Focus Keyword: NCERT Solutions Class 9 Math Chapter 7 The Mathematics of Maybe Introduction to Probability
Secondary Keywords & LSI: Class 9 Maths Chapter 7 solutions, empirical probability Class 9 questions and answers, The Mathematics of Maybe Class 9 NCERT, experimental probability Class 9, CBSE Class 9 Maths exam preparation
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H1 Title: NCERT Solutions for Class 9 Math Chapter 7: The Mathematics of Maybe: Introduction to Probability (Complete Guide)
Navigating through the CBSE Class 9 Mathematics curriculum requires an in-depth understanding of experimental trials, relative frequency ratios, empirical probability bounds, and statistical predictability. Chapter 7 of Class 9 Mathematics, “The Mathematics of Maybe: Introduction to Probability”, forms the foundation of modern data science, risk management, financial econometrics, and predictive artificial intelligence. It investigates the transition from subjective guesswork to quantitative measurement of chance; explores the empirical behavior of random experiments across repeated coin tosses, dice rolls, and real-world surveys; and details the fundamental axioms governing certainty, impossibility, and complementary outcomes. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from sports strike rates and manufacturing quality control datasets to community demographic surveys and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Probability), and clear inline geometric probability models have been highlighted to ensure students secure maximum marks in their CBSE examinations.
Chapter 7: The Mathematics of Maybe: Introduction to Probability
Master Chapter Summary & Quantitative Blueprint
In the rationalised Class 9 curriculum, Chapter 7 establishes the foundation of probability through the empirical (experimental) approach, where probabilities are deduced from actual observed data and repeated trials rather than theoretical symmetry alone.
| Conceptual Entity | Core Mathematical Definition | Governing Formula / Identity | Key Numerical Boundary | CBSE Marks Weightage |
|---|---|---|---|---|
| Random Trial | An action or experiment resulting in one of several possible outcomes | Denoted as an experiment of $n$ trials | Outcomes must be well-defined | 1 Mark |
| Event ($E$) | A collection of one or more outcomes of an experiment | Sub-collection of the total trial outcomes | Favourable outcomes $m \le n$ | 1 Mark |
| Empirical Probability | Relative frequency of an event based on actual recorded trials | $P(E) = \frac{\text{Number of trials where } E \text{ occurred}}{\text{Total number of trials}}$ | $0 \le P(E) \le 1$ | 2 to 3 Marks |
| Impossible Event | An event that has zero chance of occurring in the experiment | $P(\phi) = \frac{0}{n} = 0$ | Minimum boundary value | 1 Mark (MCQ) |
| Certain / Sure Event | An event that occurs in every single trial without exception | $P(S) = \frac{n}{n} = 1$ | Maximum boundary value | 1 Mark (MCQ) |
| Complementary Event | The non-occurrence of event $E$, denoted by $\bar{E}$ or ‘not $E$’ | $P(\bar{E}) = 1 – P(E)$ | $P(E) + P(\bar{E}) = 1$ | 2 Marks |
| Sum of Elementary Events | The combined sum of probabilities of all mutually exclusive outcomes | $\sum_{i=1}^{k} P(E_i) = 1$ | Invariant sum rule | 2 to 3 Marks |
🧠 Examiner’s Secret: In empirical probability questions, never round off intermediate fractions prematurely. Express your final probability as an unreduced fraction first (e.g., $\frac{475}{1500}$), then reduce it to its lowest terms ($\frac{19}{60}$), and finally write its decimal equivalent ($0.3167$) if asked. Showing the unreduced fraction verifies the exact count of favourable trials to the examiner.
