GANITA MANJARI | GRADE 9 | PART I

The Mathematics of Maybe

Introduction to Probability

Complete Page-wise Questions, Solutions and Answers

Chapter 7 | Textbook Pages 155-173

Important: Activity-based answers depend on the student's observations. Where no fixed numerical result exists, the correct method and a model form of the answer are provided.

Contents at a Glance

1. In-text Questions and Think and Reflect - Pages 155-169

2. Exercise Set 7.1 - Page 159

3. Exercise Set 7.2 - Pages 165-166

4. Exercise Set 7.3 - Pages 167-168

5. Exercise Set 7.4 - Page 169

6. End-of-Chapter Exercises - Pages 169-173

In-text Questions and Think and Reflect

Page 155 - Introduction

Question 1. Is it going to rain today?

Answer - Rain is a random event. Its exact occurrence cannot be predicted with complete certainty because it depends on temperature, humidity, wind and atmospheric pressure. Weather data can only estimate its probability.

Question 2. Will our school win the inter-school hockey match tomorrow?

Answer - The possible outcomes are win, lose or draw. The exact result cannot be known beforehand. Its probability may be estimated from factors such as the teams' past performances and the players' fitness.

Question 3. Will I be chosen in the monthly lucky draw to perform at the school assembly?

Answer - If there are N students, each represented by one identical slip, every student has the same chance of selection.

P(being selected) = 1/N

Question 4. Can we predict these outcomes with 100% certainty?

Answer - No. These are random events involving uncertainty. We may estimate how likely their outcomes are, but we cannot predict the exact outcome with complete certainty.

Page 156 - Think and Reflect

Question. In a cricket match, tossing a coin to decide which team will bat first is considered a fair method. Can you explain why?

Answer - A fair coin has two equally likely outcomes - Heads and Tails. Each team is assigned one outcome, so both have an equal probability of 1/2 of winning the toss. Neither team receives an unfair advantage.

Page 157 - In-text Questions

Question 1. What makes an event like rain random?

Answer - Rain depends on several complex and changing atmospheric factors, including temperature, humidity, pressure and wind patterns. Weather data can estimate the probability of rain, but its exact time and location cannot always be predicted with complete certainty.

Question 2. Ask your friend to predict the outcome of a Rs 1 coin you toss. Can your friend know the result for certain?

Answer - No. The friend can guess Heads or Tails, but cannot know the exact result beforehand. Both possible outcomes are known, but the result of an individual toss is unpredictable. This illustrates randomness.

Page 163 - Think and Reflect

Question. If a 4 has appeared on a die eight times in succession, what is the probability of rolling a 4 again?

Answer - Assuming that the die is fair, each roll is independent. The previous eight rolls do not affect the next roll.

P(rolling a 4 again) = 1/6 ≈ 0.167

Page 167 - Think and Reflect

Question. Why must a sample space have the correct level of detail for the problem being studied?

Answer - A sample space must include every outcome relevant to the question. If we only study whether it rains, {Rain, No Rain} is sufficient. If rainfall intensity matters, the sample space should also include drizzle, light rain and heavy rain.

Page 168 - In-text Question

Question. How many outcomes are possible when a fair coin is tossed twice?

S = {HH, HT, TH, TT}

Answer - There are 4 possible outcomes.

Page 169 - Think and Reflect

Question. Calculate the probability of getting one Head and one Tail when a fair coin is tossed twice.

S = {HH, HT, TH, TT}; E = {HT, TH}

P(E) = 2/4 = 1/2

Answer - The probability of getting one Head and one Tail is 1/2.

Exercise Set 7.1

Page 159

Question 1. 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 for each ranking.

(i) The next Monday will come after Sunday.

Answer - P = 1. Label: Certain. Monday always comes after Sunday in the weekly calendar.

(ii) It will snow in Mumbai in July.

Answer - P ≈ 0. Label: Impossible or nearly impossible under normal conditions. Mumbai has a warm, tropical climate, and July is part of its monsoon season.

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

Answer - P is very close to 0. Label: Less likely. Although not logically impossible everywhere, it is extremely unlikely in an ordinary classroom.

