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Class 11 · Tamil Nadu State Board

Probability Theory Exercise Solutions | KalviGavani

TN Class 11 Maths Ch12 Introduction to Probability Theory exercise solutions — classical, empirical, axiomatic probability, Bayes theorem with worked examples.

Updated 15 June 202612 min read

1. How to Use This Page

This page contains complete exercise solutions for Chapter 12 — Introduction to Probability Theory of the Tamil Nadu State Board Class 11 Mathematics textbook. Attempt all problems on your own before checking these solutions. Each section includes worked numerical examples with detailed step-by-step reasoning.

2. Key Concepts (Quick Recap)

Three Approaches to Probability

ApproachFormulaWhen to Use
ClassicalP(E) = n(E)/n(S)Equally likely outcomes
Empirical (Relative Frequency)P(E) ≈ m/nBased on experiments/observations
AxiomaticBased on Kolmogorov axiomsTheoretical foundation

Axioms of Probability (Kolmogorov)

1. P(A) ≥ 0 for any event A (non-negativity) 2. P(S) = 1 for the sample space S (normalisation) 3. If A₁, A₂, ... are mutually exclusive, P(A₁ ∪ A₂ ∪ ...) = P(A₁) + P(A₂) + ... (additivity)

Key Formulas

P(A') = 1 − P(A) (Complement rule) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (Addition rule) P(A ∩ B) = P(A) · P(B|A) (Multiplication rule) P(A|B) = P(A ∩ B)/P(B) (Conditional probability) P(A ∩ B) = P(A) · P(B) (Independent events) Bayes' Theorem: P(Aᵢ|B) = P(B|Aᵢ) · P(Aᵢ) / Σⱼ P(B|Aⱼ) · P(Aⱼ) Total Probability: P(B) = Σᵢ P(B|Aᵢ) · P(Aᵢ)

Types of Events

TypeDefinition
Impossible eventP = 0
Sure/certain eventP = 1
Mutually exclusiveA ∩ B = ∅, P(A ∩ B) = 0
IndependentP(A ∩ B) = P(A) · P(B)
ExhaustiveA₁ ∪ A₂ ∪ ... = S
ComplementaryP(A') = 1 − P(A)

3. I. Choose the Best Answer (MCQs)

Q1. If P(A) = 0.3, P(B) = 0.4, and A, B are mutually exclusive, then P(A ∪ B) = ___

(a) 0.7   (b) 0.12   (c) 0.1   (d) 0.3

Answer(a) 0.7
ExplanationMutually exclusive means P(A ∩ B) = 0.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.3 + 0.4 − 0 = 0.7
Q2. If P(A) = 0.6 and P(BA) = 0.5, then P(A ∩ B) = ___

(a) 1.1   (b) 0.3   (c) 0.5   (d) 0.1

Answer(b) 0.3
Explanation
P(A ∩ B) = P(A) · P(B|A) = 0.6 × 0.5 = 0.3

Q3. Two dice are rolled. The probability of getting a sum of 7 is:

(a) 1/6   (b) 5/36   (c) 7/36   (d) 1/36

Answer(a) 1/6
Explanation
Total outcomes = 6 × 6 = 36 Favourable outcomes for sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 P(sum = 7) = 6/36 = 1/6

Q4. If A and B are independent events with P(A) = 0.4 and P(B) = 0.5, then P(A ∩ B) = ___

(a) 0.9   (b) 0.2   (c) 0.7   (d) 0.1

Answer(b) 0.2
Explanation
P(A ∩ B) = P(A) · P(B) = 0.4 × 0.5 = 0.2

Q5. A card is drawn from a standard deck. P(king or heart) = ___

(a) 4/13   (b) 1/52   (c) 16/52   (d) 17/52

Answer(a) 4/13
Explanation
P(king) = 4/52 P(heart) = 13/52 P(king and heart) = 1/52 (king of hearts) P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13

4. II. Fill in the Blanks

1. P(S) = 1 where S is the sample space.

2. If P(A) = 0.7, then P(A') = 0.3.

3. Two events A and B are mutually exclusive if P(A ∩ B) = 0.

4. Two events A and B are independent if P(A ∩ B) = P(A) · P(B).

5. P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

6. P(AB) = P(A ∩ B) / P(B).

7. If A and B are mutually exclusive and exhaustive, then P(A) + P(B) = 1.

8. The probability of an impossible event is 0.

9. If P(A) = 0, then A is called an impossible event.

10. Bayes' theorem: P(AB) = **P(BA) · P(A) / P(B)**.

11. The range of probability is 0 ≤ P(A) ≤ 1.

12. When a coin is tossed 3 times, the sample space has 8 outcomes.

5. III. True or False (with Corrections)

1. Statement: The probability of any event is always between 0 and 1 inclusive. Answer: True. 0 ≤ P(A) ≤ 1 for any event A.