Foundational Geometric Models and Probability Scales
The Continuous Probability Scale
The probability of any event is a real number situated on a continuous scale bounded strictly between $0$ (representing absolute impossibility) and $1$ (representing absolute certainty). 0 (0%) Impossible 0.5 (50%) Even Chance 1.0 (100%) Certain / Sure
If an event $E$ can never occur under the experimental conditions, its empirical probability is zero: $P(E) = 0$. If an event is guaranteed to happen in every trial, its empirical probability is one: $P(E) = 1$. For any other real-world random event:
$$0 \le P(E) \le 1$$
Empirical vs. Theoretical Probability
Empirical probability is calculated directly from the observed results of a conducted experiment, whereas theoretical probability is calculated using logical deduction and equal likelihood assumptions without running physical trials. Empirical • Based on trials • Observed data • Value changes with trial size P = m / n Theoretical • Based on logic • Equally likely • Invariant fixed value P = n(E) / n(S)
When a coin is tossed $10$ times, heads may appear $7$ times, yielding an empirical probability of $\frac{7}{10} = 0.7$. However, by the Law of Large Numbers, as the total number of experimental trials increases into thousands or millions, the empirical probability steadily approaches the theoretical probability of $0.5$.
💡 Did You Know?: The French naturalist Comte de Buffon tossed a coin $4,040$ times and obtained $2,048$ heads, giving an empirical probability of $0.5069$. Later, English statistician Karl Pearson tossed a coin $24,000$ times, recording $12,012$ heads—an empirical probability of $0.5005$, confirming that large trial counts converge toward theoretical symmetry.
[👉 Also Read: Class 9 Math Chapter 6 Measuring Space: Perimeter and Area NCERT Solutions]
Step-by-Step Solutions: Core Textbook Exercises and Applied Problems
Question 1. [Cricket Batswoman Boundary Ratio]
In a cricket match, a batswoman hits a boundary $6\text{ times}$ out of $30\text{ balls}$ she plays. Find the probability that she did not hit a boundary.
Answer:
Step 1: Identify Given Experimental Data
- Total number of balls played (trials), $n = 30$.
- Number of times boundary is hit, $m_1 = 6$.
Step 2: Determine Favourable Trials
Number of times she did not hit a boundary:
$$m_2 = n – m_1 = 30 – 6 = 24\text{ balls}$$
Step 3: Apply Empirical Probability Formula
$$P(\text{No Boundary}) = \frac{\text{Number of balls without a boundary}}{\text{Total number of balls played}}$$
$$P(\text{No Boundary}) = \frac{24}{30}$$
Divide numerator and denominator by their greatest common factor ($6$):
$$P(\text{No Boundary}) = \frac{24 \div 6}{30 \div 6} = \frac{4}{5} = 0.8$$
(Alternative Complement Method: $P(\text{Boundary}) = \frac{6}{30} = \frac{1}{5} \implies P(\text{No Boundary}) = 1 – \frac{1}{5} = \frac{4}{5}$).
Final Answer:
The probability that she did not hit a boundary is $\frac{4}{5}$ (or $0.8$).
Question 2. [Family Demographics & Sum Rule Verification]
$1500\text{ families}$ with $2\text{ children}$ were selected randomly, and the following data were recorded:
| Number of girls in a family | 2 | 1 | 0 |
|---|---|---|---|
| Number of families | 475 | 814 | 211 |
Compute the probability of a family, chosen at random, having:
(i) $2\text{ girls}$
(ii) $1\text{ girl}$
(iii) No girl
Also check whether the sum of these probabilities is $1$.
Answer:
Step 1: Identify Total Number of Trials
$$n = 475 + 814 + 211 = 1500\text{ families}$$
Step 2: (i) Probability of a Family Having 2 Girls
- Number of families with $2$ girls, $m_1 = 475$.
$$P(\text{2 girls}) = \frac{m_1}{n} = \frac{475}{1500}$$
Divide numerator and denominator by $25$:
$$P(\text{2 girls}) = \frac{475 \div 25}{1500 \div 25} = \frac{19}{60} \approx 0.3167$$
Step 3: (ii) Probability of a Family Having 1 Girl
- Number of families with $1$ girl, $m_2 = 814$.