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

Answer - P is close to 1. Label: More likely. A student normally meets and greets friends at school, though absence means it is not absolutely certain.

Note: The values for subjective events may vary slightly. The order from least to most likely should be: snowfall in Mumbai, an elephant in the classroom, greeting a friend, and Monday following Sunday.

Exercise Set 7.2

Pages 165-166

Question 1. A teacher randomly selects a sample of 30 sweets: 10 red, 8 green, 7 yellow and 5 blue.

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

P(green) = 8/30 = 4/15 ≈ 0.267 = 26.7%

Answer - The probability is 4/15, approximately 0.267 or 26.7%.

(ii) If there are 600 sweets in the large bag, estimate how many are likely to be yellow.

Estimated yellow sweets = (7/30) x 600 = 140

Answer - Approximately 140 sweets are likely to be yellow.

Question 2. A random sample of 40 students gives these favourite clubs: Science 14, Arts 11, Sports 9 and Debate 6. Assume the school has 800 students.

(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?

P(Arts Club) = 11/40 = 0.275 = 27.5%

Answer - The probability is 0.275 or 27.5%.

(ii) Estimate how many students in the whole school are likely to prefer the Sports Club.

Estimated students = (9/40) x 800 = 180

Answer - Approximately 180 students are likely to prefer the Sports Club.

Question 3. Toss a coin 20 times and record each result as 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 Heads. (iv) If you toss once more, what is the probability of Tails?

Answer - This is activity-based. Let Heads appear h times; then Tails appears 20 - h times.

Experimental P(H) = h/20

For example, if Heads appears 11 times, Tails appears 9 times and experimental P(H) = 11/20 = 0.55.

For the next toss of a fair coin: P(T) = 1/2

Previous tosses do not change the theoretical probability of the next toss.

Question 4. Toss a paper cup 100 times. Record whether it lands on its bottom, top or side, and assign experimental probabilities.

Answer - This is activity-based. Let the counts for bottom, top and side be b, t and s respectively. Then b + t + s = 100.

P(bottom) = b/100; P(top) = t/100; P(side) = s/100

The three experimental probabilities must add up to 1.

Question 5. What is the probability of getting an even number when rolling a fair six-sided die?

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

P(E) = 3/6 = 1/2

Answer - The probability is 1/2.

Question 6. A six-sided die is rolled 12 times and a 3 appears three times.

(i) What is the experimental probability of rolling a 3?

Experimental P(3) = 3/12 = 1/4 = 0.25

(ii) What is the theoretical probability of rolling a 3?

Theoretical P(3) = 1/6 ≈ 0.167

(iii) Why might these probabilities differ? What would you expect for 60, 600 or 6,000 rolls?

Answer - Twelve rolls form a small sample, so random variation can be noticeable. With more rolls, the experimental probability should generally approach 1/6.

Note: These are expected values, not guaranteed exact results.

Exercise Set 7.3

Pages 167-168

Question 1. When a single six-sided die is rolled, what is the total number of possible outcomes in the sample space?

S = {1, 2, 3, 4, 5, 6}; n(S) = 6

Answer - There are 6 possible outcomes.

Question 2. For the following experiments, write down the sample space S.

(i) Rolling a die and tossing a coin together.

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

Answer - The sample size is n(S) = 12.

(ii) Choosing a random integer between -5 and +5.

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

Answer - Taking both endpoints as included, n(S) = 11.

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

If only colour is recorded: S = {Green, Red}

The colour outcomes are not equally likely: P(Green) = 5/12 and P(Red) = 7/12.

Question 3. At a village fair, the snacks are Samosa, Pakora and Bhaji, while the drinks are Chai and Lassi.

(i) List the sample space of all possible snack and drink combinations.

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

Answer - There are 3 x 2 = 6 combinations.

(ii) List the event 'Selecting Samosa as a snack.'

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

Exercise Set 7.4

Page 169

Question 1. Basket A has one apple and two oranges. Basket B has one banana and one mango. One fruit is randomly picked from each basket.

