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What is the probability of getting an even number when rolling a fair 6-sided die?
When rolling a standard fair 6-sided die, the sample space of possible outcomes is: S = {1, 2, 3, 4, 5, 6} Total number of possible outcomes = 6. The event E consists of getting an even number: E = {2, 4, 6} Number of favourable outcomes = 3. Using the theoretical probability formula: P(E) = NumberRead more
When rolling a standard fair 6-sided die, the sample space of possible outcomes is:
S = {1, 2, 3, 4, 5, 6}
Total number of possible outcomes = 6.
The event E consists of getting an even number:
E = {2, 4, 6}
Number of favourable outcomes = 3.
Using the theoretical probability formula:
P(E) = Number of favourable outcomes / Total possible outcomes
P(even number) = 3 / 6 = 1 / 2 = 0.5 (or 50%).
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 lessSuppose 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?
(i) Experimental probability = Number of 3s / Total rolls = 3 / 12 = 1 / 4 = 0.25 (or 25%). (ii) Theoretical probability = 1 / 6 (approximately 0.167 or 16.7%). (iii) They differ because 12 is a small number of trials where random variation is significant. By the Law of Large Numbers, as trials incrRead more
(i) Experimental probability = Number of 3s / Total rolls = 3 / 12 = 1 / 4 = 0.25 (or 25%).
(ii) Theoretical probability = 1 / 6 (approximately 0.167 or 16.7%).
(iii) They differ because 12 is a small number of trials where random variation is significant. By the Law of Large Numbers, as trials increase to 60, 600, or 6000, experimental probability gets closer and closer to theoretical probability of 1 / 6.
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 lessWhen 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 less