Class 12 Maths Chapter 13 Probability – Exercise 13.7 NCERT Solutions

Introduction

Exercise 13.7 focuses on mean and variance of probability distributions. Students learn how to calculate expected value, variance, and standard deviation for discrete random variables. This exercise is essential for connecting probability with statistics and data analysis.

Key Concepts

1. Mean (Expected Value):

$$E(X)=\sum x_{i}P(X=x_i)$$

2. Variance:

$$Var(X)=E(X^2)-[E(X){]}^2$$

3. Standard Deviation:

$$\sigma =\sqrt{Var(X)}$$

4. Probability Distribution Table: Must satisfy $\sum P(X=x_i)=1$.

Students Frequently Make Mistakes

• Forgetting to square values when calculating $E(X^2)$.

• Errors in subtracting $[E(X){]}^2$.

• Misinterpreting variance as mean.

• Skipping normalization of probabilities.

• Confusing variance with standard deviation.

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

Q1. A coin is tossed once. X = number of heads. Find mean and variance.

$$X=\{0,1\},P(0)=\frac{1}{2},P(1)=\frac{1}{2}$$

$$E(X)=0\cdot \frac{1}{2}+1\cdot \frac{1}{2}=\frac{1}{2}$$

$$E(X^2)=0^2\cdot \frac{1}{2}+1^2\cdot \frac{1}{2}=\frac{1}{2}$$

$$Var(X)=\frac{1}{2}-{\left(\frac{1}{2}\right)}^2=\frac{1}{4}$$

Q2. A coin is tossed twice. X = number of heads. Find mean and variance.

$$X=\{0,1,2\},P(0)=\frac{1}{4},P(1)=\frac{1}{2},P(2)=\frac{1}{4}$$

$$E(X)=0\cdot \frac{1}{4}+1\cdot \frac{1}{2}+2\cdot \frac{1}{4}=1$$

$$E(X^2)=0^2\cdot \frac{1}{4}+1^2\cdot \frac{1}{2}+2^2\cdot \frac{1}{4}=1.5$$

$$Var(X)=1.5-1^2=0.5$$

Q3. A die is thrown once. X = number obtained. Find mean and variance.

$$E(X)=\frac{1+2+3+4+5+6}{6}=3.5$$

$$E(X^2)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{91}{6}=15.167$$

$$Var(X)=15.167-(3.5{)}^2=2.917$$

Q4. A coin is tossed thrice. X = number of heads. Find mean and variance.

$$X=\{0,1,2,3\},P(0)=\frac{1}{8},P(1)=\frac{3}{8},P(2)=\frac{3}{8},P(3)=\frac{1}{8}$$

$$E(X)=0\cdot \frac{1}{8}+1\cdot \frac{3}{8}+2\cdot \frac{3}{8}+3\cdot \frac{1}{8}=1.5$$

$$E(X^2)=0+1\cdot \frac{3}{8}+4\cdot \frac{3}{8}+9\cdot \frac{1}{8}=3$$

$$Var(X)=3-(1.5{)}^2=0.75$$

Q5. A die is thrown twice. X = sum of numbers. Find mean.

$$E(X)=\frac{2+3+\cdots +12}{36}=7$$

Q6. A coin is tossed until head appears. X = number of tosses. Find mean.

$$E(X)=\sum _{n=1}^{\mathrm{\infty }}n\cdot {\left(\frac{1}{2}\right)}^n=2$$

Q7. A die is thrown. X = 1 if even, 0 if odd. Find mean and variance.

$$P(X=1)=\frac{1}{2},P(X=0)=\frac{1}{2}$$

$$E(X)=0\cdot \frac{1}{2}+1\cdot \frac{1}{2}=\frac{1}{2}$$

$$E(X^2)=\frac{1}{2}$$

$$Var(X)=\frac{1}{2}-{\left(\frac{1}{2}\right)}^2=\frac{1}{4}$$

Q8. A coin is tossed twice. X = 1 if at least one head, 0 otherwise. Find mean.

$$P(X=1)=\frac{3}{4},P(X=0)=\frac{1}{4}$$

$$E(X)=0\cdot \frac{1}{4}+1\cdot \frac{3}{4}=0.75$$

$$Var(X)=0.75-(0.75{)}^2=0.1875$$

Q9. A die is thrown. X = 1 if prime, 0 otherwise. Find mean. Primes: $\{2,3,5\}$.

$$P(X=1)=\frac{1}{2},P(X=0)=\frac{1}{2}$$

$$E(X)=0.5,Var(X)=0.25$$

Q10. A coin is tossed thrice. X = 1 if exactly two heads, 0 otherwise. Find mean.

$$P(X=1)=\frac{3}{8},P(X=0)=\frac{5}{8}$$

$$E(X)=0.375,Var(X)=0.375-(0.375{)}^2=0.234$$

FAQs (10)

FAQ1. What is mean of distribution? Expected value $E(X)=\sum x_{i}P(X=x_i)$.

FAQ2. What is variance formula? $Var(X)=E(X^2)-[E(X){]}^2$.

FAQ3. What is standard deviation? Square root of variance.

FAQ4. What is mean of die throw? 3.5.

FAQ5. What is variance of die throw? 2.917.

FAQ6. What is mean of coin tossed twice?

FAQ7. What is variance of coin tossed twice? 0.5.

FAQ8. What is mean of coin tossed thrice? 1.5.

FAQ9. What is variance of coin tossed thrice? 0.75.

FAQ10. Why is Exercise 13.7 important? It builds mastery of mean and variance of probability distributions.

Conclusion

Exercise 13.7 has 10 solved questions and 10 FAQs that strengthen your understanding of mean and variance of probability distributions. This completes the Probability chapter in Class 12 Maths.

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