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

Answer - Written branch structure: Basket A gives Apple with probability 1/3 or Orange with probability 2/3. From either branch, Basket B gives Banana with probability 1/2 or Mango with probability 1/2.

(ii) List the sample space.

S = {(Apple,Banana),(Apple,Mango),(Orange,Banana),(Orange,Mango)}

Note: These fruit-type outcomes are not equally likely because Basket A has two oranges but only one apple.

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

P(Apple and Banana) = 1/3 x 1/2 = 1/6

Answer - The probability is 1/6.

Question 2. A box contains 3 red pens, 4 black pens and 2 green pens. You pick a pen and replace it; then your friend does the same.

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

Let R = red, K = black and G = green. S = {RR,RK,RG,KR,KK,KG,GR,GK,GG}

At each selection: P(R) = 3/9, P(K) = 4/9 and P(G) = 2/9. Replacement makes the second selection independent with the same probabilities.

(ii) Find the probability that both people pick pens of the same colour.

P(same) = (3/9)^2 + (4/9)^2 + (2/9)^2 = (9+16+4)/81 = 29/81

Answer - The probability is 29/81.

End-of-Chapter Exercises

Pages 169-170

Question 1. Fill in the blanks.

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

Answer - 0

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

Answer - sample space

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

Answer - 1

(iv) Tossing a fair coin has a probability of ______ for getting Heads.

Answer - 1/2

Page 170

Question 2. In a survey of 50 students, 15 students said they liked football. The number of students is 15, and the ______ (frequency/relative frequency) is ______.

Relative frequency = 15/50 = 3/10 = 0.3

Answer - The first blank is relative frequency and the second is 15/50 or 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.

Answer - Not necessarily equally likely. The probabilities depend on the car's mechanical condition, battery and fuel.

(ii) Tossing a fair coin once.

Answer - Equally likely: P(H) = P(T) = 1/2.

(iii) Rolling a fair six-sided die.

Answer - Equally likely: each face has probability 1/6.

(iv) Choosing a marble randomly from a bag containing 3 red and 7 blue marbles.

Answer - The colour outcomes are not equally likely: P(Red) = 3/10 and P(Blue) = 7/10.

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

Answer - In the simplified school model, the outcomes are treated as approximately equally likely. In real population data, the exact probabilities may not be exactly 1/2. Thus, (ii) and (iii) are clearly equally likely, while (v) is approximately so in the simplified model.

Question 4. Write the sample space and calculate each probability.

(i) Two coins are tossed. Find the probability of at least one Head.

S = {HH, HT, TH, TT}; E = {HH, HT, TH}; P(E) = 3/4

(ii) Cards numbered 1 to 10 are placed in a box. Find the probability of drawing an even number.

E = {2,4,6,8,10}; P(E) = 5/10 = 1/2

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

E = {5,6}; P(E) = 2/6 = 1/3

(iv) A bag contains 3 red, 2 blue and 1 green ball. Find the probability of selecting a ball that is not red.

P(not red) = (2+1)/6 = 3/6 = 1/2

(v) Three coins are tossed. Find the probability of exactly two Heads.

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}; E = {HHT,HTH,THH}; P(E) = 3/8

Question 5. A bag has three candies: strawberry, lemon and mint. What is the probability of picking strawberry?

P(Strawberry) = 1/3

Page 171

Question 6. A child has 2 shirts - red and blue - and 3 types of pants - jeans, khakis and shorts. List all outfits containing one shirt and one pair of pants.

Total combinations = 2 x 3 = 6

Question 7. A tyre company records distances before replacement in 1,000 cases.

(i) Find the probability that a tyre lasts less than 4,000 km.

P = 20/1000 = 0.02

(ii) Find the probability that a tyre lasts between 4,000 and 14,000 km.

P = (210+325)/1000 = 535/1000 = 0.535

(iii) Find the probability that a tyre lasts more than 14,000 km.

P = 445/1000 = 0.445

Question 8. The letters of PEACE are placed on cards and one card is drawn.

(i) What is the probability that the card is P, E or C?

