---
title: "NCERT Solutions Class 9 Maths The Mathematics of Maybe: Introduction to Probability"
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# NCERT Solutions Class 9 Maths The Mathematics of Maybe: Introduction to Probability

This chapter's questions cover foundational concepts of probability, including subjective versus objective probability, experimental and theoretical probability, sample spaces, tree diagrams, and random events.

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## Exercise Set 7.1

### Question 1

*3 marks · Short answer*

Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.
(i) The next Monday will come after Sunday.
(ii) It will snow in Mumbai in July.
(iii) An elephant will walk through your classroom today.
(iv) You will greet at least one friend at school tomorrow.

**Part (i)**

1. The event is certain because the days of the week follow a fixed, invariable cyclic order where Monday always immediately follows Sunday.

Answer (i): Certain (Probability = 1).

**Part (ii)**

1. Mumbai has a tropical climate with warm temperatures throughout the year, making snowfall impossible.

Answer (ii): Impossible (Probability = 0).

**Part (iii)**

1. An elephant is a wild zoo or forest animal and cannot enter a normal school classroom, making it an event that will not happen.

Answer (iii): Impossible (Probability = 0).

**Part (iv)**

1. Since going to school involves meeting friends and greeting them daily, it is expected to happen normally.

Answer (iv): Certain or more likely depending on daily interaction, usually certain (Probability = 1).

**Answer:** Ranked each event from impossible to certain with appropriate reasons based on the probability scale.

> Common mistake: Confusing 'less likely' with 'impossible' for events that are practically not going to happen.

## 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.

## Exercise Set 7.3

### Question 1

*3 marks · Short answer*

When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

**Solution**

1. The sample space for rolling a single 6-sided die consists of all the possible numbers that can appear on the top face.
2. The possible outcomes are $1, 2, 3, 4, 5, \text{ and } 6$.
3. Therefore, the total number of possible outcomes in the sample space is $6$.

**Answer:** 6

> Common mistake: Listing outcomes instead of giving the total number of outcomes.

### Question 2

*3 marks · Short answer*

For the following experiments write down the sample space S.
(i) Rolling a die and tossing a coin together.
(ii) Choosing a random integer between – 5 and + 5.
(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.

**Part (i)**

1. Let H represent Heads and T represent Tails on the coin, and numbers $1$ to $6$ represent the die outcomes.
2. The sample space $S$ is the combination of each die outcome with both coin outcomes.
3. $S = \{(1, \text{H}), (1, \text{T}), (2, \text{H}), (2, \text{T}), (3, \text{H}), (3, \text{T}), (4, \text{H}), (4, \text{T}), (5, \text{H}), (5, \text{T}), (6, \text{H}), (6, \text{T})\}$

Answer (i): $S = \{(1, \text{H}), (1, \text{T}), (2, \text{H}), (2, \text{T}), (3, \text{H}), (3, \text{T}), (4, \text{H}), (4, \text{T}), (5, \text{H}), (5, \text{T}), (6, \text{H}), (6, \text{T})\}$

**Part (ii)**

1. An integer strictly between $-5$ and $+5$ excludes $-5$ and $+5$.
2. The integers between them are $-4, -3, -2, -1, 0, 1, 2, 3, 4$.
3. $S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

Answer (ii): $S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

**Part (iii)**

1. A box contains 5 green and 7 red balls, making 12 balls in total.
2. Let $G_1, G_2, G_3, G_4, G_5$ represent the green balls and $R_1, R_2, R_3, R_4, R_5, R_6, R_7$ represent the red balls.
3. The sample space is the set of all individual balls: $S = \{G_1, G_2, G_3, G_4, G_5, R_1, R_2, R_3, R_4, R_5, R_6, R_7\}$ (or simply representing them by colour type if indistinguishable: $\{ \text{Green}, \text{Red} \}$ depending on standard phrasing, but individual items are typically listed or grouped).

Answer (iii): $S = \{\text{Green}, \text{Red}\}$

**Answer:** Sample spaces for the given experiments are listed in the parts.

> Common mistake: Including $-5$ and $+5$ in the integer range or listing incorrect combinations.

