Ganita Manjari Class 9 Chapter 7 Solutions The Mathematics of Maybe Introduction to Probability

By using Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 7 The Mathematics of Maybe Introduction to Probability NCERT Solutions, students can improve their problem-solving skills.

The Mathematics of Maybe Introduction to Probability Class 9 Solutions

Class 9 Ganita Manjari Chapter 7 Solutions

Class 9 Maths Ganita Manjari Chapter 7 Solutions The Mathematics of Maybe Introduction to Probability

Think and Reflect (NCERT Textbook Page No. 156)

Question 1.
Such unpredictability can be useful sometimes! For example, in a cricket match, the fact that a coin is tossed to decide which team will bat first is considered to be a fair method. Can you explain why?
Solution:
A coin toss is considered a fair method because it gives both teams an equal and unbiased chance.
A coin has only two possible outcomes: Head (H) and Tail (T).
So, P(H) = \(\frac {1}{2}\), P(T) = \(\frac {1}{2}\)
Since both have probability \(\frac {1}{2}\), neither side has advantage.
Random and unbiased outcome
A properly tossed coin has no preference for head or tail.
So the outcome is unpredictable and fair.

Think and Reflect (NCERT Textbook Page No. 157)

Question 1.
Ask your friend to predict the outcome of a ₹ 1 coin you toss. Do you see that your friend could guess heads or tails but could not know for certain? That’s randomness! All possible results are known, but each try is unpredictable.
Solution:
Yes — this is exactly the idea behind randomness in probability.
When we toss a coin, there are only two possible outcomes: Head (H) & Tail (T)
So we already know all possible results.
But the important point is: In a single toss, we cannot be sure which one will come.
Even your friend can only guess, not predict with certainty.
Randomness: All outcomes are known in advance. But the actual result of a single trial is unpredictable.
For example: Coin toss
P(H) = \(\frac {1}{2}\), P(T) = \(\frac {1}{2}\)
So, heads is not guaranteed. Tail is not guaranteed. Both are equally likely.

Think and Reflect (NCERT Textbook Page No. 163)

Question 1.
If I have rolled a 4 on a die 8 times in succession, the probability of rolling a 4 again is still only ≈ 0.16 (assuming the die is fair). Probability does not tell you what will happen next but predicts what will happen in the long run.
Solution:
A die is fair, so each roll is an independent event.
That shows the result of one roll does not affect the next roll.
Even if 4 has come 8 times in a row, it does not change future outcomes.
A die has 6 faces: 1, 2, 3, 4, 5, 6
So, total possible outcomes = 6
We want the number 4.
So, favorable outcome = 1 (only one face shows 4)
P(4) = \(\frac{\text { Favourable outcomes }}{\text { Total possible outcomes }}=\frac{1}{6}\)
P(4) ≈ 0.1667 ≈ 0.16
Conclusion: Even if 4 has already come 8 times in a row, the probability of getting 4 again remains the same.
\(\frac {1}{6}\) ≈ 0.16.
This is because each roll is independent.

Think and Reflect (NCERT Textbook Page No. 169)

Question 1.
Can you calculate the probability of getting one head and one tail?
Solution:
Given: Two fair coins are tossed.
To Find: Probability of getting one head and one tail.
When two coins are tossed, the possible outcomes are:
S = {HH, HT, TH, TT} where, H = Head and T = Tail
Here, total outcomes = 4
We need one head and one tail, so favorable outcomes are: HT, TH
Number of favorable outcomes = 2
Now, P(one head and one tad) = \(\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac {2}{4}\)
= \(\frac {1}{2}\)
When two coins are tossed, getting one head and one tail can happen in two different ways (HT, TH), so the probability is \(\frac {1}{2}\).

Ex 7.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 7.1 Solutions

Exercise 7.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 7.1 Solutions

Question 1.
Rank the following events on a scale from 0 (Impossible) to I (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.
Solution:
(i) P = 1, Certain: In a week, Monday comes after Sunday. This is an event that will definitely occur.
(ii) P (0), Impossible: In July, Mumbai is exceptionally hot.
(iii) P (0), Impossible: No school permits an elephant to walk in their classrooms.
(iv) Close to 1, More likely: We usually greet our friends at school.