2. Statement: If P(A) = 0.5 and P(B) = 0.3, then P(A ∪ B) is always 0.8. Answer: False. P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Only equals 0.8 if P(A ∩ B) = 0 (mutually exclusive).

3. Statement: Mutually exclusive events are always independent. Answer: False. If A and B are mutually exclusive with P > 0, then P(A ∩ B) = 0 ≠ P(A)·P(B), so they are NOT independent. They are dependent.

4. Statement: P(A') = 1 − P(A). Answer: True. This is the complement rule.

5. Statement: If P(AB) > P(A), then A and B are positively associated (not independent).
Answer: True. P(AB) = P(A ∩ B)/P(B). If P(AB) > P(A), then P(A ∩ B) > P(A)·P(B), meaning A and B are not independent.

6. IV. Match the Following

Column AColumn B
1. P(A ∪ B)a. P(A) · P(B) for independent events
2. P(A ∩ B)b. P(A) + P(B) − P(A ∩ B)
3. P(A')c. P(A ∩ B)/P(B)
4. P(A\B)d. 1 − P(A)
5. P(S)e. Impossible event probability
6. P(∅)f. 1
Answer1→b, 2→a, 3→d, 4→c, 5→f, 6→e

7. V. Answer Briefly (2-Mark Questions)

Q1. A die is rolled. Find P(getting a number greater than 4).

S = {1, 2, 3, 4, 5, 6}, n(S) = 6 E = {5, 6}, n(E) = 2 P(E) = 2/6 = 1/3

Q2. Two coins are tossed. Find P(at least one head).

S = {HH, HT, TH, TT}, n(S) = 4 E = {HH, HT, TH}, n(E) = 3 P(E) = 3/4 Alternative: P(at least one head) = 1 − P(no heads) = 1 − P(TT) = 1 − 1/4 = 3/4

Q3. If P(A) = 0.5, P(B) = 0.3, P(A ∪ B) = 0.6, find P(A ∩ B).

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 0.6 = 0.5 + 0.3 − P(A ∩ B) P(A ∩ B) = 0.8 − 0.6 = 0.2

Q4. A bag contains 5 red and 3 blue balls. One ball is drawn at random. Find P(red).

Total balls = 5 + 3 = 8 P(red) = 5/8

8. VI. Answer in Detail (5-Mark Questions)

Q1. A bag contains 4 red, 3 green, and 5 blue balls. Two balls are drawn at random without replacement. Find the probability that both are: (i) red (ii) different colours (iii) at least one is blue.

Total balls = 4 + 3 + 5 = 12 Total ways to draw 2 balls = C(12,2) = 12×11/2 = 66 (i) Both red: Favourable = C(4,2) = 4×3/2 = 6 P(both red) = 6/66 = 1/11 (ii) Different colours — pairs of different colours: P(red and green) = C(4,1)×C(3,1)/C(12,2) = 12/66 P(red and blue) = C(4,1)×C(5,1)/C(12,2) = 20/66 P(green and blue) = C(3,1)×C(5,1)/C(12,2) = 15/66 P(different colours) = (12 + 20 + 15)/66 = 47/66 (iii) At least one is blue: P(at least one blue) = 1 − P(no blue) P(no blue) = C(7,2)/C(12,2) = 21/66 P(at least one blue) = 1 − 21/66 = 45/66 = 15/22

Q2. A factory has three machines — M₁, M₂, M₃ — producing 50%, 30%, and 20% of the items respectively. The defective rates are 2%, 3%, and 5%. An item is selected at random and found defective. Find the probability it came from M₃ using Bayes' theorem.

Step 1: Define events A₁ = item from M₁, A₂ = item from M₂, A₃ = item from M₃ B = item is defective Step 2: Given probabilities P(A₁) = 0.50, P(A₂) = 0.30, P(A₃) = 0.20 P(B|A₁) = 0.02, P(B|A₂) = 0.03, P(B|A₃) = 0.05 Step 3: Find P(B) using law of total probability P(B) = P(A₁)·P(B|A₁) + P(A₂)·P(B|A₂) + P(A₃)·P(B|A₃) = (0.50)(0.02) + (0.30)(0.03) + (0.20)(0.05) = 0.010 + 0.009 + 0.010 = 0.029 Step 4: Apply Bayes' theorem P(A₃|B) = P(B|A₃) · P(A₃) / P(B) = (0.05)(0.20) / 0.029 = 0.010 / 0.029 = 10/29 ≈ 0.345 Answer: P(item from M₃ | defective) = 10/29 ≈ 0.345 or 34.5%

Q3. Two dice are thrown. Find the probability that: (i) the sum is at most 5, (ii) the sum is a prime number, (iii) the two numbers are equal.

