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.
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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NCERT Solutions for Class 12 Mathematics — Probability. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Probability — All 8 Questions Solved
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(red card or king) = 7/13
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 (factors: 1, 2, 4, 8)
9: Not prime (factors: 1, 3, 9)
10: Not prime (factors: 1, 2, 5, 10)
Prime numbers: {2, 3, 5, 7}
Number of favorable outcomes = 4
Step 3: Calculate probability.
P(prime) = Favorable outcomes / Total outcomes
= 4/10
= 2/5
Final Answer: P(prime number) = 2/5 or 0.4
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
Q7. A jar contains 10 red, 15 blue, and 25 green marbles. If a marble is drawn at random, find the probability of drawing a marble that is either red or green.
To find P(red OR green):
Step 1: Count total marbles.
Red: 10
Blue: 15
Green: 25
Total = 10 + 15 + 25 = 50
Step 2: Count favorable outcomes.
Red or green marbles = 10 + 25 = 35
Step 3: Calculate probability.
P(red or green) = Favorable outcomes / Total outcomes
= 35/50
= 7/10
Alternative method using complement:
P(red or green) = 1 - P(blue)
= 1 - 15/50
= 1 - 3/10
= 7/10
Final Answer: P(red or green) = 7/10 or 0.7
Q8. Define conditional probability and state Bayes' theorem.
Conditional Probability and Bayes' Theorem:
1. CONDITIONAL PROBABILITY DEFINITION:
The conditional probability of event A given event B is denoted P(A|B) and is defined as:
P(A|B) = P(A ∩ B) / P(B), provided P(B) ≠ 0
Interpretation:
It represents the probability of A occurring given that B has already occurred.
Example:
If P(A ∩ B) = 0.12 and P(B) = 0.3, then
P(A|B) = 0.12 / 0.3 = 0.4
2. MULTIPLICATION RULE (derived from conditional probability):
P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
3. BAYES' THEOREM:
For events A and B:
P(A|B) = P(B|A) × P(A) / P(B)
Where:
P(A|B) = Posterior probability (probability of A given B)
P(B|A) = Likelihood (probability of B given A)
P(A) = Prior probability (initial probability of A)
P(B) = Total probability of B
4. EXTENDED BAYES' THEOREM:
If A₁, A₂, ..., Aₙ are mutually exclusive and exhaustive events, and B is any event:
P(Aᵢ|B) = P(B|Aᵢ) × P(Aᵢ) / [Σⱼ P(B|Aⱼ) × P(Aⱼ)]
Where the denominator is the total probability of B.
5. KEY POINTS:
- Conditional probability accounts for new information
- Bayes' theorem relates conditional probabilities
- Used in medical diagnosis, quality control, machine learning
- Events are independent if P(A|B) = P(A)
Final Answer: Conditional probability P(A|B) = P(A ∩ B)/P(B). Bayes' theorem: P(A|B) = P(B|A) × P(A) / P(B)
Probability — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, find P(A ∩ B).
- A. 0.24
- B. 0.48
- C. 0.32
- D. 0.6
Answer: A. 0.24 — Using the formula for conditional probability, P(A ∩ B) = P(B|A) * P(A). So, P(A ∩ B) = 0.6 * 0.4 = 0.24.
MCQ 2. If A and B are two independent events such that P(A) = 0.3 and P(B) = 0.5, find P(A U B).
- A. 0.8
- B. 0.65
- C. 0.15
- D. 0.5
Answer: B. 0.65 — For independent events, P(A ∩ B) = P(A)P(B) = 0.3 * 0.5 = 0.15. Then, P(A U B) = P(A) + P(B) - P(A ∩ B) = 0.3 + 0.5 - 0.15 = 0.65.
MCQ 3. A bag contains 5 red and 3 blue balls. If two balls are drawn at random without replacement, what is the probability that the first ball is red and the second ball is blue?
- A. 15/56
- B. 15/64
- C. 5/8
- D. 3/8
Answer: A. 15/56 — The probability of drawing a red ball first is 5/8. After drawing one red ball, there are 7 balls left, with 3 blue. So, the probability of drawing a blue ball second (given the first was red) is 3/7. The combined probability is (5/8) * (3/7) = 15/56.