Ex 7.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 7.2 Solutions

Exercise 7.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 7.2 Solutions

Question 1.
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.
Solution:
(i) Using the experimental probability
P(Green) = \(\frac{\text { Number of green sweets }}{\text { Total no. of sweets }}\)
= \(\frac {8}{30}\)
= \(\frac {4}{15}\)

(ii) From the sample
P(Yellow) = \(\frac{\text { No. of yellow sweets }}{\text { Total no. of sweets }}\) = \(\frac {7}{30}\)
Now estimated yellow sweets = P(Yellow) × Total sweets
= \(\frac {7}{30}\) × 600
= 7 × 20
= 140 sweets

Question 2.
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.
Solution:
(i) P(Arts club) = \(\frac{\text { No. of students preferring Arts club }}{\text { Total no. of students }}\) = \(\frac {11}{40}\)

(ii) P(Sports club) = \(\frac{\text { No. of students preferring Sports club }}{\text { Total no. of students }}\)
Estimated students preferring sports club = \(\frac {9}{40}\) × 800
= 9 × 20
= 180

Question 3.
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?
Solution:
(i) It is a practical experiment. Results will vary for each student.
Assume that we get a ‘head’ 13 times.

(ii) We get a tail 7 times (20 – 13).

(iii) Experimental Probability = \(\frac{\text { No. of times heads appeared }}{\text { Total no. of tosses }}=\frac{13}{20}\)

(iv) On tossing the coin once more, we may get a ‘head’ or a ‘tail’.
Case 1. When ‘head’ appears
Probability of getting tails = \(\frac {7}{21}\)
Case 2. When ‘tail’ appears
Probability of getting tails = \(\frac {8}{21}\)

Question 4.
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. below). Assign probabilities to the outcomes by using experimental probability.

Solution:
This is a practical experiment.
Results will vary for each student.
Assume that the cup lands on its bottom 38 times, and lands on its top 30 times.
This means the cup lands on its side 32 times (100 – 38 – 30).
Therefore, P(cup lands on its bottom) = \(\frac{\text { No. of times cup lands on bottom }}{\text { Total no. of tosses }}\) = \(\frac {38}{100}\)
P(cup lands on its top) = \(\frac{\text { No. of times cup lands on top }}{\text { Total no. of tosses }}\) = \(\frac {38}{100}\)
P(cup lands on its side) = \(\frac {32}{100}\)

Question 5.
What is the probability of getting an even number when rolling a fair 6-sided die?
Solution:
Here, the sample space S = {1, 2, 3, 4, 5, 6}
On a 6-sided die, there are three even numbers: 2, 4, and 6.
So, P(an even number) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of possible outcomes }}\)
= \(\frac {3}{6}\)
= \(\frac {1}{2}\)

Question 6.
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?
Solution:
Here, on rolling a 6-sided die 12 times, you get a ‘3’ three times.
(i) Number of all outcomes = 12 (Total rolls in experiment)
Number of favourable outcomes = 3 (times ‘3’ actually appeared)
P(rolling a 3) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all outcomes }}\)
= \(\frac {3}{12}\)
= 0.25 or 25%

(ii) Number of all possible outcomes = 6 (numbers 1 through 6)
Number of favourable outcomes = 1 (only the number 3)
P(rolling a 3) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac {1}{6}\)
= 0.167 or 16.7%

(iii) Difference between Experimental and Theoretical probabilities: The difference exists because experimental probability is based on evidence from a limited sample, while theoretical probability is based on the ideal mathematical outcome. As the number of trials increases, the experimental probability will likely get closer to the theoretical one. This is why a larger sample size makes our estimates more reliable.

Ex 7.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 7.3 Solutions

Exercise 7.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 7.3 Solutions

Question 1.
When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Solution:
Here, sample space: S = {1, 2, 3, 4, 5, 6}
And sample size: n(S) = 6
Therefore, there are 6 possible outcomes in the sample space.

