Call : 94779 94779
GANITA MANJARI | Chapter 7 - Probability | Verbatim Q&A
CHAPTER 7

The Mathematics of Maybe: Introduction to Probability

GANITA MANJARI | Class 9 Maths | Part I

Think and Reflect

Page 156
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?
Answer
Because a fair coin has two equally likely outcomes (heads or tails), so neither team has an advantage.
Page 157
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 individual try is unpredictable.
Answer
Yes, randomness means individual outcomes are uncertain even though all possibilities are known.
Page 163
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.
Answer
Each roll is independent. Past outcomes do not affect future outcomes.
Page 164
Gambler's Fallacy explanation
Answer
The coin/die has no memory. Each trial is independent.
Page 167
When we used the sample space {Rain, No Rain} in Example 1, we focused only on whether it will rain or not. However, if we want to include different amounts of rainfall like drizzle, light rain or heavy rain, we need to expand the sample space to {No Rain, Drizzle, Light Rain, Heavy Rain} so that it better matches the level of detail required for the question. It is important to ensure the sample space is detailed enough to suit the specific problem being studied.
Answer
Sample space must be chosen appropriately for the level of detail needed.
Page 169
Can you calculate the probability of getting one head and one tail?
Answer
For two coins, outcomes: HH, HT, TH, TT. One head and one tail: HT, TH → Probability = 2/4 = 1/2.

EXERCISE SET 7.1 (Page 159)

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 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) Certain (1) - Monday always follows Sunday.
(ii) Impossible (0) - It never snows in Mumbai in July.
(iii) Impossible (0) - Extremely unlikely.
(iv) More likely (close to 1) - Most students greet friends daily.

EXERCISE SET 7.2 (Page 165)

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) P(green) = 8/30 = 4/15 ≈ 0.267
(ii) Estimated yellow = (7/30) × 600 = 140 sweets.
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) = 11/40 = 0.275
(ii) Estimated Sports = (9/40) × 800 = 180 students.
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
Answers vary. Theoretical probability of tails = 1/2.
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. 7.5). Assign probabilities to the outcomes by using experimental probability.
Solution
Answers depend on experimental results.
5.
What is the probability of getting an even number when rolling a fair 6-sided die?
Solution
Even numbers: 2,4,6 → 3 outcomes. P = 3/6 = 1/2.
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
(i) Experimental = 3/12 = 1/4 = 0.25
(ii) Theoretical = 1/6 ≈ 0.167
(iii) Small sample size causes difference. As trials increase, experimental probability approaches theoretical probability.

EXERCISE SET 7.3 (Pages 167-168)

1.
When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Solution
6 outcomes.
2.
For the following experiments write down the sample space S. (i) Rolling a die and tossing a coin together. (ii) ???
Solution
(i) S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T} (12 outcomes)
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) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)} (6 outcomes)
(ii) E = {(Samosa, Chai), (Samosa, Lassi)}

EXERCISE SET 7.4 (Page 169)

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) Tree diagram: A: Apple (1/3) or Orange (2/3); B: Banana (1/2) or Mango (1/2)
(ii) S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}
(iii) P(Apple and Banana) = 1/3 × 1/2 = 1/6
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) Possible outcomes: (R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)
(ii) P(same colour) = P(R,R)+P(B,B)+P(G,G) = (3/9×3/9)+(4/9×4/9)+(2/9×2/9) = (9+16+4)/81 = 29/81

END-OF-CHAPTER EXERCISES (Pages 169-173)