Cards: P,E,A,C,E; favourable cards: P,E,E,C; P = 4/5

(ii) What is the probability that the card is not E?

Favourable cards: P,A,C; P(not E) = 3/5

Question 9. A spinner has the equally likely numbers 1 to 8. Find the probability that it points at:

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

Pages 171-172

Question 10. A basket contains 4 red and 5 blue balls. One ball is drawn and laid aside, then a second ball is drawn. Represent the outcomes and probabilities.

Answer - The draws are without replacement.

(i) What is the probability of drawing a red ball and then a blue ball?

P(RB) = 4/9 x 5/8 = 20/72 = 5/18

(ii) What is the probability of drawing two blue balls?

P(BB) = 5/9 x 4/8 = 20/72 = 5/18

Page 172

Question 11. A pair of six-sided dice is thrown. Give an event with probability 0 and an event with probability 1.

Answer - Probability 0: obtaining a sum of 1, because the smallest possible sum is 2.

P(sum = 1) = 0

Answer - Probability 1: obtaining a sum from 2 to 12, because every possible throw has a sum in this range.

P(2 <= sum <= 12) = 1

Note: No single ordered outcome of two fair dice has probability 1. Probability 1 belongs to the certain event containing all possible outcomes.

Question 12. Write the sample space and calculate each probability.

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

S = {(i,j): i,j are in {1,2,3,4,5,6}}; n(S) = 36

Possible prime sums greater than 5 are 7 and 11. Sum 7 has six outcomes, while sum 11 has two outcomes.

P = (6+2)/36 = 8/36 = 2/9

(ii) A bag contains 4 red, 3 green and 2 blue balls. Two balls are drawn without replacement. What is the probability that they have different colours?

Colour sequences: S = {RR,RG,RB,GR,GG,GB,BR,BG,BB}

P(different) = (12+8+12+6+8+6)/72 = 52/72 = 13/18

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

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}; E = {HHT,HTH}; P(E) = 2/8 = 1/4

(iv) A four-digit number is formed using 1, 2, 3 and 4 without repetition. What is the probability that the number is even?

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}

There are 4! = 24 numbers. An even number must end in 2 or 4. For each final digit, the remaining digits can be arranged in 3! = 6 ways.

P(even) = (2 x 6)/24 = 12/24 = 1/2

(v) A student guesses answers to 3 multiple-choice questions, each with 4 options and one correct option. What is the probability of exactly 2 correct answers?

There are 4^3 = 64 equally likely answer sequences. Choose which two questions are correct in 3 ways, and choose one of 3 wrong options for the remaining question.

P(exactly 2 correct) = (3 x 3)/64 = 9/64

Question 13. A box contains balls numbered 1 to 4. Record sample spaces for two successive draws.

(i) The first ball is replaced before the second draw.

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) The first ball is not replaced before the second draw.

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) What are the sizes of the two sample spaces?

n(S1) = 4 x 4 = 16; n(S2) = 4 x 3 = 12

Question 14. List the sample space for tossing a coin and simultaneously drawing one card from cards numbered 1 to 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 - The sample size is n(S) = 12.

Pages 172-173

Question 15. Three coins are tossed and the number of Heads is recorded. Which list is a sample space: (i) {1,2,3}, (ii) {0,1,2}, (iii) {0,1,2,3,4}, or (iv) {0,1,2,3}? Why do the others fail?

Answer - The possible numbers of Heads are 0, 1, 2 and 3. Therefore, option (iv), {0,1,2,3}, is correct.

7. Option (i) omits 0 Heads.

8. Option (ii) omits 3 Heads.

9. Option (iii) includes 4 Heads, which is impossible with only three coins.

Page 173

Question 16. A point of dye is dropped randomly on a 3 m by 2 m rectangle containing a circle of diameter 1 m. What is the probability that it lands inside the circle?

Area of rectangle = 3 x 2 = 6 m². The circle has radius 1/2 m, so its area is π(1/2)² = π/4 m².

P(inside circle) = (π/4)/6 = π/24 ≈ 0.131 = 13.1%

Answer - The probability is π/24, approximately 0.131 or 13.1%.