Question 2.
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.
Solution:
(i) Rolling a die and tossing a coin together:
S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}
where the number represents the die outcome, and H/T represents the coin outcome.

(ii) Choosing a random integer between -5 and +5:
S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

(iii) A box containing 5 green and 7 red balls.
One ball is drawn at random:
S = {Green, Red}

Question 3.
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’.
Solution:
(i) Sample space of all possible combinations:
S = {Samosa-Chai, Samosa-Lassi, Pakora-Chai, Pakora-Lassi, Bhaji-Chai, Bhaji-Lassi}
(We can also write this as: S = {(Sa, C), (Sa, L), (P, C), (P, L), (B, C), (B, L)})

(ii) Event ‘Selecting Samosa as a snack’:
E = {Samosa-Chai, Samosa-Lassi}
or
E = {(Sa, C), (Sa, L)}
This event contains all combinations where Samosa is selected with either Chai or Lassi.

Ex 7.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 7.4 Solutions

Exercise 7.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 7.4 Solutions

Question 1.
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?
Solution:
(i)

(ii) Here, sample space S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}
∴ n(S) = 4
(iii) P(Apple, Banana) = \(\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)

Question 2.
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?
Solution:
(i) S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}
No. of pens = 3 Red + 4 Black + 2 Green = 9

(ii) Here, P(Red pen) = \(\frac {3}{9}\)
P(Black pen) = \(\frac {4}{9}\)
P(Green pen) = \(\frac {2}{9}\)

So, P(Both pick up pens of the same colour) = P(R, R) + P(B, B) + P(G, G)
= \(\frac{3}{9} \times \frac{3}{9}+\frac{4}{9} \times \frac{4}{9}+\frac{2}{9} \times \frac{2}{9}\)
= \(\frac{9}{81}+\frac{16}{81}+\frac{4}{81}\)
= \(\frac {29}{81}\)

Ganita Manjari Class 9 Maths Chapter 7 End of Chapter Exercise Solutions

The Mathematics of Maybe Introduction to Probability End of Chapter Exercise Solutions

Question 1.
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.
Explanation: If something cannot happen (like rolling a 7 on a standard 6-sided die), there is a 0% chance of it occurring.
In probability, we represent this as 0.

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

(iii) The probability of an event that is certain to happen is 1.
Explanation: If an event is guaranteed to happen (like the sun rising or pulling a red ball from a bag of only red balls), the probability is 100%, represented as 1.

(iv) Tossing a fair coin has a probability of \(\frac {1}{2}\) (or 0.5 or 50%) for getting heads.
Explanation: Heads and tails are equally likely outcomes, so each has probability \(\frac {1}{2}\).

Question 2.
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:
The number of students who like football is 15,
and the relative frequency is \(\frac {15}{50}\) = \(\frac {3}{10}\) = 0.3 (or 30%).
Explanation:
Frequency: The count of occurrences = 15
Relative Frequency: The proportion or probability = \(\frac{\text { Frequency }}{\text { Total }}\)
= \(\frac {15}{50}\)
= 0.3

Question 3.
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) Does NOT have equally likely outcomes.
Explanation: A well-maintained car is much more likely to start than not start.
These outcomes are not equally likely.
The probability of starting is much higher than the probability of not starting.

(ii) It has equally likely outcomes.
Explanation: A fair coin is symmetrical, so heads and tails have equal probability.
P(Heads) = P(Tails) = \(\frac {1}{2}\)

(iii) It has equally likely outcomes.
Explanation: A fair die is symmetrical, so each face has equal probability.
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = \(\frac {1}{6}\)

(iv) Does NOT have equally likely outcomes.
Explanation: Blue marbles are more numerous than red marbles.
P(Blue) = \(\frac {7}{10}\) = 0.7
while P(Red) = \(\frac {3}{10}\) = 0.3
These probabilities are not equal.