$$P(\text{1 girl}) = \frac{m_2}{n} = \frac{814}{1500}$$
Divide numerator and denominator by $2$:
$$P(\text{1 girl}) = \frac{814 \div 2}{1500 \div 2} = \frac{407}{750} \approx 0.5427$$
Step 4: (iii) Probability of a Family Having No Girl
- Number of families with no girl, $m_3 = 211$.
$$P(\text{0 girls}) = \frac{m_3}{n} = \frac{211}{1500} \approx 0.1407$$
Step 5: Verify the Sum of Probabilities
$$\text{Sum} = P(\text{2 girls}) + P(\text{1 girl}) + P(\text{0 girls})$$
$$\text{Sum} = \frac{475}{1500} + \frac{814}{1500} + \frac{211}{1500} = \frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = 1$$
Final Answer:
(i) $P(\text{2 girls}) = \mathbf{\frac{19}{60}}$, (ii) $P(\text{1 girl}) = \mathbf{\frac{407}{750}}$, (iii) $P(\text{no girl}) = \mathbf{\frac{211}{1500}}$.
The sum of the probabilities is strictly verified to be $1$.
Question 3. [Classroom Birth Month Frequency Analysis]
In a particular section of Class IX, $40\text{ students}$ were asked about the month of their birth and the following graph/table was prepared for the data so obtained:
| Birth Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Number of Students | 3 | 4 | 2 | 2 | 5 | 1 | 2 | 6 | 3 | 4 | 4 | 4 |
Find the probability that a student of the class was born in August.
Answer:
Step 1: Identify Total Outcomes
Total number of students surveyed (trials), $n = 40$.
Step 2: Identify Favourable Outcomes
Number of students born in the month of August, $m = 6$.
Step 3: Compute Empirical Probability
$$P(\text{Born in August}) = \frac{\text{Number of students born in August}}{\text{Total number of students}}$$
$$P(\text{Born in August}) = \frac{6}{40}$$
Divide numerator and denominator by $2$:
$$P(\text{Born in August}) = \frac{6 \div 2}{40 \div 2} = \frac{3}{20} = 0.15$$
Final Answer:
The probability that a student was born in August is $\frac{3}{20}$ (or $0.15$).
Question 4. [Three-Coin Toss Experiment]
Three coins were tossed simultaneously $200\text{ times}$ with the following frequencies of different outcomes:
| Outcome | 3 heads | 2 heads | 1 head | No head |
|---|---|---|---|---|
| Frequency | 23 | 72 | 77 | 28 |
If the three coins are simultaneously tossed again, compute the probability of $2\text{ heads}$ coming up.
Answer:
Step 1: Identify Total Number of Trials
$$n = 23 + 72 + 77 + 28 = 200\text{ trials}$$
Step 2: Identify Favourable Outcomes
Frequency of obtaining exactly $2$ heads, $m = 72$.
Step 3: Compute Empirical Probability
$$P(\text{2 heads}) = \frac{\text{Frequency of 2 heads}}{\text{Total number of tosses}}$$
$$P(\text{2 heads}) = \frac{72}{200}$$
Divide numerator and denominator by $8$:
$$P(\text{2 heads}) = \frac{72 \div 8}{200 \div 8} = \frac{9}{25} = 0.36$$
Final Answer:
The probability of getting 2 heads is $\frac{9}{25}$ (or $0.36$).
Question 5. [Household Vehicle Ownership Survey]
An organisation selected $2400\text{ families}$ at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:
| Monthly Income (in ₹) | 0 vehicles | 1 vehicle | 2 vehicles | Above 2 vehicles |
|---|---|---|---|---|
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000 – 10000 | 0 | 305 | 27 | 2 |
| 10000 – 13000 | 1 | 535 | 29 | 1 |
| 13000 – 16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
Suppose a family is chosen at random. Find the probability that the family chosen is:
(i) earning ₹$10000 – 13000\text{ per month}$ and owning exactly $2\text{ vehicles}$.
(ii) earning ₹$16000\text{ or more per month}$ and owning exactly $1\text{ vehicle}$.
(iii) earning less than ₹$7000\text{ per month}$ and does not own any vehicle.
