Probability — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Probability" — 8 important questions with detailed answers for CBSE board exam preparation.
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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Probability" — 8 important questions with detailed answers for CBSE board exam pre…
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Key Questions Covered:
- Two dice are thrown simultaneously. Find the probability of getting a sum of 7.
- A card is drawn from a standard deck of 52 cards. Find the probability that i…
- In a class of 30 students, 18 study Mathematics, 12 study Physics, and 8 stud…
- A box contains 5 red balls and 3 blue balls. Two balls are drawn one after an…
- A number is selected from 1 to 10. Find the probability that it is a prime nu…
- The probability that a student passes Mathematics is 0.8 and the probability …
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Two dice are thrown simultaneously. Find the probability … | ✓ Solved |
| A card is drawn from a standard deck of 52 cards. Find th… | ✓ Solved |
| In a class of 30 students, 18 study Mathematics, 12 study… | ✓ Solved |
| A box contains 5 red balls and 3 blue balls. Two balls ar… | ✓ Solved |
| A number is selected from 1 to 10. Find the probability t… | ✓ Solved |
| The probability that a student passes Mathematics is 0.8 … | ✓ Solved |
Showing 6 of 8 questions
Q1: Two dice are thrown simultaneously. Find the probability of getting a sum of 7.
To find P(sum = 7):
Step 1: Find total possible outcomes.
When two dice are thrown: 6 × 6 = 36 outcomes
Step 2: Find favorable outcomes (sum = 7).
List all pairs (a, b) where a + b = 7:
(1, 6): 1 + 6 = 7
(2, 5): 2 + 5 = 7
(3, 4): 3 + 4 = 7
(4, 3): 4 + 3 = 7
(5, 2): 5 + 2 = 7
(6, 1): 6 + 1 = 7
Number of favorable outcomes = 6
Step 3: Calculate probability.
P(sum = 7) = Number of favorable outcomes / Total outcomes
= 6/36
= 1/6
Final Answer: P(sum = 7) = 1/6
Q2: A card is drawn from a standard deck of 52 cards. Find the probability that it is a red card or a king.
To find P(red card OR king):
Step 1: Identify the events.
Event A: Card is red
Event B: Card is a king
Step 2: Count favorable outcomes.
Red cards: 26 (13 hearts + 13 diamonds)
Kings: 4 (one in each suit)
Red kings: 2 (king of hearts and king of diamonds)
Step 3: Use inclusion-exclusion principle.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Where:
P(A) = 26/52 = 1/2
P(B) = 4/52 = 1/13
P(A ∩ B) = 2/52 = 1/26 (red kings)
Step 4: Calculate.
P(A ∪ B) = 26/52 + 4/52 - 2/52
= 28/52
= 7/13
Final Answer: P(...
Q3: In a class of 30 students, 18 study Mathematics, 12 study Physics, and 8 study both. If a student is selected at random, find the probability that the student studies either Mathematics or Physics.
To find P(Mathematics OR Physics):
Step 1: Identify given information.
n(M) = 18 (students studying Mathematics)
n(P) = 12 (students studying Physics)
n(M ∩ P) = 8 (students studying both)
n(Total) = 30
Step 2: Use inclusion-exclusion principle.
n(M ∪ P) = n(M) + n(P) - n(M ∩ P)
= 18 + 12 - 8
= 22
Step 3: Calculate probability.
P(M ∪ P) = n(M ∪ P) / n(Total)
= 22/30
= 11/15
Final Answer: P(Mathematics or Physics) = 11/15
Q4: A box contains 5 red balls and 3 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are red.
To find P(both red) without replacement:
Step 1: Find probability of first ball being red.
P(1st red) = 5/8
Step 2: Find probability of second ball being red given first is red.
After drawing one red ball:
Remaining red balls = 4
Total remaining balls = 7
P(2nd red | 1st red) = 4/7
Step 3: Apply multiplication rule for dependent events.
P(both red) = P(1st red) × P(2nd red | 1st red)
= 5/8 × 4/7
= 20/56
= 5/14
Final Answer: P(both red) = 5/14
Q5: A number is selected from 1 to 10. Find the probability that it is a prime number.
To find P(prime number from 1 to 10):
Step 1: Identify total outcomes.
Numbers from 1 to 10: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Total outcomes = 10
Step 2: Identify prime numbers.
Prime numbers are natural numbers greater than 1 with exactly two factors (1 and itself).
From 1 to 10:
1: Not prime (only one factor)
2: Prime (factors: 1, 2)
3: Prime (factors: 1, 3)
4: Not prime (factors: 1, 2, 4)
5: Prime (factors: 1, 5)
6: Not prime (factors: 1, 2, 3, 6)
7: Prime (factors: 1, 7)
8: Not prime (fact...
Q6: The probability that a student passes Mathematics is 0.8 and the probability that the student passes Physics is 0.7. If the events are independent, find the probability that the student passes both subjects.
To find P(both Mathematics and Physics):
Step 1: Identify given information.
P(M) = Probability of passing Mathematics = 0.8
P(P) = Probability of passing Physics = 0.7
Events are independent
Step 2: Recall definition of independent events.
Two events A and B are independent if:
P(A ∩ B) = P(A) × P(B)
Step 3: Apply the formula.
P(M and P) = P(M) × P(P)
= 0.8 × 0.7
= 0.56
Final Answer: P(both Mathematics and Physics) = 0.56
Showing 6 of 8 questions. Visit the full page for complete solutions.
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