(v) Approximately has equally likely outcomes.
Explanation: Biologically, the probability of a baby being a boy or a girl is approximately equal, though slightly more boys are born.
We typically assume P(Boy) ≈ P(Girl) ≈ 0.5 for practical purposes.

Question 4.
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?
Solution:
(i) Here, Sample space:
S = {HH, HT, TH, TT}
∴ n(S) = 4
Event “At least one head” = {HH, HT, TH}
∴ Number of favourable outcomes n(A) = 3
n(A) = 3
P(A) = \(\frac{n(A)}{n(S)}=\frac{3}{4}\)

(ii) Here, Sample space:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
n(S) = 10
Even numbers = {2, 4, 6, 8, 10}
Number of favourable outcomes
n(A) = 5
P(B) = \(\frac{5}{10}=\frac{1}{2}\)

(iii) Here, Sample space:
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Numbers greater than 4 = {5, 6}
Number of favourable outcomes
n(A) = 2
∴ P(C) = \(\frac{2}{6}=\frac{1}{3}\)

(iv) Here, Sample space:
S = {Red, Blue, Green}
Total balls n(S) = 3 + 2 + 1 = 6
Balls that are not red = {Blue, Green}
Number of non-red balls
n(A) = 2 + 1 = 3
P(D) = \(\frac{3}{6}=\frac{1}{2}\)

(v) Here, Sample space:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ n(S) = 8
Outcomes with exactly two heads = {HHT, HTH, THH}
Number of favourable outcomes
n(A) = 3
∴ P(E) = \(\frac {3}{8}\)

Question 5.
A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
Solution:
The bag contains 3 distinct types of candy: strawberry, lemon, and mint.
Since one is picked at random, there are 3 possible outcomes in total.
The question asks specifically for the probability of picking a strawberry candy.
Since there is only 1 strawberry candy in the bag, there is 1 favorable outcome.
∴ P(strawberry) = \(\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac {1}{3}\)
= 0.333 or 33.3%

Question 6.
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:

Shirt	Pants	Outfit Combination
Red	Jeans	Red shirt with Jeans
Red	Khakis	Red shirt with Khakis
Red	Shorts	Red shirt with Shorts
Blue	Jeans	Blue shirt with Jeans
Blue	Khakis	Blue shirt with Khakis
Blue	Shorts	Blue shirt with Shorts

Total number of possible outfit combinations = 2 × 3 = 6

Question 7.
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 number of cases = 20 + 210 + 325 + 445 = 1000
(i) Here, P(Less than 4000) = \(\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac {20}{1000}\)
= 0.02

(ii) Here, Cases between 4000 and 14000 km = 210 + 325 = 535
P(Between 4000 and 14000) = \(\frac {535}{1000}\) = 0.535

(iii) Here, P(More than 14000) = \(\frac {445}{1000}\) = 0.445

Question 8.
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:
The word ‘PEACE’ has 5 letters: P, E, A, C, E
Sample space: S = {P, E, A, C, E} with n(S) = 5
(i) Probability of drawing P, E, or C: To find this, we count how many times these specific letters appear in the word:
P: 1 time
E: 2 times
C: 1 time
Number of favourable outcomes = 1 + 2 + 1 = 4
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of cards }}\)
= \(\frac {4}{5}\)
= 0.8

(ii) Probability of NOT drawing an E:
(Counting non-E letters): The letters that are not E are P, A, and C (3 letters).
Number of favourable outcomes = 3
Probability = \(\frac {3}{5}\) = 0.6

Question 9.
A game of chance consists of spinning an arrow (see figure) 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:
(i) 8; There is only one ‘8’.
Probability = \(\frac {1}{8}\)

(ii) The odd numbers are 1, 3, 5, and 7 (4 total).
Probability = \(\frac{4}{8}=\frac{1}{2}\)

(iii) A number greater than 2: These are 3, 4, 5, 6, 7, and 8 (6 total).
Probability = \(\frac{6}{8}=\frac{3}{4}\)