(iv) earning ₹$13000 – 16000\text{ per month}$ and owning more than $2\text{ vehicles}$.
(v) owning not more than $1\text{ vehicle}$.
Answer:
Total Number of Families (Trials): $n = 2400$.
(i) Income ₹10000 – 13000 and exactly 2 vehicles:
- Favourable families from table, $m_1 = 29$.
$$P(E_1) = \frac{29}{2400}$$
(ii) Income ₹16000 or more and exactly 1 vehicle:
- Favourable families from table, $m_2 = 579$.
$$P(E_2) = \frac{579}{2400} = \frac{193}{800}$$
(iii) Income less than ₹7000 and 0 vehicles:
- Favourable families from table, $m_3 = 10$.
$$P(E_3) = \frac{10}{2400} = \frac{1}{240}$$
(iv) Income ₹13000 – 16000 and more than 2 vehicles:
- Favourable families from table, $m_4 = 25$.
$$P(E_4) = \frac{25}{2400} = \frac{1}{96}$$
(v) Owning not more than 1 vehicle (0 or 1 vehicle across all income levels):
- Sum of 0 vehicles column: $10 + 0 + 1 + 2 + 1 = 14$
- Sum of 1 vehicle column: $160 + 305 + 535 + 469 + 579 = 2048$
- Total favourable families: $m_5 = 14 + 2048 = 2062$.
$$P(E_5) = \frac{2062}{2400} = \frac{1031}{1200}$$
Final Answer:
(i) $\mathbf{\frac{29}{2400}}$, (ii) $\mathbf{\frac{193}{800}}$, (iii) $\mathbf{\frac{1}{240}}$, (iv) $\mathbf{\frac{1}{96}}$, (v) $\mathbf{\frac{1031}{1200}}$.
Question 6. [Mathematics Test Performance Distribution]
A teacher analysed the performance of $70\text{ students}$ of a mathematics test of $100\text{ marks}$ given in the following table:
| Marks Interval | 0 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 100 |
|---|---|---|---|---|---|---|---|
| Number of Students | 7 | 10 | 10 | 20 | 20 | 15 | 8 |
(Note: Summing standard textbook distribution yields $n = 90$ students: $7 + 10 + 10 + 20 + 20 + 15 + 8 = 90$).
Find the probability that a student:
(i) obtained less than $20%\text{ marks}$ in the mathematics test.
(ii) obtained marks $60\text{ or above}$.
Answer:
Step 1: Identify Total Number of Students
$$n = 7 + 10 + 10 + 20 + 20 + 15 + 8 = 90\text{ students}$$
Step 2: (i) Probability of Marks Less than 20%
Less than $20%$ corresponds to the interval $0 – 20$.
Number of students, $m_1 = 7$.
$$P(\text{Marks } < 20) = \frac{7}{90}$$
Step 3: (ii) Probability of Marks 60 or Above
Marks $60$ or above includes intervals $60 – 70$ and $70 – 100$.
Number of students, $m_2 = 15 + 8 = 23$.
$$P(\text{Marks } \ge 60) = \frac{23}{90}$$
Final Answer:
(i) $P(\text{marks } < 20%) = \mathbf{\frac{7}{90}}$, (ii) $P(\text{marks } \ge 60) = \mathbf{\frac{23}{90}}$.
Question 7. [Student Subject Preference Survey]
To know the opinion of the students about the subject statistics, a survey of $200\text{ students}$ was conducted. The data is recorded in the following table:
| Opinion | Number of Students |
|---|---|
| Like | 135 |
| Dislike | 65 |
Find the probability that a student chosen at random:
(i) likes statistics,
(ii) does not like it.
Answer:
Step 1: Identify Total Trials
Total students surveyed, $n = 200$.
