A standard 6-sided die has six distinct faces numbered 1, 2, 3, 4, 5, and 6. When it is rolled, any one of these faces can land face-up. The sample space S is the set of all possible outcomes: S = {1, 2, 3, 4, 5, 6}. The total number of possible outcomes (or sample size) is given by: n(S) = 6. ThereRead more
A standard 6-sided die has six distinct faces numbered 1, 2, 3, 4, 5, and 6.
When it is rolled, any one of these faces can land face-up.
The sample space S is the set of all possible outcomes:
S = {1, 2, 3, 4, 5, 6}.
The total number of possible outcomes (or sample size) is given by:
n(S) = 6.
Therefore, there are 6 possible outcomes in the sample space.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
(i) Die gives {1, 2, 3, 4, 5, 6} and coin gives {H, T}: S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}. Size n(S) = 6 x 2 = 12. (ii) The integers lying strictly between -5 and +5 are: S = {-4, -3, -2, -1, 0, 1, 2, 3, 4}. (iii) By outcome of color, the sample space is: S = {Green, Red}. If eachRead more
(i) Die gives {1, 2, 3, 4, 5, 6} and coin gives {H, T}:
Given snacks: {Samosa, Pakora, Bhaji} and drinks: {Chai, Lassi}. (i) The sample space S contains all pairs combining one snack with one drink: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. The sample size is n(S) = 3 x 2 = 6 combinations. (ii)Read more
Given snacks: {Samosa, Pakora, Bhaji} and drinks: {Chai, Lassi}.
(i) The sample space S contains all pairs combining one snack with one drink:
Basket A contains 3 fruits: 1 Apple (A), 2 Oranges (O1, O2). Basket B contains 2 fruits: 1 Banana (B), 1 Mango (M). (i) Tree Diagram: Start -> A -> B => (A, B) Start -> A -> M => (A, M) Start -> O1 -> B => (O1, B) Start -> O1 -> M => (O1, M) Start -> O2Read more
Basket A contains 3 fruits: 1 Apple (A), 2 Oranges (O1, O2). Basket B contains 2 fruits: 1 Banana (B), 1 Mango (M).
(i) Tree Diagram:
Start -> A -> B => (A, B)
Start -> A -> M => (A, M)
Start -> O1 -> B => (O1, B)
Start -> O1 -> M => (O1, M)
Start -> O2 -> B => (O2, B)
Start -> O2 -> M => (O2, M)
(ii) Sample Space:
S = {(A, B), (A, M), (O1, B), (O1, M), (O2, B), (O2, M)}, so n(S) = 6.
(iii) Favourable outcome = {(A, B)}, which is 1.
P(Apple and Banana) = 1 / 6 (or about 0.167).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
For a neutral atom of element X, the number of protons equals the number of electrons, which is 31. The superscript 70 represents the mass number, which is the total count of protons and neutrons combined. Subtracting 31 protons from the mass number 70 yields 39 neutrons. Therefore, there are 39 neuRead more
For a neutral atom of element X, the number of protons equals the number of electrons, which is 31. The superscript 70 represents the mass number, which is the total count of protons and neutrons combined. Subtracting 31 protons from the mass number 70 yields 39 neutrons. Therefore, there are 39 neutrons in its nucleus.
For more NCERT Solutions of Class 9 Science Exploration Chapter 8 Journey Inside the Atom Question Answer (2026-27)
When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
A standard 6-sided die has six distinct faces numbered 1, 2, 3, 4, 5, and 6. When it is rolled, any one of these faces can land face-up. The sample space S is the set of all possible outcomes: S = {1, 2, 3, 4, 5, 6}. The total number of possible outcomes (or sample size) is given by: n(S) = 6. ThereRead more
A standard 6-sided die has six distinct faces numbered 1, 2, 3, 4, 5, and 6.
When it is rolled, any one of these faces can land face-up.
The sample space S is the set of all possible outcomes:
S = {1, 2, 3, 4, 5, 6}.
The total number of possible outcomes (or sample size) is given by:
n(S) = 6.
Therefore, there are 6 possible outcomes in the sample space.
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 lessFor 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.
(i) Die gives {1, 2, 3, 4, 5, 6} and coin gives {H, T}: S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}. Size n(S) = 6 x 2 = 12. (ii) The integers lying strictly between -5 and +5 are: S = {-4, -3, -2, -1, 0, 1, 2, 3, 4}. (iii) By outcome of color, the sample space is: S = {Green, Red}. If eachRead more
(i) Die gives {1, 2, 3, 4, 5, 6} and coin gives {H, T}:
S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}.
Size n(S) = 6 x 2 = 12.
(ii) The integers lying strictly between -5 and +5 are:
S = {-4, -3, -2, -1, 0, 1, 2, 3, 4}.
(iii) By outcome of color, the sample space is:
S = {Green, Red}.
If each ball is treated distinctly, S = {G1, G2, G3, G4, G5, R1, R2, R3, R4, R5, R6, R7}, where n(S) = 12.
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 lessIn 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.’
Given snacks: {Samosa, Pakora, Bhaji} and drinks: {Chai, Lassi}. (i) The sample space S contains all pairs combining one snack with one drink: S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. The sample size is n(S) = 3 x 2 = 6 combinations. (ii)Read more
Given snacks: {Samosa, Pakora, Bhaji} and drinks: {Chai, Lassi}.
(i) The sample space S contains all pairs combining one snack with one drink:
S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}.
The sample size is n(S) = 3 x 2 = 6 combinations.
(ii) The event E representing ‘Selecting Samosa as a snack’ contains only pairs with Samosa:
E = {(Samosa, Chai), (Samosa, Lassi)}
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 lessThere 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?
Basket A contains 3 fruits: 1 Apple (A), 2 Oranges (O1, O2). Basket B contains 2 fruits: 1 Banana (B), 1 Mango (M). (i) Tree Diagram: Start -> A -> B => (A, B) Start -> A -> M => (A, M) Start -> O1 -> B => (O1, B) Start -> O1 -> M => (O1, M) Start -> O2Read more
Basket A contains 3 fruits: 1 Apple (A), 2 Oranges (O1, O2). Basket B contains 2 fruits: 1 Banana (B), 1 Mango (M).
(i) Tree Diagram:
Start -> A -> B => (A, B)
Start -> A -> M => (A, M)
Start -> O1 -> B => (O1, B)
Start -> O1 -> M => (O1, M)
Start -> O2 -> B => (O2, B)
Start -> O2 -> M => (O2, M)
(ii) Sample Space:
S = {(A, B), (A, M), (O1, B), (O1, M), (O2, B), (O2, M)}, so n(S) = 6.
(iii) Favourable outcome = {(A, B)}, which is 1.
P(Apple and Banana) = 1 / 6 (or about 0.167).
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 lessAn atom 70 X has 31 electrons. How many neutrons are there in its nucleus?
For a neutral atom of element X, the number of protons equals the number of electrons, which is 31. The superscript 70 represents the mass number, which is the total count of protons and neutrons combined. Subtracting 31 protons from the mass number 70 yields 39 neutrons. Therefore, there are 39 neuRead more
For a neutral atom of element X, the number of protons equals the number of electrons, which is 31. The superscript 70 represents the mass number, which is the total count of protons and neutrons combined. Subtracting 31 protons from the mass number 70 yields 39 neutrons. Therefore, there are 39 neutrons in its nucleus.
For more NCERT Solutions of Class 9 Science Exploration Chapter 8 Journey Inside the Atom Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-8/
See less