(iv) A number less than 9: All the numbers (1 through 8) are less than 9 (8 total).
Probability = \(\frac {8}{8}\) = 1

(v) A multiple of 3: These are 3 and 6 (2 total).
Probability = \(\frac{2}{8}=\frac{1}{4}\)

Question 10.
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 initially = 4 Red + 5 Blue = 9 balls
Tree Diagram:

(i) First draw: P(Red) = \(\frac {4}{9}\)
After removing a Red ball, 8 balls remain: 3 Red and 5 Blue
Second draw: P(Blue | Red first) = \(\frac {5}{8}\)
P(Red then Blue) = \(\frac{4}{9} \times \frac{5}{8}\)
= \(\frac {20}{72}\)
= \(\frac {5}{18}\)
= 0.278 or 27.8%

(ii) First draw: P(Blue) = \(\frac {5}{9}\)
After removing a Blue ball, 8 balls remain: 4 Red and 4 Blue
Second draw: P(Blue | Blued first) = \(\frac {4}{8}\)
P(Blue then Blue) = \(\frac{5}{9} \times \frac{4}{8}\)
= \(\frac{5}{9} \times \frac{1}{2}\)
= \(\frac {5}{18}\)
= 0.278 or 27.8%

Question 11.
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 event): The event of getting a sum of 1.
Since the minimum value on each die is 1, the minimum possible sum is 1 + 1 = 2.
A sum of 1 can never occur.
P(sum = 1) = 0
Event with probability 1 (certain event): The event of getting a sum between 2 and 12 (inclusive).
Since each die shows a value from 1 to 6, the sum of two dice is always between 2 and 12.
This is guaranteed to happen on every throw.
P(2 ≤ sum ≤ 12) = 1

Question 12.
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?
Solution:
(i) Two dice are rolled.
Sample space (S) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Sum is a prime number greater than 5.
Possible sums: 7 and 1
Favourable outcomes = {(2, 5), (1, 6), (3, 4), (4, 4), (6, 1), (5, 2), (5, 6), (6, 5)}
Number of favourable outcomes = 8
Total number of outcomes = 36
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac {8}{36}\)
= \(\frac {2}{9}\)
= 0.222

(ii) 4 Red, 3 Green, 2 Blue (without replacement): different colours
Since balls are distinct, we label them R₁, R₂, R₃, R₄, G₁, G₂, G₃, B₁, B₂.
Sample space (S): Set of all combinations of 2 balls from the 9 available
{(R₁, R₂), (R₁, R₃), (R₁, R₄), (R₁, G₁), (R₁, G₂), (R₁, G₃), (R₁, B₁), (R₁, B₂), (R₂, R₃), (R₂, R₄), (R₂, G₁), (G₂, G₃), (G₂, B₁), (G₂, B₂), (G₃, B₁), (G₃, B₂), (B₁, B₂)}
Total outcomes = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36
P(same colour) = P(R, R) + P(G, G) + P(B, B)
= \(\left(\frac{4}{9} \times \frac{3}{8}\right)+\left(\frac{3}{9} \times \frac{2}{8}\right)+\left(\frac{2}{9} \times \frac{1}{8}\right)\)
= \(\frac {20}{72}\)
= \(\frac {5}{18}\)
P(different colours) = 1 – P(same colour)
= 1 – \(\frac {5}{18}\)
= \(\frac {13}{18}\)
= 0.722

(iii) Three coins are tossed.
Sample space (S) = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Event: First is Heads and exactly two heads
Favourable outcomes = {HHT, HTH}
No. of favourable outcomes = 2
Total no. of outcomes = 8
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac {2}{8}\)
= \(\frac {1}{4}\)
= 0.25

(iv) Four-digit numbers are formed using 1, 2, 3, 4 without repetition.
Event: Number is even
Sample space (S) = {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4213, 4231, 4312, 4321}
Total number of outcomes = 24
An even number must end in 2 or 4.
Favourable outcomes = 12
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
= \(\frac {12}{24}\)
= \(\frac {1}{2}\)
= 0.5

