Class 9 Maths NCERT Solutions Chapter 7 – The Mathematics of Maybe: Introduction to Probability

Exercise – 7.1

Q1. Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event and give reasons.


(i) The next Monday will come after Sunday.

Solution:
Probability = 1
Label = Certain

Reason:
Days of the week follow a fixed order. Monday always comes after Sunday.



(ii) It will snow in Mumbai in July.

Solution:
Probability = 0
Label = Impossible

Reason:
Mumbai has a hot and humid climate. Snowfall does not occur there, especially in July.



(iii) An elephant will walk through your classroom today.

Solution:
Probability = 0
Label = Impossible

Reason:
An elephant cannot enter a classroom in normal situations, so this event cannot happen.



(iv) You will greet at least one friend at school tomorrow.

Solution:
Probability ≈ 0.8 (or close to 1)
Label = More likely

Reason:
We usually meet and greet friends at school, so this event is very likely to happen.

Exercise – 7.2

Q1. A teacher mixes a large bag of sweets and selects a sample of 30 sweets: 10 red, 8 green, 7 yellow, 5 blue.

(i) Calculate the probability that a randomly picked sweet is green.

Solution:
Total sweets = 30
Green sweets = 8

Probability = favourable outcomestotal outcomes\frac{\text{favourable outcomes}}{\text{total outcomes}}total outcomesfavourable outcomes​

=830=415= \frac{8}{30} = \frac{4}{15}=308​=154​

Answer: 415\frac{4}{15}154​


(ii) Estimate how many sweets are yellow out of 600 sweets.

Solution:
Yellow sweets in sample = 7 out of 30

Estimated yellow sweets:
=730×600= \frac{7}{30} \times 600=307​×600

=140= 140=140

Answer: 140 sweets



Q2. A survey of 40 students: 14 Science, 11 Arts, 9 Sports, 6 Debate. Total students = 800.

(i) Find the probability that a student prefers Arts Club.

Solution:
Arts students = 11
Total = 40

Probability = 1140\frac{11}{40}4011​

Answer: 1140\frac{11}{40}4011​


(ii) Estimate number of students who prefer Sports Club.

Solution:
Sports students = 9 out of 40

Estimated number:
=940×800= \frac{9}{40} \times 800=409​×800

=180= 180=180

Answer: 180 students



Q3. Toss a coin 20 times.

(i) Number of heads

(ii) Number of tails

Solution:
👉 This depends on your experiment.
Example: Heads = 11, Tails = 9


(iii) Experimental probability of heads

Solution:
=Number of headsTotal tosses= \frac{\text{Number of heads}}{\text{Total tosses}}=Total tossesNumber of heads​

=1120= \frac{11}{20}=2011​

Answer: 1120\frac{11}{20}2011​


(iv) Probability of getting tails in next toss

Solution:
For a fair coin:

Probability = 12\frac{1}{2}21​

Answer: 12\frac{1}{2}21​



Q4. Toss a paper cup 100 times and record outcomes.

Solution:
👉 Count results like this (example):

Bottom = 40
Top = 30
Side = 30

Probability =
Bottom = 40100=0.4\frac{40}{100} = 0.410040​=0.4
Top = 30100=0.3\frac{30}{100} = 0.310030​=0.3
Side = 30100=0.3\frac{30}{100} = 0.310030​=0.3

Answer:
Assign probabilities based on your experiment results.



Q5. What is the probability of getting an even number when rolling a fair die?

Solution:
Even numbers = 2, 4, 6 → total = 3
Total outcomes = 6

Probability = 36=12\frac{3}{6} = \frac{1}{2}63​=21​

Answer: 12\frac{1}{2}21​

Exercise – 7.3

Q1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Solution:
A die has numbers: 1, 2, 3, 4, 5, 6

Sample space:
S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}S={1,2,3,4,5,6}

Total number of outcomes = 6

Answer: 6



Q2. For the following experiments, write down the sample space SSS.

(i) Rolling a die and tossing a coin together

Solution:
Coin outcomes = H (Head), T (Tail)
Die outcomes = 1, 2, 3, 4, 5, 6

Sample space:
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)}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)\}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: 12 outcomes


(ii) Choosing a random integer between –5 and +5

Solution:
Integers include both negative and positive numbers:

S={5,4,3,2,1,0,1,2,3,4,5}S = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}S={−5,−4,−3,−2,−1,0,1,2,3,4,5}

Answer: 11 outcomes


(iii) A box containing 5 green and 7 red balls. One ball is drawn at random

Solution:
Possible outcomes are based on colour:

S={Green,Red}S = \{\text{Green}, \text{Red}\}S={Green,Red}

Answer: 2 outcomes



Q3. In a village fair, snacks: Samosa, Pakora, Bhaji; drinks: Chai, Lassi

(i) List the sample space of all possible combinations

Solution:
Combine each snack with each drink:

S={S = \{S={
(Samosa, Chai),
(Samosa, Lassi),
(Pakora, Chai),
(Pakora, Lassi),
(Bhaji, Chai),
(Bhaji, Lassi)
}\}}

Answer: 6 outcomes


(ii) List the event ‘Selecting Samosa as a snack’

Solution:
Event includes all outcomes where snack is Samosa:

E={(Samosa,Chai),(Samosa,Lassi)}E = \{(Samosa, Chai), (Samosa, Lassi)\}E={(Samosa,Chai),(Samosa,Lassi)}

Answer: Event with 2 outcomes

Exercise – 7.4

Q1. There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango.

