Chapter 7: The Mathematics of Maybe: Introduction to ProbabilityPage 169 – 172All 16 Questions on This Page

Class 9 Maths Chapter 7 End-of-Chapter Exercises Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 7 End-of-Chapter Exercises (Questions 1–16, Page 169–172). Covers sample spaces, tree diagrams with and without replacement, ordered pairs, coin-card combinations, number of heads, and geometric area probability.

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Key Concepts for End-of-Chapter Exercises

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Sample Space

The set of all possible outcomes of a random experiment.

Tree Diagram

A branching diagram that systematically lists all possible outcomes of a multi-step experiment in sequence.

With Replacement

When an object is returned before the next draw, all choices remain available, so total outcomes equal n × n.

Without Replacement

When an object is not returned, the number of choices decreases by 1 for the next draw, making repeated outcomes impossible.

Ordered Pair

A pair (x, y) recording outcomes in sequence, where the first element is the first draw and the second is the second draw.

Geometric Probability

Probability determined by comparing geometric measures (such as areas or lengths) rather than counting discrete outcomes.

P(E) = Area of Favourable Region / Area of Total Region

Area of a Region

Standard geometric formulas: Area of rectangle = Length × Breadth (l × b), and Area of circle = πr².

A_{rectangle} = l × b, A_{circle} = πr²

Favourable Region and Total Region

For a point chosen randomly within a total region, the probability that it falls in a specific favourable target is the ratio of their areas.

P(E) = Area of Target / Total Area

Question 1

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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 ______:

An impossible event can never occur.

Therefore,

(ii) The set of all possible outcomes of a random experiment is called the ______:

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

Therefore,

sample space

(iii) The probability of an event that is certain to happen is ______:

A certain event always occurs.

Therefore,

(iv) Tossing a fair coin has a probability of ______ for getting heads:

A fair coin has two equally likely outcomes:

H, T

Only one of them is heads.

Therefore,

P(Heads) = 12

Hence,

Answer:(i) 0 | (ii) sample space | (iii) 1 | (iv) 12

Question 2

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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:

Number of students who like football:

We know,

Relative Frequency = Number of times the event occurred / Total number of observations

Therefore,

Relative Frequency = 1550

= 310

= 0.3

Hence,

3/10 or 0.3

is the required frequency/relative frequency.

Answer:310 or 0.3

Question 3

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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) A driver attempts to start a car. The car starts or does not start:

The two outcomes are:

Starts, Does not start

These outcomes are not necessarily equally likely because the chance of the car starting may be different from the chance of it not starting.

Therefore,

the outcomes are not necessarily equally likely.

(ii) Tossing a fair coin once:

The possible outcomes are:

H, T

Since the coin is fair, heads and tails are equally likely.

Therefore,

the outcomes are equally likely.

(iii) Rolling a fair 6-sided die:

The possible outcomes are:

1, 2, 3, 4, 5, 6

Since the die is fair, each number has the same chance of occurring.

Therefore,

the outcomes are equally likely.

(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles:

The possible colour outcomes are:

Red, Blue

But there are 3 red marbles and 7 blue marbles, so the two colours do not have the same chance of being selected.

Therefore,

the outcomes are not equally likely.

(v) A baby is born. It is a boy or a girl:

The two possible outcomes are:

Boy, Girl

In this simple textbook model, these are treated as equally likely outcomes.

Therefore,

the outcomes are equally likely.

Hence,

the experiments with equally likely outcomes are:

(ii), (iii) and (v)

Answer:(ii), (iii) and (v)

Question 4

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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) Two coins are tossed at the same time (Probability of getting at least one head):

The sample space is:

S = {HH, HT, TH, TT}

There are 4 possible outcomes.

Let E be the event of getting at least one head.

The favourable outcomes are:

E = {HH, HT, TH}

There are 3 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(at least one head) = 34

Hence,

is the probability of getting at least one head.

