}, \quad n(E)=3

\]

P(E) = \frac{3}{8}

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

Q9. A card is drawn. Find probability of getting king. Kings = 4.

$$P(E)=\frac{4}{52}=\frac{1}{13}$$

Q10. A coin is tossed once. Find probability of not getting head.

$$P(E^{\mathrm{\prime }})=1-P(E)=1-\frac{1}{2}=\frac{1}{2}$$

FAQs (10)

FAQ1. What is random experiment? Experiment with uncertain outcome.

FAQ2. What is sample space? Set of all possible outcomes.

FAQ3. What is event? Subset of sample space.

FAQ4. What is probability formula? $P(E)=\frac{n(E)}{n(S)}$.

FAQ5. What is complement of event? $P(E)+P(E^{\mathrm{\prime }})=1$.

FAQ6. What is probability of getting ace from 52 cards? $\frac{1}{13}$.

FAQ7. What is probability of getting red card? $\frac{1}{2}$.

FAQ8. What is probability of getting two heads in two tosses? $\frac{1}{4}$.

FAQ9. What is probability of getting prime number on die? $\frac{1}{2}$.

FAQ10. Why is Exercise 13.1 important? It builds foundation for advanced probability concepts.

Conclusion

Exercise 13.1 has 10 solved questions and 10 FAQs that strengthen your understanding of basic probability concepts. This begins the Probability chapter in Class 12 Maths.

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