MCQ 4. If A and B are mutually exclusive events, then P(A ∩ B) is:
- A. P(A)P(B)
- B. 0
- C. P(A) + P(B)
- D. 1
Answer: B. 0 — Mutually exclusive events cannot occur at the same time, meaning their intersection is an empty set, and its probability is 0.
MCQ 5. A test for a certain disease has a 99% accuracy rate, meaning it correctly identifies 99% of people with the disease and 99% of people without the disease. If 0.5% of the population has the disease, what is the probability that a person who tests positive actually has the disease?
- A. Approx 0.33
- B. Approx 0.005
- C. Approx 0.99
- D. Approx 0.01
Answer: A. Approx 0.33 — Let D be having the disease and T be testing positive. P(D)=0.005, P(T|D)=0.99, P(T|D')=0.01. Using Bayes' Theorem, P(D|T) = (P(T|D)P(D)) / (P(T|D)P(D) + P(T|D')P(D')) = (0.99*0.005) / (0.99*0.005 + 0.01*0.995) = 0.00495 / (0.00495 + 0.00995) = 0.00495 / 0.0149 ≈ 0.3322.
MCQ 6. A speaks truth in 75% cases and B in 80% cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
- A. 35%
- B. 20%
- C. 60%
- D. 15%
Answer: A. 35% — Contradiction occurs if A speaks truth and B lies, or A lies and B speaks truth. P(A lies) = 0.25, P(B lies) = 0.20. So, the probability is (0.75 * 0.20) + (0.25 * 0.80) = 0.15 + 0.20 = 0.35 or 35%.
MCQ 7. For a discrete random variable X, the sum of all probabilities P(X=x) must be equal to:
- A. 0
- B. 1
- C. -1
- D. P(X=0)
Answer: B. 1 — According to the axioms of probability, the sum of probabilities for all possible outcomes in any probability distribution must always equal 1.
MCQ 8. A fair coin is tossed three times. Let X be the number of heads obtained. Find the expected value E(X).
- A. 1.5
- B. 1
- C. 2
- D. 0.5
Answer: A. 1.5 — This is a binomial distribution B(n=3, p=0.5). The expected value E(X) for a binomial distribution is given by np. So, E(X) = 3 * 0.5 = 1.5.
MCQ 9. Let X be a random variable with the following probability distribution: X: 0 1 2 P(X): 0.2 0.5 0.3 Find the variance Var(X).
- A. 0.49
- B. 0.51
- C. 0.61
- D. 0.79
Answer: A. 0.49 — First, find E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 0.5 + 0.6 = 1.1. Then find E(X^2) = 0^2(0.2) + 1^2(0.5) + 2^2(0.3) = 0.5 + 1.2 = 1.7. Var(X) = E(X^2) - [E(X)]^2 = 1.7 - (1.1)^2 = 1.7 - 1.21 = 0.49.
MCQ 10. Which of the following is NOT a condition for a Bernoulli trial?
- A. There are only two possible outcomes
- B. The trials are independent
- C. The probability of success remains constant
- D. The number of trials is infinite
Answer: D. The number of trials is infinite — A Bernoulli trial is a single experiment with two outcomes. A sequence of Bernoulli trials (Binomial distribution) requires a fixed, finite number of trials, not infinite.
MCQ 11. A die is thrown 5 times. What is the probability of getting exactly 3 fours?
- A. 5C3 * (1/6)^3 * (5/6)^2
- B. 5C3 * (1/6)^3 * (1/6)^2
- C. 5C3 * (1/4)^3 * (3/4)^2
- D. (1/6)^3
Answer: A. 5C3 * (1/6)^3 * (5/6)^2 — This is a binomial distribution with n=5 trials and probability of success (getting a four) p=1/6. The probability of exactly 3 successes is 5C3 * p^3 * (1-p)^(5-3) = 5C3 * (1/6)^3 * (5/6)^2.
MCQ 12. A coin is tossed 6 times. What is the probability of getting at least 5 heads?
- A. 7/64
- B. 1/64
- C. 6/64
- D. 1/2
Answer: A. 7/64 — This is a binomial distribution B(n=6, p=0.5). P(X≥5) = P(X=5) + P(X=6). P(X=5) = 6C5(0.5)^5(0.5)^1 = 6/64. P(X=6) = 6C6(0.5)^6(0.5)^0 = 1/64. So, P(X≥5) = 6/64 + 1/64 = 7/64.