Step 2: (i) Probability that Student Likes Statistics
Number of students who like statistics, $m_1 = 135$.
$$P(\text{Likes}) = \frac{135}{200}$$
Divide by $5$:
$$P(\text{Likes}) = \frac{135 \div 5}{200 \div 5} = \frac{27}{40} = 0.675$$
Step 3: (ii) Probability that Student Does Not Like Statistics
Number of students who dislike statistics, $m_2 = 65$.
$$P(\text{Dislikes}) = \frac{65}{200}$$
Divide by $5$:
$$P(\text{Dislikes}) = \frac{65 \div 5}{200 \div 5} = \frac{13}{40} = 0.325$$
(Notice: $\frac{27}{40} + \frac{13}{40} = 1$).
Final Answer:
(i) $P(\text{likes}) = \mathbf{\frac{27}{40}}$ (or $0.675$), (ii) $P(\text{dislikes}) = \mathbf{\frac{13}{40}}$ (or $0.325$).
Question 8. [Workplace Commuting Distances]
The distance (in $\text{km}$) of $40\text{ female engineers}$ from their residence to their place of work were found as follows:
$$5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2,$$
$$7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12$$
What is the empirical probability that an engineer lives:
(i) less than $7\text{ km}$ from her place of work?
(ii) more than or equal to $7\text{ km}$ from her place of work?
(iii) within $\frac{1}{2}\text{ km}$ from her place of work?
Answer:
Step 1: Identify Total Number of Engineers
$$n = 40\text{ engineers}$$
Step 2: (i) Lives Less than 7 km from Workplace
Distances strictly $< 7\text{ km}$:
Values: $5, 3, 2, 3, 5, 3, 2, 6, 6$.
Count of favourable trials, $m_1 = 9$.
$$P(< 7\text{ km}) = \frac{9}{40} = 0.225$$
Step 3: (ii) Lives More than or Equal to 7 km from Workplace
This is the complement of living less than $7\text{ km}$:
Count of favourable trials, $m_2 = 40 – 9 = 31$.
$$P(\ge 7\text{ km}) = \frac{31}{40} = 0.775$$
Step 4: (iii) Lives Within 1/2 km (0.5 km) from Workplace
Inspecting the dataset, the minimum recorded distance is $2\text{ km}$.
There is no distance $\le 0.5\text{ km}$.
Count of favourable trials, $m_3 = 0$.
$$P(\le 0.5\text{ km}) = \frac{0}{40} = 0 \quad (\text{Impossible Event})$$
Final Answer:
(i) $P(< 7\text{ km}) = \mathbf{\frac{9}{40}}$, (ii) $P(\ge 7\text{ km}) = \mathbf{\frac{31}{40}}$, (iii) $P(\le 0.5\text{ km}) = \mathbf{0}$.
Question 9. [Tyre Lifespan Quality Control Record]
A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of $1000\text{ cases}$:
| Distance (in km) | Less than 4000 | 4000 to 9000 | 9001 to 14000 | More than 14000 |
|---|---|---|---|---|
| Frequency | 20 | 210 | 325 | 445 |
If you buy a tyre of this company, what is the probability that:
(i) it will need to be replaced before it has covered $4000\text{ km}$?
(ii) it will last more than $9000\text{ km}$?
(iii) it will need to be replaced after it has covered somewhere between $4000\text{ km}$ and $14000\text{ km}$?
Answer:
Total Number of Trials: $n = 1000\text{ tyres}$.
(i) Replaced before covering 4000 km:
- Favourable count, $m_1 = 20$.
$$P(E_1) = \frac{20}{1000} = \frac{2}{100} = \frac{1}{50} = 0.02$$
(ii) Lasts more than 9000 km:
- This includes the categories “$9001 – 14000\text{ km}$” and “More than $14000\text{ km}$”:
$$m_2 = 325 + 445 = 770$$
$$P(E_2) = \frac{770}{1000} = \frac{77}{100} = 0.77$$
(iii) Replaced between 4000 km and 14000 km:
- This includes the categories “$4000 – 9000\text{ km}$” and “$9001 – 14000\text{ km}$”:
$$m_3 = 210 + 325 = 535$$
$$P(E_3) = \frac{535}{1000} = 0.535 = \frac{107}{200}$$
Final Answer:
(i) $\mathbf{0.02}$ (or $\mathbf{\frac{1}{50}}$), (ii) $\mathbf{0.77}$ (or $\mathbf{\frac{77}{100}}$), (iii) $\mathbf{0.535}$ (or $\mathbf{\frac{107}{200}}$).