### Question 3

*3 marks · Short answer*

In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.
(i) List the sample space of all possible snack and drink combinations a person could choose at the fair.
(ii) List the event 'Selecting Samosa as a snack.'

**Part (i)**

1. There are 3 snacks (Samosa, Pakora, Bhaji) and 2 drinks (Chai, Lassi).
2. Each snack can be paired with each drink to form combinations.
3. $S = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi}), (\text{Pakora, Chai}), (\text{Pakora, Lassi}), (\text{Bhaji, Chai}), (\text{Bhaji, Lassi})\}$

Answer (i): $S = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi}), (\text{Pakora, Chai}), (\text{Pakora, Lassi}), (\text{Bhaji, Chai}), (\text{Bhaji, Lassi})\}$

**Part (ii)**

1. The event requires selecting Samosa as a snack with any of the available drinks.
2. The outcomes containing Samosa are (Samosa, Chai) and (Samosa, Lassi).
3. $E = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi})\}$

Answer (ii): $E = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi})\}$

**Answer:** Sample space and event are listed in the parts.

> Common mistake: Forgetting to pair each snack with both drinks.

## Exercise Set 7.4

### Question 1

*3 marks · Short answer*

There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.
(i) Draw a tree diagram showing all possible pairs of fruits.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?

**Part (i)**

1. Diagram: Draw a starting point with branches to Apple, Orange (from basket A), and from each fruit, branch out to Banana and Mango (from basket B).
2. The branches represent all possible pairs of fruits chosen from the two baskets.

Answer (i): Tree diagram showing pairs from baskets A and B.

**Part (ii)**

1. List all possible pairs from the branches of the tree diagram.
2. Sample space $S = \{\text{(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}\}$.

Answer (ii): $S = \{\text{(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}\}$

**Part (iii)**

1. Identify the total number of possible outcomes, which is $4$.
2. Identify the number of favourable outcomes for picking one apple and one banana, which is $1$ (Apple, Banana).
3. Calculate the probability: $\text{P}(\text{Apple and Banana}) = \frac{1}{4} = 0.25$ or $25\%$. $\blacksquare$

Answer (iii): $\frac{1}{4}$ or $0.25$ or $25\%$

**Answer:** Tree diagram drawn, sample space listed, and probability calculated.

> Common mistake: Listing outcomes in the sample space without considering the order or missing one of the combinations.

### Question 2

*3 marks · Short answer*

Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

**Part (i)**

1. The possible colours of the pens are Red (R), Black (BK), and Green (G).
2. Diagram: Draw a starting point with 3 main branches for R, BK, and G (your pick), and from each branch, draw 3 further branches for R, BK, and G (your friend's pick).
3. There are $3 \times 3 = 9$ possible outcomes in total.

Answer (i): Outcomes are Red, Black, Green; tree diagram with 9 branches drawn.

**Part (ii)**

1. Identify the outcomes where both pick pens of the same colour: (Red, Red), (Black, Black), and (Green, Green).
2. Count the number of favourable outcomes, which is $3$. Total possible outcomes is $9$.
3. Calculate the probability: $\text{P}(\text{same colour}) = \frac{3}{9} = \frac{1}{3} \approx 0.333$ or $33.3\%$. $\blacksquare$

Answer (ii): $\frac{1}{3}$ or $33.3\%$

**Answer:** Possible outcomes identified, tree diagram described, and probability calculated.

> Common mistake: Forgetting that pens are replaced, so the total number of options remains the same for the friend.

## End-of-Chapter Exercises

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks.
(i) The probability of an impossible event is _______.
(ii) The set of all possible outcomes of a random experiment is called the __________.
(iii) The probability of an event that is certain to happen is _______.
(iv) Tossing a fair coin has a probability of ______ for getting heads.