(i) Draw a tree diagram showing all possible pairs of fruits.

Solution:
Basket A → Apple (A), Orange (O)
Basket B → Banana (B), Mango (M)

Tree structure (text form):

  • Apple → Banana → (A, B)
  • Apple → Mango → (A, M)
  • Orange → Banana → (O, B)
  • Orange → Mango → (O, M)

(ii) List the sample space

Solution:
S={(Apple,Banana),(Apple,Mango),(Orange,Banana),(Orange,Mango)}S = \{(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)\}S={(Apple,Banana),(Apple,Mango),(Orange,Banana),(Orange,Mango)}

Answer: 4 outcomes


(iii) What is the probability of picking one apple and one banana?

Solution:
Favourable outcome = (Apple, Banana)

Total outcomes = 4

Probability = 14\frac{1}{4}41​

Answer: 14\frac{1}{4}41​



Q2. A box contains 3 red pens, 4 black pens, and 2 green pens. You pick one pen and replace it, then your friend picks.

(i) What are the possible outcomes? Draw tree diagram.

Solution:
Possible colours = Red (R), Black (B), Green (G)

Tree (text form):

  • You pick R → Friend picks R, B, G → (R,R), (R,B), (R,G)
  • You pick B → Friend picks R, B, G → (B,R), (B,B), (B,G)
  • You pick G → Friend picks R, B, G → (G,R), (G,B), (G,G)

Sample space:
S={(R,R),(R,B),(R,G),(B,R),(B,B),(B,G),(G,R),(G,B),(G,G)}S = \{(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)\}S={(R,R),(R,B),(R,G),(B,R),(B,B),(B,G),(G,R),(G,B),(G,G)}

Answer: 9 outcomes


(ii) Probability that both pick pens of the same colour

Solution:
Total pens = 3+4+2=93 + 4 + 2 = 93+4+2=9

Probability of Red = 39=13\frac{3}{9} = \frac{1}{3}93​=31​
Probability of Black = 49\frac{4}{9}94​
Probability of Green = 29\frac{2}{9}92​

Same colour cases:
(R,R), (B,B), (G,G)

Probability:
=P(R,R)+P(B,B)+P(G,G)= P(R,R) + P(B,B) + P(G,G)=P(R,R)+P(B,B)+P(G,G)

=(13×13)+(49×49)+(29×29)= \left(\frac{1}{3} \times \frac{1}{3}\right) + \left(\frac{4}{9} \times \frac{4}{9}\right) + \left(\frac{2}{9} \times \frac{2}{9}\right)=(31​×31​)+(94​×94​)+(92​×92​)

=19+1681+481= \frac{1}{9} + \frac{16}{81} + \frac{4}{81}=91​+8116​+814​

=9+16+481=2981= \frac{9 + 16 + 4}{81} = \frac{29}{81}=819+16+4​=8129​

Answer: 2981\frac{29}{81}8129​

End-of-Chapter Exercises

End-of-Chapter Exercises

Q1. 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.

Solution:

(i) The probability of an impossible event is 0.

(ii) The set of all possible outcomes of a random experiment is called the sample space.

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

(iv) Tossing a fair coin has a probability of 1/2 for getting heads.



Q2. 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:

Total number of students = 50
Number of students who like football = 15

Relative frequency = 1550=310=0.3\frac{15}{50} = \frac{3}{10} = 0.35015​=103​=0.3

So, the correct answer is:
Relative frequency = 3/10 or 0.3



Q3. 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.

Solution:

(i) Not equally likely (car may fail more often)
(ii) Equally likely (Head and Tail have same chance)
(iii) Equally likely (all numbers 1–6 have equal chance)
(iv) Not equally likely (3 red and 7 blue → unequal chances)
(v) Equally likely (assumed equal probability of boy and girl)



Q4. 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?

Solution:

Sample space:
S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}S={HH,HT,TH,TT}

Favourable outcomes (at least one head):
HH, HT, TH → 3

Probability = 34\frac{3}{4}43​


(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?

Solution:

Even numbers = 2, 4, 6, 8, 10 → total = 5

Total cards = 10

Probability = 510=12\frac{5}{10} = \frac{1}{2}105​=21​


(iii) A die is rolled once. What is the probability of getting a number greater than 4?

Solution:

Numbers greater than 4 = 5, 6 → total = 2

Total outcomes = 6

Probability = 26=13\frac{2}{6} = \frac{1}{3}62​=31​


(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?