1.
Fill in the blanks. (i) The probability of an impossible event is ______.
Solution
0
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
Relative frequency = 15/50 = 3/10 = 0.3
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) Not equally likely (car more likely to start)
(ii) Equally likely (1/2 each)
(iii) Equally likely (1/6 each)
(iv) Not equally likely (3/10 red, 7/10 blue)
(v) Approximately equally likely (≈ 1/2 each)
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) S = {HH, HT, TH, TT}, P(at least one head) = 3/4
(ii) S = {1,...,10}, P(even) = 5/10 = 1/2
(iii) S = {1,2,3,4,5,6}, P(>4) = 2/6 = 1/3
(iv) Total balls = 6, non-red = 3, P(not red) = 3/6 = 1/2
(v) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, P(exactly 2 heads) = 3/8
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
P = 1/3
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
Table: (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis), (Blue, Shorts)
7.
A tyre company records distances before replacement in 1000 cases.
Distance (km): Less than 4000 | 4001-9000 | 9001-14000 | 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.
Solution
(i) P = 20/1000 = 0.02
(ii) P = (210+325)/1000 = 535/1000 = 0.535
(iii) P = 445/1000 = 0.445
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
Total letters = 5. (i) P, E, C appear: P(1), E(2), C(1) → total 4. P = 4/5
(ii) Not E = letters other than E: 3 letters. P = 3/5
*9.
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?
Solution
(i) P(8) = 1/8
(ii) Odd numbers: 1,3,5,7 → P = 4/8 = 1/2
(iii) Numbers >2: 3,4,5,6,7,8 → P = 6/8 = 3/4
(iv) Numbers <9: all → P = 1
(v) Multiples of 3: 3,6 → P = 2/8 = 1/4
*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.
Solution
Tree diagram: First draw: R(4/9), B(5/9). Second draw depends on first. Probabilities can be calculated accordingly.
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: sum of dice = 13. Outcome with probability 1: sum of dice between 2 and 12.
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) Prime numbers >5: 7,11. Sum 7: 6 outcomes; sum 11: 2 outcomes. Total = 8/36 = 2/9
(ii) Total ways = 9C2 = 36. Same colour: red-red=6, green-green=3, blue-blue=1 → total 10. Different colour = 26, P = 26/36 = 13/18
(iii) Possibilities: HHT, HTH → 2 outcomes. Total outcomes = 8. P = 2/8 = 1/4
(iv) Total 4-digit numbers = 24. Even numbers (ending with 2 or 4): 2 × 3! = 12. P = 12/24 = 1/2
(v) P(correct) = 1/4, P(wrong) = 3/4. Exactly 2 correct: C(3,2) × (1/4)² × (3/4) = 3 × 1/16 × 3/4 = 9/64
13.
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) With replacement: 4×4 = 16 outcomes
(ii) Without replacement: 4×3 = 12 outcomes
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.
Solution
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} (12 outcomes)
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? {1,2,3} {0,1,2} {0,1,2,3,4} {0,1,2,3}
Solution
{0,1,2,3} is the correct sample space. {1,2,3} misses 0 heads. {0,1,2} misses 3 heads. {0,1,2,3,4} includes 4 heads which is impossible.
16.
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
Probability = (area of circle)/(area of rectangle) = π(0.5)² / (area of rectangle). Depends on rectangle dimensions.

Chapter Summary (Page 173)

  • Probability is a measurement of the likelihood of an event, expressed on a scale from 0 to 1.
  • Experimental probability = (Number of times event occurred)/(Total number of trials).
  • Theoretical probability = (Number of favourable outcomes)/(Total number of possible outcomes).
  • Sample space is the list of all possible outcomes of a random experiment.
  • An event is any subset of the sample space.
  • Tree diagrams help list and visualise all possible outcomes of a random experiment.
  • The Law of Large Numbers: As the number of trials increases, experimental probability approaches theoretical probability.

Also Get

Class 9- NCERT- Science Solⁿ.

Class 9-NCERT- English Solⁿ

Class 9-NCERT- Maths
Solⁿ

Class 9-NCERT- Social Science Solⁿ

Class 9 CBSE - SYLLABUS

Study materials

Send Us A Message






    Latest posts

    Facebook Twitter Youtube Icon-instagram-1

    Thank You For Registering !

    Our Team Will Reach You Soon

    Enter Your Contact Details To Download the PDF

      This form uses Akismet to reduce spam. Learn how your data is processed.

      REGISTER NOW

      👉 (Limited Seats Available) 👈