**Part (i)**

1. The probability of an impossible event cannot occur.

Answer (i): 0

**Part (ii)**

1. The set of all possible outcomes of a random experiment is defined as the sample space.

Answer (ii): sample space

**Part (iii)**

1. The probability of an event that is certain to happen is 1.

Answer (iii): 1

**Part (iv)**

1. For a fair coin, the probability of getting heads is 1/2.

Answer (iv): $1/2$

**Answer:** Fill in the blanks with 0, sample space, 1, and 1/2 respectively.

> Common mistake: Confusing experimental probability with theoretical certainty.

### Question 2

*3 marks · Short answer*

In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).

**Solution**

1. Total number of students = 50
2. Number of students who like football = 15
3. The measure is relative frequency and its value is 15/50 = 3/10 or 0.3.

**Answer:** relative frequency, 3/10 (or 0.3)

> Common mistake: Writing frequency instead of relative frequency.

### Question 3

*3 marks · Short answer*

Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) Tossing a fair coin once.
(iii) Rolling a fair 6-sided die.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
(v) A baby is born. It is a boy or a girl.

**Part (i)**

1. A car starting depends on mechanical condition and fuel, so outcomes are not equally likely.

Answer (i): Not equally likely

**Part (ii)**

1. Tossing a fair coin has two symmetric outcomes, heads and tails, which are equally likely.

Answer (ii): Equally likely

**Part (iii)**

1. A fair 6-sided die has six faces with identical chances, making outcomes equally likely.

Answer (iii): Equally likely

**Part (iv)**

1. The bag has 3 red and 7 blue marbles, so drawing a red marble and blue marble do not have equal chances.

Answer (iv): Not equally likely

**Part (v)**

1. Assuming standard biological births, a newborn being a boy or a girl are equally likely outcomes.

Answer (v): Equally likely

**Answer:** Only (ii), (iii), and (v) have equally likely outcomes.

> Common mistake: Assuming every experiment with two outcomes is equally likely.

### Question 4

*3 marks · Short answer*

Write the sample space and calculate the probability based on the given information.
(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

**Part (i)**

1. Sample space S = {HH, HT, TH, TT}, n(S) = 4.
2. Favorable outcomes for at least one head = {HH, HT, TH}, number = 3.
3. Probability = 3/4.

Answer (i): $3/4$

**Part (ii)**

1. Sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10.
2. Even numbers = {2, 4, 6, 8, 10}, number of favorable outcomes = 5.
3. Probability = 5/10 = 1/2.

Answer (ii): $1/2$

**Part (iii)**

1. Sample space S = {1, 2, 3, 4, 5, 6}, n(S) = 6.
2. Numbers greater than 4 = {5, 6}, number of favorable outcomes = 2.
3. Probability = 2/6 = 1/3.

Answer (iii): $1/3$

**Part (iv)**

1. Total balls = 3 red + 2 blue + 1 green = 6.
2. Favorable outcomes of not being red = 2 blue + 1 green = 3.
3. Probability = 3/6 = 1/2.

Answer (iv): $1/2$

**Part (v)**

1. Sample space for three coins has n(S) = 8 outcomes.
2. Favorable outcomes for exactly two heads = {HHT, HTH, THH}, number = 3.
3. Probability = 3/8.

Answer (v): $3/8$

**Answer:** Probabilities are calculated for each part using sample spaces and favorable outcomes.

> Common mistake: Miscounting the sample size or favorable outcomes.

### Question 5

*3 marks · Short answer*

A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

**Solution**

1. Total number of possible outcomes = 3 (strawberry, lemon, and mint)
2. Number of favorable outcomes for strawberry = 1
3. Probability of picking a strawberry candy = 1/3

**Answer:** $1/3$

> Common mistake: Dividing by an incorrect total number of candies.

### Question 6

*3 marks · Short answer*

A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

**Solution**

1. Shirts = {Red, Blue}, Pants = {Jeans, Khakis, Shorts}
2. Possible combinations consist of pairing each shirt with each type of pants.
3. Table of combinations: (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis), (Blue, Shorts).

**Answer:** 6 possible combinations displayed in a table format.

> Common mistake: Missing some combinations of shirts and pants.

### Question 7

*3 marks · Short answer*

A tyre company records distances before replacement in 1000 cases.
Distance (km): Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000
Number of cases: 20 | 210 | 325 | 445
Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.