(ii) Ten cards numbered 1 to 10 in a box (Probability of drawing an even number):

The sample space is:

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

There are 10 possible outcomes.

Let E be the event of drawing an even-numbered card.

The favourable outcomes are:

E = {2, 4, 6, 8, 10}

There are 5 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(E) = 510

= 12

Hence,

is the probability of drawing a card with an even number.

(iii) A die is rolled once (Probability of getting a number greater than 4):

The sample space is:

S = {1, 2, 3, 4, 5, 6}

There are 6 possible outcomes.

Let E be the event of getting a number greater than 4.

The favourable outcomes are:

E = {5, 6}

There are 2 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(E) = 26

= 13

Hence,

is the probability of getting a number greater than 4.

(iv) Bag of 3 red, 2 blue, and 1 green ball (Probability that the ball is not red):

Let the individual balls be:

R₁, R₂, R₃ (Red)

B₁, B₂ (Blue), G₁ (Green)

The sample space is:

S = {R₁, R₂, R₃, B₁, B₂, G₁}

There are 6 possible outcomes.

Let E be the event that the ball picked is not red.

The favourable outcomes are:

E = {B₁, B₂, G₁}

There are 3 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(E) = 36

= 12

Hence,

is the probability that the ball picked is not red.

(v) Three coins tossed simultaneously (Probability of getting exactly two heads):

The sample space is:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

There are 8 possible outcomes.

Let E be the event of getting exactly two heads.

The favourable outcomes are:

E = {HHT, HTH, THH}

There are 3 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(exactly two heads) = 38

Hence,

is the probability of getting exactly two heads.

Answer:(i) P(at least one head) = 34 | (ii) P(even number) = 12 | (iii) P(number > 4) = 13 | (iv) P(not red) = 12 | (v) P(exactly two heads) = 38

Question 5

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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 possible outcomes are:

S = {Strawberry, Lemon, Mint}

There are 3 possible outcomes.

The favourable outcome is:

Strawberry

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

Therefore,

P(Strawberry) = 13

Hence,

is the probability of picking a strawberry candy.

Answer:13

Question 6

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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

The shirts are:

Red, Blue

The types of pants are:

Jeans, Khakis, Shorts

Each shirt can be combined with each type of pants.

Therefore, the possible combinations are:

Shirt	Pants
Red	Jeans
Red	Khakis
Red	Shorts
Blue	Jeans
Blue	Khakis
Blue	Shorts

Hence, there are:

possible outfit combinations.

Answer:6 possible outfit combinations (see table above)

Question 7

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A tyre company records distances before replacement in 1000 cases. Table: Distance (km) | Number of cases Less than 4000 | 20 4001 to 9000 | 210 9001 to 14000 | 325 More than 14000 | 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.

Solution

Question data:

Distance (km)	Number of cases
Less than 4000	20
4001 to 9000	210
9001 to 14000	325
More than 14000	445
Total	1000

Total number of cases:

1000

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

(i) Less than 4000 km:

Number of tyres lasting less than 4000 km:

Therefore,

P(less than 4000 km) = 201000

= 150

Hence,

150

is the probability that a randomly chosen tyre lasts less than 4000 km.

(ii) Between 4000 and 14000 km:

The relevant categories are:

4001 to 9000 km and 9001 to 14000 km

Number of tyres:

210 + 325 = 535

Therefore,

P(between 4000 and 14000 km) = 5351000

= 107200

Hence,

107200

is the probability that a randomly chosen tyre lasts between 4000 and 14000 km.

(iii) More than 14000 km:

Number of tyres lasting more than 14000 km:

445

Therefore,

P(more than 14000 km) = 4451000

= 89200

Hence,

89200

is the probability that a randomly chosen tyre lasts more than 14000 km.

Answer:(i) P(less than 4000 km) = 150 | (ii) P(between 4000 and 14000 km) = 107200 | (iii) P(more than 14000 km) = 89200

Question 8

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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

There are 5 cards in total.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

(i) Probability that it is a P, E or C:

The favourable cards are:

P, E, C, E

There are 4 favourable cards because E appears twice.