(v) Data: 3 questions, 4 options (1 correct (C), 3 wrong (W)).
For 3 questions, Sample space (S) = {CCC, CCW, CWC, WCC, WCW, WWC, WWW}
For each question, P(C) = \(\frac {1}{4}\), P(W) = \(\frac {3}{4}\)
Favourable (outcomes) for exactly 2 correct answers = {CCW, CWC, WCC}
Calculation for one path (for example, CCW):
\(\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4}=\frac{3}{64}\)
Total Probability = (sum of all 3 paths):
P(E) = 3 × \(\frac {3}{64}\)
= \(\frac {9}{64}\)
= 0.141 or 14.1%

Question 13.
A box contains 4 balls numbered l 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?
Solution:
(i) With Replacement: A ball is drawn, recorded, and returned before the second draw.
Since the ball is returned, the options for the second draw are the same as the first draw: {1, 2, 3, 4}.

Sample Space (S₁):
S₁ = {(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)}

(ii) Without Replacement: A ball is drawn and recorded. The second ball is drawn without replacing the first.
If we pick ball 1 first, it is no longer in the box.
Therefore, the second draw can only be {2, 3, 4}.
So, we cannot have outcomes like (1, 1) or (2, 1).

Sample Space (S₂):
S₂ = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
(iii) Sizes of the Sample Spaces
The size of a sample space is denoted by n(S).
1. Size of S₁ (With Replacement)
Calculation:
4 (first pick) × 4 (second pick) = 16
∴ n(S₁) = 16
2. Size of S₂ (Without Replacement)
Calculation:
4 (first pick) × 3 (remaining options) = 12
∴ n(S₂) = 12

Question 14.
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.
The coin gives outcomes {H, T} and the card gives outcomes {1, 2, 3, 4, 5, 6}.
Solution:
Sample Space:

Coin	Card	Outcome
H	1	(H, 1)
H	2	(H, 2)
H	3	(H, 3)
H	4	(H, 4)
H	5	(H, 5)
H	6	(H, 6)
T	1	(T, 1)
T	2	(T, 2)
T	3	(T, 3)
T	4	(T, 4)
T	5	(T, 5)
T	6	(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)}
∴ n(S) = 12

Question 15.
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:
When three coins are tossed, the possible number of heads is 0, 1, 2, or 3 – no more, no less.
A valid sample space must list all possible outcomes and only possible outcomes.
(i) {1, 2, 3} – NOT a valid sample space
This list is missing 0 (getting no heads, i.e., all tails – TTT – is a valid outcome).
Since 0 heads is possible and excluded, this is not a complete sample space.

(ii) {0, 1, 2} – NOT a valid sample space
This list is missing 3 (getting all three heads – HHH – is a valid outcome).
Since 3 heads is possible and excluded, this is not a complete sample space.

(iii) {0, 1, 2, 3, 4} – NOT a valid sample space
This list includes 4, which is impossible when only three coins are tossed (maximum heads = 3).
A sample space must contain only possible outcomes.

(iv) {0, 1, 2, 3} – VALID sample space
This list includes all possible values (0, 1, 2, 3 heads).
Every element corresponds to a genuinely possible outcome, and no possible outcome is omitted.

Question 16.
Suppose you drop a dye at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with a diameter of 1 m?

Solution:
The rectangular region has a length of 3 m and a breadth of 2 m.
Area of rectangle = length × breadth
= 3 m × 2 m
= 6 m²
The circle has a diameter of 1 m, meaning its radius is 0.5 m or \(\frac {1}{2}\) m.
Area of circle = πr²
= \(\pi \times\left(\frac{1}{2}\right)^2\)
= \(\frac{\pi}{4} \mathrm{~m}^2\)
The probability is the ratio of the favourable area (the circle) to the total area (The rectangle).
Probability = \(\frac{\text { Area of circle }}{\text { Area of rectangle }}\)
= \(\frac{\frac{\pi}{4}}{6}\)
= \(\frac{\pi}{4 \times 6}\)
= \(\frac{\pi}{24}\)