**Part (i)**

1. Total number of cases = $1000$.
2. Number of tyres lasting less than $4000\text{ km} = 20$.
3. Probability = $\frac{20}{1000} = 0.02$.

Answer (i): $0.02$

**Part (ii)**

1. Total number of cases = $1000$.
2. Number of tyres lasting between $4001$ and $14000\text{ km} = 210 + 325 = 535$.
3. Probability = $\frac{535}{1000} = 0.535$.

Answer (ii): $0.535$

**Part (iii)**

1. Total number of cases = $1000$.
2. Number of tyres lasting more than $14000\text{ km} = 445$.
3. Probability = $\frac{445}{1000} = 0.445$.

Answer (iii): $0.445$

**Answer:** Probabilities are (i) $0.02$, (ii) $0.535$, (iii) $0.445$

> Common mistake: Adding incorrect frequency ranges for part (ii).

### Question 8

*3 marks · Short answer*

The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.
(i) What is the probability that it is a P, E or C?
(ii) What is the probability that it is not an E?

**Part (i)**

1. Total number of letters in the word PEACE is $5$.
2. Number of letters that are P, E, or C is $4$ (P, E, A, C, E contains P, E, C).
3. Probability = $\frac{4}{5}$.

Answer (i): $\frac{4}{5}$

**Part (ii)**

1. Total number of cards is $5$.
2. Number of cards that are not E is $3$ (P, A, C).
3. Probability = $\frac{3}{5}$.

Answer (ii): $\frac{3}{5}$

**Answer:** Probabilities are (i) $\frac{4}{5}$, (ii) $\frac{3}{5}$

> Common mistake: Counting duplicate letters incorrectly in total outcomes.

### Question 9

*3 marks · Short answer*

A game of chance consists of spinning an arrow (see Fig. 7.7) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?

**Part (i)**

1. Total possible outcomes = $8$.
2. Number of favourable outcomes for getting $8$ is $1$.
3. Probability = $\frac{1}{8}$.

Answer (i): $\frac{1}{8}$

**Part (ii)**

1. Odd numbers from $1$ to $8$ are $1, 3, 5, 7$ (total $4$).
2. Probability = $\frac{4}{8} = \frac{1}{2}$.

Answer (ii): $\frac{1}{2}$

**Part (iii)**

1. Numbers greater than $2$ are $3, 4, 5, 6, 7, 8$ (total $6$).
2. Probability = $\frac{6}{8} = \frac{3}{4}$.

Answer (iii): $\frac{3}{4}$

**Part (iv)**

1. All numbers from $1$ to $8$ are less than $9$ (total $8$).
2. Probability = $\frac{8}{8} = 1$.

Answer (iv): $1$

**Part (v)**

1. Multiples of $3$ between $1$ and $8$ are $3, 6$ (total $2$).
2. Probability = $\frac{2}{8} = \frac{1}{4}$.

Answer (v): $\frac{1}{4}$

**Answer:** Probabilities are (i) $\frac{1}{8}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{4}$, (iv) $1$, (v) $\frac{3}{8}$

> Common mistake: Including $9$ in the count for part (iv) or miscounting multiples.

### Question 10

*3 marks · Short answer*

A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.
(i) What is the probability of drawing a red ball and then a blue ball?
(ii) What is the probability of drawing 2 blue balls?

**Part (i)**

1. Total balls = $4$ red and $5$ blue = $9$ balls.
2. Probability of drawing red then blue = $\frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18}$.

Answer (i): $\frac{5}{18}$

**Part (ii)**

1. Probability of drawing two blue balls = $\frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18}$.

Answer (ii): $\frac{5}{18}$

**Answer:** Probabilities are (i) $\frac{5}{18}$, (ii) $\frac{5}{18}$

> Common mistake: Failing to reduce the denominator for the second draw without replacement.

### Question 11

*3 marks · Short answer*

I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.

**Solution**

1. Total possible outcomes when throwing a pair of 6-sided dice is $36$.
2. An event with probability $0$: getting a sum equal to $13$ (since the maximum sum is $12$).
3. An outcome with probability $1$: getting a sum less than or equal to $12$.

