When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12. Event with probability 0: 'Getting a sum of 13' (or getting a negative sum). This is an impossible event because the highest possible total is 12. Event with probability 1: 'Getting a sum between 2 and 12 inclusiRead more
When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12.
Event with probability 0: ‘Getting a sum of 13’ (or getting a negative sum).
This is an impossible event because the highest possible total is 12.
Event with probability 1: ‘Getting a sum between 2 and 12 inclusive’ (or getting a sum less than 13). This is a certain event because every possible outcome satisfies it.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6. Outfit 1 is (Red Shirt, Jeans), Outfit 2 is (Red Shirt, Khakis), Outfit 3 is (Red Shirt, Shorts), Outfit 4 is (Blue Shirt, Jeans), Outfit 5 is (Blue Shirt, Khakis) and Outfit 6 is (BRead more
The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6.
Outfit 1 is (Red Shirt, Jeans),
Outfit 2 is (Red Shirt, Khakis),
Outfit 3 is (Red Shirt, Shorts),
Outfit 4 is (Blue Shirt, Jeans),
Outfit 5 is (Blue Shirt, Khakis) and
Outfit 6 is (Blue Shirt, Shorts).
Thus, the sample space comprises exactly these 6 outfit combinations.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
The sample space of candies in the bag is: S = {strawberry, lemon, mint} Total number of possible outcomes = 3. The event E consists of picking a strawberry candy: E = {strawberry} Number of favourable outcomes = 1. Using the formula for theoretical probability: P(E) = Number of favourable outcomesRead more
The sample space of candies in the bag is:
S = {strawberry, lemon, mint}
Total number of possible outcomes = 3.
The event E consists of picking a strawberry candy:
E = {strawberry}
Number of favourable outcomes = 1.
Using the formula for theoretical probability:
P(E) = Number of favourable outcomes / Total possible outcomes
P(strawberry candy) = 1 / 3 (which is approximately 0.333 or 33.3%).
Hence, the probability is 1 / 3.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
(i) Not equally likely: A normal working car usually starts, depending on mechanical condition rather than pure chance. (ii) Equally likely: A fair coin is balanced, giving heads and tails an equal chance of 1/2 each. (iii) Equally likely: A fair 6-sided die gives each number from 1 to 6 an identicaRead more
(i) Not equally likely: A normal working car usually starts, depending on mechanical condition rather than pure chance.
(ii) Equally likely: A fair coin is balanced, giving heads and tails an equal chance of 1/2 each.
(iii) Equally likely: A fair 6-sided die gives each number from 1 to 6 an identical probability of 1/6.
(iv) Not equally likely: Blue marbles (7) outnumber red marbles (3), making blue more likely.
(v) Equally likely: Biologically, a newborn being a boy or a girl is equally likely with probability 1/2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
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.
When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12. Event with probability 0: 'Getting a sum of 13' (or getting a negative sum). This is an impossible event because the highest possible total is 12. Event with probability 1: 'Getting a sum between 2 and 12 inclusiRead more
When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12.
Event with probability 0: ‘Getting a sum of 13’ (or getting a negative sum).
This is an impossible event because the highest possible total is 12.
Event with probability 1: ‘Getting a sum between 2 and 12 inclusive’ (or getting a sum less than 13). This is a certain event because every possible outcome satisfies it.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessA 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.
The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6. Outfit 1 is (Red Shirt, Jeans), Outfit 2 is (Red Shirt, Khakis), Outfit 3 is (Red Shirt, Shorts), Outfit 4 is (Blue Shirt, Jeans), Outfit 5 is (Blue Shirt, Khakis) and Outfit 6 is (BRead more
The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6.
Outfit 1 is (Red Shirt, Jeans),
Outfit 2 is (Red Shirt, Khakis),
Outfit 3 is (Red Shirt, Shorts),
Outfit 4 is (Blue Shirt, Jeans),
Outfit 5 is (Blue Shirt, Khakis) and
Outfit 6 is (Blue Shirt, Shorts).
Thus, the sample space comprises exactly these 6 outfit combinations.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessA bag has 3 candies: strawberry, lemon and mint. One is picked at random. What is the probability of picking a strawberry candy?
The sample space of candies in the bag is: S = {strawberry, lemon, mint} Total number of possible outcomes = 3. The event E consists of picking a strawberry candy: E = {strawberry} Number of favourable outcomes = 1. Using the formula for theoretical probability: P(E) = Number of favourable outcomesRead more
The sample space of candies in the bag is:
S = {strawberry, lemon, mint}
Total number of possible outcomes = 3.
The event E consists of picking a strawberry candy:
E = {strawberry}
Number of favourable outcomes = 1.
Using the formula for theoretical probability:
P(E) = Number of favourable outcomes / Total possible outcomes
P(strawberry candy) = 1 / 3 (which is approximately 0.333 or 33.3%).
Hence, the probability is 1 / 3.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessWrite 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?
(i) S = {HH, HT, TH, TT}. Favourable = {HH, HT, TH}. P(at least 1 head) = 3/4 = 0.75. (ii) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Even = {2, 4, 6, 8, 10}. P(even) = 5/10 = 1/2 = 0.5. (iii) S = {1, 2, 3, 4, 5, 6}. Favourable = {5, 6}. P(> 4) = 2/6 = 1/3. (iv) Total balls = 3 + 2 + 1 = 6. Not red = 2Read more
(i) S = {HH, HT, TH, TT}. Favourable = {HH, HT, TH}. P(at least 1 head) = 3/4 = 0.75.
(ii) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Even = {2, 4, 6, 8, 10}. P(even) = 5/10 = 1/2 = 0.5.
(iii) S = {1, 2, 3, 4, 5, 6}. Favourable = {5, 6}. P(> 4) = 2/6 = 1/3.
(iv) Total balls = 3 + 2 + 1 = 6. Not red = 2 blue + 1 green = 3. P(not red) = 3/6 = 1/2 = 0.5.
(v) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (8 outcomes). Favourable = {HHT, HTH, THH}. P(exactly 2 heads) = 3/8 = 0.375.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessWhich 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.
(i) Not equally likely: A normal working car usually starts, depending on mechanical condition rather than pure chance. (ii) Equally likely: A fair coin is balanced, giving heads and tails an equal chance of 1/2 each. (iii) Equally likely: A fair 6-sided die gives each number from 1 to 6 an identicaRead more
(i) Not equally likely: A normal working car usually starts, depending on mechanical condition rather than pure chance.
(ii) Equally likely: A fair coin is balanced, giving heads and tails an equal chance of 1/2 each.
(iii) Equally likely: A fair 6-sided die gives each number from 1 to 6 an identical probability of 1/6.
(iv) Not equally likely: Blue marbles (7) outnumber red marbles (3), making blue more likely.
(v) Equally likely: Biologically, a newborn being a boy or a girl is equally likely with probability 1/2.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See less