Therefore,

P(P, E or C) = 45

Hence,

is the probability that the drawn card is a P, E or C.

(ii) Probability that it is not an E:

The cards that are not E are:

P, A, C

There are 3 favourable cards.

Therefore,

P(not E) = 35

Hence,

is the probability that the drawn card is not an E.

Answer:(i) P(P, E or C) = 45 | (ii) P(not E) = 35

Question 9

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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?

Game of Chance SpinnerFig. 7.7

Fig. 7.7: An 8-sector game of chance spinner with equally likely outcomes 1, 2, 3, 4, 5, 6, 7, 8.

Solution

The spinner can point to any of the numbers:

1, 2, 3, 4, 5, 6, 7, 8

Since all 8 outcomes are equally likely,

Total number of possible outcomes = 8

We use,

P(E) = Number of favourable outcomes / Number of possible outcomes

(i) Probability of pointing at 8:

The favourable outcome is:

Therefore,

P(8) = 18

Hence,

is the probability of pointing at 8.

(ii) Probability of pointing at an odd number:

The odd numbers are:

1, 3, 5, 7

So, there are 4 favourable outcomes.

Therefore,

P(odd number) = 48

= 12

Hence,

is the probability of pointing at an odd number.

(iii) Probability of pointing at a number greater than 2:

The numbers greater than 2 are:

3, 4, 5, 6, 7, 8

So, there are 6 favourable outcomes.

Therefore,

P(number greater than 2) = 68

= 34

Hence,

is the probability of pointing at a number greater than 2.

(iv) Probability of pointing at a number less than 9:

All the possible numbers are less than 9.

So, all 8 outcomes are favourable.

Therefore,

P(number less than 9) = 88

= 1

Hence,

is the probability of pointing at a number less than 9.

(v) Probability of pointing at a multiple of 3:

The multiples of 3 are:

3, 6

So, there are 2 favourable outcomes.

Therefore,

P(multiple of 3) = 28

= 14

Hence,

is the probability of pointing at a multiple of 3.

Answer:(i) P(8) = 18 | (ii) P(odd number) = 12 | (iii) P(number > 2) = 34 | (iv) P(number < 9) = 1 | (v) P(multiple of 3) = 14

Question 10

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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

The basket contains:

4 red balls and 5 blue balls

So, the total number of balls is:

4 + 5 = 9

Since the first ball is laid aside, it is not replaced before the second draw. Therefore, the probabilities for the second draw change.

Tree diagram

The first draw can be:

P(R) = 49, P(B) = 59

If the first ball is red, 3 red and 5 blue balls remain:

P(R) = 38, P(B) = 58

If the first ball is blue, 4 red and 4 blue balls remain:

P(R) = 48, P(B) = 48

Use the tree diagram to follow each required path.

For a complete path, multiply the probabilities along the branches:

P(Path) = P(Branch 1) × P(Branch 2)

(i) Probability of drawing a red ball and then a blue ball

From the tree diagram, the required path is:

R → B

Therefore,

P(R then B)

= 49 × 58

= 2072

= 518

Hence,

518

is the probability of drawing a red ball and then a blue ball.

(ii) Probability of drawing 2 blue balls

From the tree diagram, the required path is:

B → B

Therefore,

P(B then B)

= 59 × 48

= 2072

= 518

Hence,

518

is the probability of drawing 2 blue balls.

Answer:(i) P(R then B) = 518 | (ii) P(B then B) = 518

Question 11

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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

When two 6-sided dice are thrown, the smallest possible sum is:

1 + 1 = 2

and the largest possible sum is:

6 + 6 = 12

Event with probability 0

Getting a sum of 13 is impossible because the maximum possible sum is 12.

Therefore,

P(sum of 13) = 0

Hence, an event with probability 0 is:

Getting a sum of 13.

Event with probability 1

The sum of the two dice must always lie between 2 and 12.