**Answer:** Event with probability $0$: getting a sum of $13$. Outcome with probability $1$: getting a sum $\le 12$.

> Common mistake: Confusing an event with a single outcome.

### Question 12

*3 marks · Short answer*

Write the sample space and calculate the probability based on the given information.
(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

**Part (i)**

1. Total outcomes for two dice = $36$.
2. Prime sums greater than $5$ are $7$ and $11$. Favourable outcomes = $6$ (for sum $7$) $+ 2$ (for sum $11$) $= 8$.
3. Probability = $\frac{8}{36} = \frac{2}{9}$.

Answer (i): $\frac{2}{9}$

**Part (ii)**

1. Total balls = $4 + 3 + 2 = 9$.
2. Probability of both same colour = $P(\text{RR}) + P(\text{GG}) + P(\text{BB}) = \frac{4}{9}\times\frac{3}{8} + \frac{3}{9}\times\frac{2}{8} + \frac{2}{9}\times\frac{1}{8} = \frac{12 + 6 + 2}{72} = \frac{20}{72}$.
3. Probability of different colours = $1 - \frac{20}{72} = \frac{52}{72} = \frac{13}{18}$.

Answer (ii): $\frac{13}{18}$

**Part (iii)**

1. Sample space for three coins has $8$ outcomes: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
2. Favourable outcomes (first is H and exactly two H in total): {HHT, HTH} (total $2$).
3. Probability = $\frac{2}{8} = \frac{1}{4}$.

Answer (iii): $\frac{1}{4}$

**Part (iv)**

1. Total four-digit numbers without repetition using $1, 2, 3, 4$ is $4! = 24$.
2. An even number must end in $2$ or $4$ ($2$ choices for the last digit, $3! = 6$ ways for the remaining digits, giving $12$ even numbers).
3. Probability = $\frac{12}{24} = \frac{1}{2}$.

Answer (iv): $\frac{1}{2}$

**Part (v)**

1. Total possible guessing outcomes for $3$ questions with $4$ options each = $4^3 = 64$.
2. Number of ways to get exactly $2$ correct = $^3C_2 \times 3^1 = 9$.
3. Probability = $\frac{9}{64}$.

Answer (v): $\frac{9}{64}$

**Answer:** Probabilities are (i) $\frac{5}{18}$, (ii) $\frac{29}{36}$, (iii) $\frac{3}{8}$, (iv) $\frac{1}{2}$, (v) $\frac{27}{64}$

> Common mistake: Forgetting to subtract from $1$ when finding different colours in part (ii).

### Question 13

*3 marks · Short answer*

A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:
(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
(iii) What are the sizes of these two sample spaces?

**Part (i)**

1. First draw has 4 possible outcomes: 1, 2, 3, 4.
2. Since the ball is returned, the second draw also has 4 possible outcomes: 1, 2, 3, 4.
3. The sample space for drawing with replacement is S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}.

Answer (i): S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}

**Part (ii)**

1. First draw has 4 possible outcomes: 1, 2, 3, 4.
2. Since the ball is not replaced, the second draw has 3 possible outcomes remaining for each first draw.
3. The sample space for drawing without replacement is S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}.

Answer (ii): S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

**Part (iii)**

1. The size of the first sample space with replacement is $4 \times 4 = 16$.
2. The size of the second sample space without replacement is $4 \times 3 = 12$.

Answer (iii): 16 and 12

**Answer:** Sample spaces and sizes for experiments with and without replacement.

> Common mistake: Forgetting that without replacement means identical numbers like (1,1) cannot occur.

### Question 14

*3 marks · Short answer*

List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.

**Solution**

1. A coin toss has two possible outcomes: Heads (H) and Tails (T).
2. A set of 6 cards numbered 1 through 6 has 6 possible outcomes: 1, 2, 3, 4, 5, 6.
3. The sample space for the simultaneous tossing of a coin and drawing of a card is the set of all possible pairs combining each coin outcome with each card outcome.
4. S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

**Answer:** S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

> Common mistake: Listing outcomes as single elements instead of ordered pairs.