MCQ 13. If X follows a binomial distribution with mean 4 and variance 2, find the number of trials (n).
- A. 8
- B. 4
- C. 16
- D. 2
Answer: A. 8 — For a binomial distribution, mean = np and variance = npq. Given np=4 and npq=2. Dividing variance by mean, q = (npq)/(np) = 2/4 = 0.5. Since p = 1-q, p = 1-0.5 = 0.5. Substituting p into np=4, we get n * 0.5 = 4, so n=8.
MCQ 14. Given P(A) = 0.5, P(B) = 0.6, and P(A ∩ B) = 0.3. Find P(A U B).
- A. 0.8
- B. 0.9
- C. 0.2
- D. 0.5
Answer: A. 0.8 — Using the Addition Theorem of Probability, P(A U B) = P(A) + P(B) - P(A ∩ B) = 0.5 + 0.6 - 0.3 = 1.1 - 0.3 = 0.8.
MCQ 15. Two fair dice are rolled. What is the probability that the sum is 8, given that the sum is an even number?
- A. 5/18
- B. 5/36
- C. 1/6
- D. 1/9
Answer: A. 5/18 — The total number of outcomes for two dice is 36. The number of outcomes where the sum is an even number is 18 (e.g., (1,1), (1,3)...). The outcomes where the sum is 8 are (2,6), (3,5), (4,4), (5,3), (6,2), which are 5 outcomes. So, P(Sum=8 | Sum is Even) = 5/18.
MCQ 16. Bag A contains 2 white and 3 red balls. Bag B contains 4 white and 5 red balls. One bag is chosen at random and one ball is drawn. If the ball drawn is white, what is the probability that it came from Bag A?
- A. 9/19
- B. 10/19
- C. 2/5
- D. 4/9
Answer: A. 9/19 — Using Bayes' Theorem, P(Bag A | White) = [P(White | Bag A)P(Bag A)] / [P(White | Bag A)P(Bag A) + P(White | Bag B)P(Bag B)]. This is ((2/5)*(1/2)) / ((2/5)*(1/2) + (4/9)*(1/2)) = (1/5) / (1/5 + 2/9) = (1/5) / (19/45) = 9/19.
MCQ 17. A random variable X takes values 0, 1, 2. If P(X=0) = 0.3, P(X=1) = 0.4, then P(X=2) is:
- A. 0.3
- B. 0.4
- C. 0.5
- D. 0.2
Answer: A. 0.3 — The sum of probabilities for all possible values of a random variable must be 1. Therefore, P(X=2) = 1 - P(X=0) - P(X=1) = 1 - 0.3 - 0.4 = 0.3.
MCQ 18. If A and B are independent events, then P(A' ∩ B') is equal to:
- A. P(A') + P(B')
- B. P(A'|B')
- C. P(A')P(B')
- D. P(A' U B')
Answer: C. P(A')P(B') — If two events A and B are independent, then their complements A' and B' are also independent. For independent events, the probability of their intersection is the product of their individual probabilities.
MCQ 19. A game consists of tossing a fair coin 3 times. A player wins ₹10 for each head and loses ₹5 for each tail. Find the expected amount won/lost by the player.
- A. ₹7.50
- B. ₹5
- C. ₹2.50
- D. ₹0
Answer: A. ₹7.50 — Let X be the number of heads. E(X) = np = 3 * (1/2) = 1.5. The amount won is Y = 10X - 5(3-X) = 15X - 15. E(Y) = E(15X - 15) = 15E(X) - 15 = 15(1.5) - 15 = 22.5 - 15 = ₹7.50.
MCQ 20. A student passes if they score at least 50% marks. The probability of passing an exam is 0.7. If the student studies, the probability of passing is 0.9. If they don't study, the probability of passing is 0.3. What is the probability that the student studied?
- A. 2/3
- B. 1/2
- C. 0.7
- D. 0.9
Answer: A. 2/3 — Let P be 'passes' and S be 'studies'. P(P)=0.7, P(P|S)=0.9, P(P|S')=0.3. Using the Law of Total Probability: P(P) = P(P|S)P(S) + P(P|S')P(S'). So, 0.7 = 0.9P(S) + 0.3(1-P(S)). Solving for P(S) gives 0.7 = 0.9P(S) + 0.3 - 0.3P(S) => 0.4 = 0.6P(S) => P(S) = 0.4/0.6 = 2/3.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.