Therefore,

P(2 ≤ sum ≤ 12) = 1

Hence, an event with probability 1 is:

Getting a sum between 2 and 12.

Answer:Probability 0: Getting a sum of 13 | Probability 1: Getting a sum between 2 and 12

Question 12

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Find the probability in each of the following cases: (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, 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 (Probability that the sum is a prime number greater than 5):

The sample space for rolling two 6-sided dice consists of all 36 ordered pairs:

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)}

Total number of possible outcomes:

The sum of the numbers on the two dice ranges from 2 to 12.

The prime numbers greater than 5 in this range are:

7 and 11

The favourable outcomes are:

For sum = 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

For sum = 11:

(5, 6), (6, 5)

So, the favourable outcomes are:

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}

There are 6 + 2 = 8 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(sum is a prime number > 5) = 8/36

= 29

Hence,

is the probability that the sum is a prime number greater than 5.

(ii) Two balls are drawn without replacement (Probability that both are of different colours):

The bag contains:

4 red, 3 green, and 2 blue balls

Total number of balls:

4 + 3 + 2 = 9

Since two balls are drawn without replacement, there are 8 balls remaining for the second draw.

The pairs of different colours are:

RG, RB, GR, GB, BR, BG

Using the path-probability rule:

P(Path) = P(Branch 1) × P(Branch 2)

For RG,

P(RG) = 49 × 38

= 1272

For RB,

P(RB) = 49 × 28

= 872

For GR,

P(GR) = 39 × 48

= 1272

For GB,

P(GB) = 39 × 28

= 672

For BR,

P(BR) = 29 × 48

= 872

For BG,

P(BG) = 29 × 38

= 672

Therefore,

P(different colours) = (12 + 8 + 12 + 6 + 8 + 6) / 72

= 5272

= 1318

Hence,

1318

is the probability that both balls are of different colours.

(iii) Three coins are tossed (Probability that the first coin shows heads and exactly two heads occur in total):

The sample space is:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

There are 8 possible outcomes.

We need the first coin to show heads and exactly two heads in total.

The favourable outcomes are:

E = {HHT, HTH}

So, there are 2 favourable outcomes.

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(first coin heads and exactly two heads) = 2/8

= 14

Hence,

is the probability that the first coin shows heads and exactly two heads occur in total.

(iv) Four-digit number without repetition (Probability that the number is even):

The digits are:

1, 2, 3, 4

Since there is no repetition, the sample space consists of all four-digit numbers formed by arranging these four digits.

The complete sample space is:

S = {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}

Thus, the total number of possible outcomes is:

4 × 3 × 2 × 1 = 24

A four-digit number is even if its last digit is even.

The possible even last digits are:

2, 4

When the last digit is 2:

The remaining digits 1, 3, 4 can be arranged in:

3 × 2 × 1 = 6

ways.

The corresponding even numbers are:

1342, 1432, 3142, 3412, 4132, 4312

When the last digit is 4:

The remaining digits 1, 2, 3 can be arranged in:

3 × 2 × 1 = 6

ways.

The corresponding even numbers are:

1234, 1324, 2134, 2314, 3124, 3214

Therefore, the total number of favourable outcomes is:

6 + 6 = 12

Using,

P(E) = Number of favourable outcomes / Number of possible outcomes

we get,

P(even number) = 1224

= 12

Hence,

is the probability that the four-digit number is even.

(v) Multiple-choice test (Probability that the student guesses and gets exactly 2 answers correct):

Each question has 4 options, with only 1 correct answer.

Therefore,

P(Correct) = 14

and

P(Wrong) = 34

We need exactly 2 correct answers.

The possible arrangements are:

CCW, CWC, WCC

For one arrangement, for example CCW:

P(CCW) = 14 × 14 × 34

= 364

Similarly,

P(CWC) = 364

and

P(WCC) = 364

Therefore,

P(exactly 2 correct) = 3/64 + 3/64 + 3/64

= 964

Hence,

964

is the probability that the student gets exactly 2 answers correct.

