--- title: "NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2" url: https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-2 dateModified: 2026-10-07T15:46:13+00:00 --- # NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 Chapter 7: The Mathematics of Maybe: Introduction to Probability. Every question from Exercise 7.2, with full working and the final answer. Free PDF (20 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-7-the-mathematics-of-maybe-introduction-to-probability-1d9ff48bee.pdf ## Exercise Set 7.2 ### Question 1 *3 marks · Short answer* A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour: 10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets (i) Calculate the probability that a randomly picked sweet from the sample is green. (ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results. **Part (i)** 1. Total number of sweets in the sample = $30$ 2. Number of green sweets = $8$ 3. Probability that a picked sweet is green = $\frac{8}{30} = \frac{4}{15}$ Answer (i): $\frac{4}{15}$ **Part (ii)** 1. Number of yellow sweets in the sample = $7$, so the probability of picking a yellow sweet is $\frac{7}{30}$ 2. Total number of sweets in the large bag = $600$ 3. Estimated number of yellow sweets = $\frac{7}{30} \times 600 = 140$ Answer (ii): $140$ **Answer:** (i) $\frac{4}{15}$ (ii) $140$ sweets > Common mistake: Using the wrong total number of sweets for probability or estimation. ### Question 2 *3 marks · Short answer* A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are: 14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club Assume there are 800 students in the whole school. (i) What is the probability that a randomly chosen student from the sample prefers the Arts Club? (ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club. **Part (i)** 1. Total number of students in the sample = $40$ 2. Number of students who prefer the Arts Club = $11$ 3. Probability that a randomly chosen student prefers the Arts Club = $\frac{11}{40}$ Answer (i): $\frac{11}{40}$ **Part (ii)** 1. Number of students in the sample who prefer the Sports Club = $9$, so the probability is $\frac{9}{40}$ 2. Total number of students in the school = $800$ 3. Estimated number of students who prefer the Sports Club = $\frac{9}{40} \times 800 = 180$ Answer (ii): $180$ students **Answer:** (i) $\frac{11}{40}$ (ii) $180$ students > Common mistake: Confusing sample size with the total population when estimating. ### Question 3 *3 marks · Short answer* Toss a coin 20 times and record the result each time (heads or tails). (i) How many times did you get heads? (ii) How many times did you get tails? (iii) Calculate the experimental probability of getting heads. (iv) If you toss the coin once more, what is the probability of getting tails? **Part (i)** 1. Perform the experiment of tossing a coin 20 times. 2. Record the number of times heads appear, which is typically around 10 times. Answer (i): $10$ times **Part (ii)** 1. Record the number of times tails appear in the 20 tosses, which is typically around 10 times. Answer (ii): $10$ times **Part (iii)** 1. Use the formula for experimental probability: $\text{Experimental Probability} = \frac{\text{Number of times the event occurred}}{\text{Total Number of trials}}$. 2. Substitute the values: $\frac{10}{20} = \frac{1}{2}$. Answer (iii): $\frac{1}{2}$ **Part (iv)** 1. Recognise that each coin toss is an independent event with no memory of past results. 2. The theoretical probability of getting tails on a fair coin toss is always $\frac{1}{2}$. Answer (iv): $\frac{1}{2}$ **Answer:** (i) $10$ times, (ii) $10$ times, (iii) $\frac{1}{2}$, (iv) $\frac{1}{2}$ > Common mistake: Assuming that past tosses affect the probability of the next independent toss due to the gambler's fallacy. ### Question 4 *3 marks · Short answer* Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top or on its side (See Fig. 7.5). Assign probabilities to the outcomes by using experimental probability. **Solution** 1. Conduct the experiment by tossing a paper cup $100$ times and recording whether it lands on its bottom, top, or side. 2. Count the number of times the cup lands in each position. 3. Calculate the experimental probability for each outcome by dividing the number of times the outcome occurred by $100$. **Answer:** Probabilities are $\frac{\text{Number of bottom landings}}{100}$, $\frac{\text{Number of top landings}}{100}$, and $\frac{\text{Number of side landings}}{100}$ > Common mistake: Not ensuring the sum of all experimental probabilities equals $1$. ### Question 5 *3 marks · Short answer* What is the probability of getting an even number when rolling a fair 6-sided die? **Solution** 1. Number of all possible outcomes when rolling a standard $6$-sided die = $6$ ($\{1, 2, 3, 4, 5, 6\}$) 2. Number of favourable outcomes for an even number = $3$ ($\{2, 4, 6\}$) 3. Theoretical probability = $\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$ **Answer:** $\frac{1}{2}$ > Common mistake: Listing the number of even outcomes incorrectly. ### Question 6 *3 marks · Short answer* Suppose you roll a 6-sided die 12 times and get a '3' three times. (i) What is the experimental probability of rolling a '3'? (ii) What is the theoretical probability of rolling a '3'? (iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times? **Part (i)** 1. Number of trials = $12$ 2. Number of times '3' occurred = $3$ 3. Experimental probability = $\frac{3}{12} = \frac{1}{4}$ Answer (i): $\frac{1}{4}$ **Part (ii)** 1. Number of favourable outcomes for '3' = $1$ 2. Number of possible outcomes = $6$ 3. Theoretical probability = $\frac{1}{6}$ Answer (ii): $\frac{1}{6}$ **Part (iii)** 1. Experimental probability can differ from theoretical probability when the number of trials is small. 2. According to the Law of Large Numbers, as the number of trials increases ($60, 600, 6000$), the experimental probability tends to get closer to the theoretical probability. Answer (iii): Probabilities differ due to small sample size; larger trials bring experimental probability closer to theoretical probability. **Answer:** (i) $\frac{1}{4}$ (ii) $\frac{1}{6}$ (iii) Due to small sample size; as trials increase, experimental probability gets closer to theoretical probability. > Common mistake: Confusing experimental probability with theoretical probability. ## Related pages - [All Chapter 7 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions) - [Exercise 7.1](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-1) - [Exercise 7.3](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-3) - [Exercise 7.4](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-4) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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