Sample space: 6 × 6 = 36 outcomes (i) Sum at most 5 (sum ≤ 5): Sum = 2: (1,1) → 1 outcome Sum = 3: (1,2), (2,1) → 2 outcomes Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes Favourable = 1 + 2 + 3 + 4 = 10 P(sum ≤ 5) = 10/36 = 5/18 (ii) Sum is a prime number (primes from 2 to 12: 2, 3, 5, 7, 11): Sum = 2: 1 outcome Sum = 3: 2 outcomes Sum = 5: 4 outcomes Sum = 7: 6 outcomes Sum = 11: 2 outcomes Favourable = 1 + 2 + 4 + 6 + 2 = 15 P(prime sum) = 15/36 = 5/12 (iii) Two numbers are equal (doubles): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes P(doubles) = 6/36 = 1/6

Q4. A box contains 6 white and 4 black balls. Three balls are drawn at random without replacement. Find the probability that: (i) all are white, (ii) exactly 2 are white, (iii) at least one is black.

Total balls = 6 + 4 = 10 Total ways to draw 3 balls = C(10,3) = 120 (i) All white: Favourable = C(6,3) = 20 P(all white) = 20/120 = 1/6 (ii) Exactly 2 white (and 1 black): Favourable = C(6,2) × C(4,1) = 15 × 4 = 60 P(exactly 2 white) = 60/120 = 1/2 (iii) At least one black: P(at least one black) = 1 − P(no black) P(no black) = P(all white) = 1/6 P(at least one black) = 1 − 1/6 = 5/6 Alternative for (iii): P(1 black) + P(2 black) + P(3 black) = C(4,1)C(6,2)/C(10,3) + C(4,2)C(6,1)/C(10,3) + C(4,3)/C(10,3) = 60/120 + 36/120 + 4/120 = 100/120 = 5/6 ✓

Q5. Verify whether A and B are independent given P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7.

Step 1: Find P(A ∩ B) using addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 0.7 = 0.5 + 0.4 − P(A ∩ B) P(A ∩ B) = 0.9 − 0.7 = 0.2 Step 2: Check independence For independence: P(A ∩ B) = P(A) · P(B) P(A) · P(B) = 0.5 × 0.4 = 0.2 Step 3: Compare P(A ∩ B) = 0.2 = P(A) · P(B) Since P(A ∩ B) = P(A) · P(B), events A and B ARE independent. Additional verification — check P(A|B): P(A|B) = P(A ∩ B)/P(B) = 0.2/0.4 = 0.5 = P(A) ✓ Since P(A|B) = P(A), the events are independent.

9. Common Mistakes to Avoid

1. Forgetting to subtract P(A ∩ B): P(A ∪ B) = P(A) + P(B) − P(A ∩ B), not P(A) + P(B). If you don't subtract the intersection, you're double-counting it.

2. Confusing mutually exclusive with independent: Mutually exclusive means P(A ∩ B) = 0. Independent means P(A ∩ B) = P(A)·P(B). Two events with positive probability cannot be both mutually exclusive AND independent.

3. Wrong sample space for drawing without replacement: When drawing multiple items without replacement, use combinations C(n,r), not simple multiplication. The total number of outcomes changes after each draw.

4. Probability outside [0, 1]: All probabilities must be between 0 and 1. If your calculation gives a value outside this range, you've made an error.

5. **Confusing P(AB) with P(BA):** P(AB) = P(A ∩ B)/P(B) is NOT the same as P(BA) = P(A ∩ B)/P(A). These have different denominators. Bayes' theorem relates the two.

6. Not considering all cases in "at least one" problems: Use the complement: P(at least one) = 1 − P(none). This is often much easier than adding all individual cases.

7. Assuming equally likely outcomes when they aren't: The classical formula P(E) = n(E)/n(S) only works when outcomes are equally likely. In real-world problems, this may not be the case.

10. Exam Checklist

  • [ ] Know three approaches to probability: classical, empirical, axiomatic
  • [ ] Can apply the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
  • [ ] Can apply the multiplication rule: P(A ∩ B) = P(A) · P(B|A)
  • [ ] Understand conditional probability: P(A|B) = P(A ∩ B)/P(B)
  • [ ] Can apply Bayes' theorem to find posterior probabilities
  • [ ] Know the law of total probability: P(B) = Σ P(B|Aᵢ)P(Aᵢ)
  • [ ] Can distinguish mutually exclusive (P(A∩B) = 0) from independent (P(A∩B) = P(A)·P(B))
  • [ ] Can solve problems with cards, dice, coins, and balls
  • [ ] Use complement rule for "at least one" problems: P(at least one) = 1 − P(none)
  • [ ] Know that P(A') = 1 − P(A) and can use it to simplify calculations
  • [ ] Can count using combinations C(n,r) = n!/(r!(n−r)!)

Frequently Asked Questions

P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
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