Question 10. [Flour Bag Weight Sampling]
Eleven bags of wheat flour, each marked $5\text{ kg}$, actually contained the following weights of flour (in $\text{kg}$):
$$4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 4.98, 5.04, 5.07, 5.00, 5.12$$
Find the probability that any of these bags chosen at random contains more than $5\text{ kg}$ of flour.
Answer:
Step 1: Identify Total Outcomes
Total number of flour bags, $n = 11$.
Step 2: Identify Favourable Outcomes
Bags containing strictly more than $5\text{ kg}$ (weights $> 5.00\text{ kg}$):
$$5.05, 5.08, 5.03, 5.06, 5.04, 5.07, 5.12$$
(Note: Bags marked exactly $5.00\text{ kg}$ are not strictly greater than $5\text{ kg}$).
Count of favourable bags, $m = 7$.
Step 3: Compute Empirical Probability
$$P(\text{Weight } > 5\text{ kg}) = \frac{m}{n} = \frac{7}{11}$$
Final Answer:
The probability that a chosen bag contains more than $5\text{ kg}$ of flour is $\frac{7}{11}$.
Question 11. [Environmental Atmospheric Pollutant Monitoring]
The concentration of sulphur dioxide ($\text{SO}_2$) in the air (in parts per million, i.e., $\text{ppm}$) for a certain city was recorded over $30\text{ days}$ as follows:
| Concentration Interval (ppm) | 0.00 – 0.04 | 0.04 – 0.08 | 0.08 – 0.12 | 0.12 – 0.16 | 0.16 – 0.20 | 0.20 – 0.24 |
|---|---|---|---|---|---|---|
| Number of Days (Frequency) | 4 | 9 | 9 | 2 | 4 | 2 |
Using this table, find the probability of the concentration of sulphur dioxide in the interval $0.12 – 0.16\text{ ppm}$ on any randomly selected day.
Answer:
Step 1: Identify Total Days (Trials)
$$n = 4 + 9 + 9 + 2 + 4 + 2 = 30\text{ days}$$
Step 2: Identify Favourable Days
Number of days in the interval $0.12 – 0.16\text{ ppm}$, $m = 2$.
Step 3: Compute Probability
$$P(0.12 – 0.16\text{ ppm}) = \frac{2}{30} = \frac{1}{15} \approx 0.067$$
Final Answer:
The probability that the concentration is in the interval $0.12 – 0.16\text{ ppm}$ is $\frac{1}{15}$.
Question 12. [Blood Group Distribution in Students]
The blood groups of $30\text{ students}$ of Class VIII are recorded as follows:
$$\text{A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,}$$
$$\text{A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O}$$
Use this table to determine the probability that a student of this class, selected at random, has blood group $\text{AB}$.
Answer:
Step 1: Compile Frequency Table
- Blood group A: $9$ students
- Blood group B: $6$ students
- Blood group O: $12$ students
- Blood group AB: $3$ students
- Total students, $n = 9 + 6 + 12 + 3 = 30$.
Step 2: Identify Favourable Outcomes
Number of students with blood group AB, $m = 3$.
Step 3: Compute Empirical Probability
$$P(\text{Blood Group AB}) = \frac{3}{30} = \frac{1}{10} = 0.1$$
Final Answer:
The probability that a student has blood group AB is $\frac{1}{10}$ (or $0.1$).
[👉 Also Read: Class 9 Math Chapter 1 Number Systems NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is the fundamental difference between empirical probability and theoretical probability?
Empirical probability is calculated from actual observed data and physical experimental trials ($P(E) = \frac{\text{favourable trials}}{\text{total trials}}$), so its value can change between different sets of trials. Theoretical probability is based on deductive reasoning and assumptions of equal likelihood ($P(E) = \frac{n(E)}{n(S)}$), remaining constant regardless of physical trials.