### Question 15

*3 marks · Short answer*

Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?
(i) {1, 2, 3}
(ii) {0, 1, 2}
(iii) {0, 1, 2, 3, 4}
(iv) {0, 1, 2, 3}

**Solution**

1. When three coins are tossed, the number of heads obtained can be 0 (if all are tails), 1, 2, or 3 (if all are heads).
2. Therefore, the set of all possible outcomes for the number of heads is {0, 1, 2, 3}.
3. Thus, list (iv) is the correct sample space.
4. Lists (i), (ii), and (iii) fail to qualify because list (i) misses 0, list (ii) misses 3, and list (iii) includes 4 which is impossible when only three coins are tossed.

**Answer:** (iv) {0, 1, 2, 3} is the correct sample space because the number of heads when three coins are tossed can range from 0 to 3.

> Common mistake: Choosing a list based on the number of coins rather than the possible values of the recorded variable.

### Question 16

*3 marks · Short answer*

Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?

**Solution**

1. Area of the rectangular region = $\text{length} \times \text{breadth} = 3\text{ m} \times 2\text{ m} = 6\text{ m}^2$.
2. Diameter of the circle = $1\text{ m}$, so the radius $r = 0.5\text{ m}$.
3. Area of the circular region = $\pi r^2 = \pi (0.5)^2 = 0.25\pi\text{ m}^2 = \frac{\pi}{4}\text{ m}^2$.
4. Probability that the dye lands inside the circle = $\frac{\text{Area of the circle}}{\text{Area of the rectangle}} = \frac{\pi / 4}{6} = \frac{\pi}{24}$.

**Answer:** $\frac{\pi}{24}$

> Common mistake: Dividing the rectangle's area by the circle's area instead of finding the ratio of the favorable area to the total area.

## Frequently asked questions

### How many exercises and total questions are there in NCERT Solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability?

This chapter has 4 exercise sets and one end-of-chapter exercise, following the new NCERT book for the 2026-27 session. Exercise Set 7.1 has 1 question, Exercise Set 7.2 has 6 questions, Exercise Set 7.3 has 3 questions, Exercise Set 7.4 has 2 questions, and the End-of-Chapter Exercises contain 16 questions. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### Which topics are covered in the questions of Class 9 Maths Chapter 7?

The questions cover concepts like the probability scale, ranking events, experimental and theoretical probability, statistical estimation from sample data, and paper cup experiments. Other covered topics include sample spaces for single dice, snack and drink combinations, tree diagrams for independent trials with replacement, geometrical probability, and drawing with and without replacement. SwaVid's free PDF and step-by-step solutions for mastering these topics are available on this page only.

### What are the hardest question types in this chapter and how should we approach them?

The most challenging questions usually involve multi-step experiments, tree diagrams for independent trials, and geometrical probability. To approach them, you should carefully list out the sample space, draw clear tree diagrams, and apply the correct probability formulas step by step. SwaVid's free PDF and step-by-step solutions on this page only provide detailed breakdowns for these tricky questions.

### How should I write answers to score full marks in Class 9 Maths Chapter 7 examinations?

To score full marks, always clearly define your sample space, state the relevant formula before substitution, and show all calculation steps explicitly. For experimental and theoretical probability questions, write down the given values and final fractions in their simplest forms. You can refer to SwaVid's free PDF and step-by-step solutions available on this page only to understand the ideal presentation format.

### Is the free PDF for Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability available?

Yes, the complete and updated study material for the 2026-27 session is ready for students. SwaVid's free PDF and step-by-step solutions covering fill-in-the-blanks and short answer questions are on this page only to help you prepare effectively.

## Related pages

- [Exercise 7.1 solutions](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-1)
- [Exercise 7.2 solutions](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-2)
- [Exercise 7.3 solutions](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-3)
- [Exercise 7.4 solutions](https://www.swavid.com/maths/class/9/chapter/the-mathematics-of-maybe-introduction-to-probability/ncert-solutions/exercise-7-4)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student