Answer:(i) P(sum is prime > 5) = 29 | (ii) P(different colours) = 1318 | (iii) P(first H and exactly two H) = 14 | (iv) P(even number) = 12 | (v) P(exactly 2 correct) = 964

Question 13

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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?

Solution

(i) Ball drawn, recorded, and returned (with replacement) — Tree diagram and sample space:

The balls are numbered:

1, 2, 3, 4

The first draw has 4 possible outcomes.

Since the first ball is returned, all 4 balls are available again for the second draw.

Therefore, the complete sample space 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)}

Number of possible outcomes:

4 × 4 = 16

Therefore, the size of this sample space is:

Tree diagram for drawing 2 balls numbered 1 to 4 with replacement

(ii) Ball drawn and recorded without replacement — Tree diagram and sample space:

After the first ball is drawn, it is not replaced.

Therefore, only 3 balls remain for the second draw.

The complete sample space 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)}

Repeated pairs such as (1, 1), (2, 2), (3, 3), (4, 4) are impossible because the first ball is not replaced.

Number of possible outcomes:

4 × 3 = 12

Therefore, the size of this sample space is:

Tree diagram for drawing 2 balls numbered 1 to 4 without replacement

(iii) Sizes of these two sample spaces:

With replacement:

4 choices for first draw × 4 choices for second draw

= 4 × 4 = 16

Without replacement:

4 choices for first draw × 3 choices for second draw

= 4 × 3 = 12

Therefore:

With replacement → 16 outcomes

Without replacement → 12 outcomes

Answer:(i) S has 16 outcomes (see tree diagram above) | (ii) S has 12 outcomes (see tree diagram above) | (iii) With replacement = 16, Without replacement = 12

Question 14

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A coin is tossed and a card numbered 1 to 6 is drawn. Record a sample space for the experiment.

Solution

The coin has two possible outcomes:

H, T

The card has six possible outcomes:

1, 2, 3, 4, 5, 6

Therefore, the complete sample space is:

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)}

There are:

2 × 6 = 12

possible outcomes.

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)} (12 possible outcomes)

Question 15

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Three coins are tossed and the number of heads is recorded. Which of the following could be the sample space? (i) {1, 2, 3} (ii) {0, 1, 2} (iii) {0, 1, 2, 3, 4} (iv) {0, 1, 2, 3}

Solution

When 3 coins are tossed, the number of heads can be:

0, 1, 2 or 3

It cannot be 4 because only 3 coins are tossed.

Therefore, the sample space must contain all possible values from 0 to 3, and no impossible values.

Hence:

(iv) {0, 1, 2, 3}

is the correct sample space.

Answer:(iv) {0, 1, 2, 3}

Question 16

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Use the given figure to determine the probability of the event that a point chosen at random from the rectangular region falls inside the circle.

Rectangle with Inscribed CircleFig. 7.8

A 3 m by 2 m rectangular region containing a circular region with a diameter of 1 m

Fig. 7.8: Rectangular region (3 m × 2 m) containing a circular region (diameter = 1 m).

Solution

Area of rectangle:

= 3 × 2

= 6 m²

Radius of circle:

r = 12 m

Area of circle:

= πr²

= π(12)²

= π/4 m²

Therefore:

P(E) = Number of favourable outcomes / Number of possible outcomes

For this geometric probability:

P(E) = Area of circle / Area of rectangle

= π/46

= π/24

Hence, the required probability is:

π/24

Answer:P(falling inside circle) = π/24

Need to practice other sections in Chapter 7?

Return to Chapter 7 (The Mathematics of Maybe: Introduction to Probability)→

💡Key Concepts for End-of-Chapter Exercises

Sample Space

Tree Diagram

With Replacement

Without Replacement

Ordered Pair

Geometric Probability

Area of a Region

Favourable Region and Total Region

Key Concepts for End-of-Chapter Exercises