Solution:

Total balls = 6
Non-red balls = 3

Probability = 36=12\frac{3}{6} = \frac{1}{2}63​=21​


(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Solution:

Sample space:
S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Favourable outcomes = HHT, HTH, THH → 3

Probability = 38\frac{3}{8}83​



Q5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Solution:

Total outcomes = 3

Probability of strawberry = 13\frac{1}{3}31​



Q6. 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:

ShirtPants
RedJeans
RedKhakis
RedShorts
BlueJeans
BlueKhakis
BlueShorts

Total combinations = 6



Q7. A tyre company records distances before replacement in 1000 cases. 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

Solution:

Total cases = 1000

(i) Less than 4000 km:
=201000=0.02= \frac{20}{1000} = 0.02=100020​=0.02

(ii) Between 4000 and 14000 km:
=210+3251000=5351000=0.535= \frac{210 + 325}{1000} = \frac{535}{1000} = 0.535=1000210+325​=1000535​=0.535

(iii) More than 14000 km:
=4451000=0.445= \frac{445}{1000} = 0.445=1000445​=0.445



Q8. 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?

Solution:

Total letters = 5

(i) P, E, C → total = 4

Probability = 45\frac{4}{5}54​

(ii) Not E → total = 3

Probability = 35\frac{3}{5}53​

Q9. A game of chance consists of spinning an arrow 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?

Solution:

Total outcomes = 8

(i) Favourable outcome = {8} → 1
Probability = 18\frac{1}{8}81​

(ii) Odd numbers = {1, 3, 5, 7} → 4
Probability = 48=12\frac{4}{8} = \frac{1}{2}84​=21​

(iii) Numbers greater than 2 = {3, 4, 5, 6, 7, 8} → 6
Probability = 68=34\frac{6}{8} = \frac{3}{4}86​=43​

(iv) Numbers less than 9 = all numbers (1–8) → 8
Probability = 88=1\frac{8}{8} = 188​=1

(v) Multiples of 3 = {3, 6} → 2
Probability = 28=14\frac{2}{8} = \frac{1}{4}82​=41​



Q10. 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?

Solution:

Total balls = 9

(i) Red then Blue:
=49×58=2072=518= \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18}=94​×85​=7220​=185​

(ii) Blue then Blue:
=59×48=2072=518= \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18}=95​×84​=7220​=185​



Q11. 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:

Event with probability 0 (impossible):
Getting sum = 13

Event with probability 1 (certain):
Getting sum less than or equal to 12



Q12. 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?

Solution:

Prime numbers > 5 = 7, 11

Number of favourable outcomes = 8
Total outcomes = 36

Probability = 836=29\frac{8}{36} = \frac{2}{9}368​=92​


(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?

Solution:

Total balls = 9

Probability (same colour):
Red = 49×38\frac{4}{9} \times \frac{3}{8}94​×83​
Green = 39×28\frac{3}{9} \times \frac{2}{8}93​×82​
Blue = 29×18\frac{2}{9} \times \frac{1}{8}92​×81​

Total same = 12+6+272=2072=518\frac{12 + 6 + 2}{72} = \frac{20}{72} = \frac{5}{18}7212+6+2​=7220​=185​

Different colours = 1518=13181 – \frac{5}{18} = \frac{13}{18}1−185​=1813​


(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

Solution:

Favourable outcomes = HHT, HTH → 2

Total outcomes = 8

Probability = 28=14\frac{2}{8} = \frac{1}{4}82​=41​


(iv) A four-digit number is formed using digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

Solution:

Total numbers = 4! = 24

Even numbers → last digit = 2 or 4 → 2 choices

Favourable = 2×3!=122 \times 3! = 122×3!=12

Probability = 1224=12\frac{12}{24} = \frac{1}{2}2412​=21​


(v) A student takes a multiple-choice test with 3 questions, each having 4 options. What is the probability that the student guesses and gets exactly 2 answers correct?

Solution:

Probability = 3×(14)2×343 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4}3×(41​)2×43​

=964= \frac{9}{64}=649​



Q13. A box contains 4 balls numbered 1 to 4.

(i) A ball is drawn and replaced, then drawn again.
(ii) A ball is drawn and not replaced, then drawn again.
(iii) What are the sizes of these sample spaces?

Solution:

(i) With replacement:
Total outcomes = 4×4=164 \times 4 = 164×4=16

(ii) Without replacement:
Total outcomes = 4×3=124 \times 3 = 124×3=12

(iii) Sizes = 16 and 12



Q14. List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from 1 to 6.

Solution:

S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),S = \{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),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)}(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

Total outcomes = 12



Q15. Three coins are tossed, and the number of heads is recorded. Which list is correct sample space?

Options:
(i) {1,2,3}
(ii) {0,1,2}
(iii) {0,1,2,3,4}
(iv) {0,1,2,3}

Solution:

Possible number of heads = 0, 1, 2, 3

Correct answer = (iv) {0,1,2,3}