What is the Law of Large Numbers in probability?
The Law of Large Numbers states that as the number of repetitions of a random experiment increases, the relative frequency (empirical probability) of an event gets closer and closer to its true theoretical probability.
Can the probability of an event be negative or greater than 1?
No. Probability is defined as the ratio of a non-negative part of trials to the total trials. Because the number of favourable trials $m$ must satisfy $0 \le m \le n$, dividing by $n$ ensures that $0 \le P(E) \le 1$. Values outside this range are mathematically invalid.
What is an impossible event and what is its probability?
An impossible event is an outcome that cannot occur under any circumstances within the defined experiment (such as rolling a $7$ on a standard $6$-sided die). Its probability is always $0$.
What is a sure or certain event and what is its probability?
A sure or certain event is an outcome that is guaranteed to happen on every single trial of the experiment (such as rolling a number less than $7$ on a standard die). Its probability is always $1$.
Why must the sum of probabilities of all elementary outcomes equal 1?
The elementary outcomes of an experiment are mutually exclusive and together cover all possibilities in the sample space. Because every trial must produce one of these outcomes, the sum of their individual probabilities equals $\frac{\text{Total Trials}}{\text{Total Trials}} = 1$.
What is a complementary event in Class 9 probability?
A complementary event represents the non-occurrence of an event $E$, denoted as $\bar{E}$ or ‘not $E$’. Its probability is calculated by subtracting the probability of $E$ from 1: $P(\bar{E}) = 1 – P(E)$.
In Question 1, why was she did not hit a boundary calculated as 24/30?
The batswoman played $30$ balls and hit boundaries on $6$ of them. The remaining $30 – 6 = 24$ balls did not result in a boundary. The empirical probability of not hitting a boundary is therefore $\frac{24}{30} = \frac{4}{5} = 0.8$.
Why are bags with exactly 5.00 kg excluded when finding bags with more than 5 kg?
The condition “more than $5\text{ kg}$” requires the weight to be strictly greater than $5$ ($w > 5.00$). A bag weighing exactly $5.00\text{ kg}$ satisfies $w = 5$, not $w > 5$, so it cannot be counted as a favourable outcome.
How do you calculate the probability of not more than 1 vehicle in Question 5?
“Not more than 1 vehicle” means owning either $0$ vehicles or $1$ vehicle. Sum the frequencies across all income brackets for both the $0$-vehicle and $1$-vehicle columns ($14 + 2048 = 2062$), then divide by the total number of families ($2400$) to get $\frac{2062}{2400} = \frac{1031}{1200}$.
How is empirical probability applied in quality control manufacturing?
Manufacturing plants inspect a sample of produced items (e.g., $1000$ tyres or lightbulbs) and record how many are defective or reach a given lifespan. The resulting relative frequency gives the empirical probability of defect rates, helping engineers maintain quality standards.
Does empirical probability allow us to predict the exact outcome of the next single trial?
No. Empirical probability measures long-term likelihood across many trials, but it cannot predict the specific result of the next individual trial, which remains random and uncertain.
How should probability answers be expressed in CBSE board examinations?
Answers should be written as simplified fractions in lowest terms (e.g., $\frac{3}{20}$) or as exact terminating decimals (e.g., $0.15$). Unreduced fractions should be shown during working steps to indicate the raw trial counts clearly.
What is a trial versus an outcome?
A trial is a single performance of a random experiment (such as tossing a coin once or taking a survey). An outcome is the specific result produced by that trial (such as getting a Head or choosing a family with 2 children).
What are the key presentation steps to secure full marks in Class 9 probability questions?
To secure full marks: (1) state the total number of trials $n$, (2) clearly define the event $E$ and list the count of favourable trials $m$, (3) write the formula $P(E) = \frac{m}{n}$ before substituting numbers, and (4) state the final probability in simplest fractional form and boxed decimal form.