Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
This page provides complete solutions for the Probability Theory chapter from the Samacheer Kalvi 11th Standard Business Mathematics textbook (Tamil Nadu State Board, Commerce stream).
How the solutions are organised:
- MCQs come with the correct option and a short reason.
- Fill in the Blanks give the exact answer with the formula or logic behind it.
- True or False statements include corrections for false ones.
- Match the Following is presented as a clear table.
- Brief Answers show key steps; skip the steps only if the calculation is trivial.
- Detail Answers include full worked-out numerical examples in code blocks so you can follow every calculation.
Tips for exam preparation:
1. Memorise the key formulas first, then practise applying them. 2. Work through the detail section examples on paper before checking the solution. 3. Pay special attention to the Common Mistakes section — those errors cost the most marks. 4. Use the Exam Checklist as a final review before your exam.
Key Concepts (Quick Recap)
Sample Space and Events
| Concept | Definition | Example |
|---|---|---|
| Random Experiment | An experiment whose outcome cannot be predicted in advance | Tossing a coin, rolling a die |
| Sample Space (S) | Set of all possible outcomes of a random experiment | Rolling a die: S = {1, 2, 3, 4, 5, 6} |
| Sample Point | Each individual outcome in the sample space | Rolling a 4 |
| n(S) | Number of elements in the sample space | n(S) = 6 for a single die |
Types of Events
| Type | Definition | Condition |
|---|---|---|
| Simple (Elementary) Event | An event with exactly one outcome | A = {4} when rolling a die |
| Compound Event | An event with two or more outcomes | A = {2, 4, 6} (even numbers) |
| Sure (Certain) Event | Event equal to the entire sample space | P(S) = 1 |
| Impossible Event | Event with no outcomes, equals empty set | P(∅) = 0 |
| Mutually Exclusive Events | Events that cannot occur simultaneously | A ∩ B = ∅ |
| Exhaustive Events | Events whose union equals the sample space | A ∪ B = S |
| Independent Events | Occurrence of one does not affect the other | P(A ∩ B) = P(A) · P(B) |
Core Probability Formulas
| Theorem / Definition | Formula | |||
|---|---|---|---|---|
| Classical Definition | P(A) = m/n, where m = favourable outcomes, n = total outcomes | |||
| Axiomatic | 0 ≤ P(A) ≤ 1; P(S) = 1; P(A ∪ B) = P(A) + P(B) if A ∩ B = ∅ | |||
| Complementary | P(A′) = 1 − P(A) | |||
| Addition Theorem | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | |||
| Addition (Mutually Exclusive) | P(A ∪ B) = P(A) + P(B) | |||
| Multiplication Theorem | P(A ∩ B) = P(A) · P(B \ | A) = P(B) · P(A \ | B) | |
| Multiplication (Independent) | P(A ∩ B) = P(A) · P(B) | |||
| Conditional Probability | P(A \ | B) = P(A ∩ B) / P(B), provided P(B) ≠ 0 | ||
| Bayes' Theorem | P(Ei \ | A) = [P(Ei) · P(A \ | Ei)] / Σ[P(Ej) · P(A \ | Ej)] |
| Total Probability Theorem | P(A) = Σ P(Ei) · P(A \ | Ei), where E1, E2, …, En are mutually exclusive and exhaustive |
I. Choose the Best Answer (MCQs)
1. If P(A) = 0.3, P(B) = 0.4 and A, B are mutually exclusive, then P(A ∪ B) is
(a) 0.7   (b) 0.12   (c) 0.1   (d) 0.3
Since A and B are mutually exclusive, P(A ∩ B) = 0.
2. If A and B are independent events with P(A) = 1/2 and P(B) = 1/3, then P(A ∩ B) is
(a) 5/6   (b) 1/6   (c) 2/3   (d) 1/3
For independent events, P(A ∩ B) = P(A) · P(B).
3. If P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.9, then P(A ∩ B) is
(a) 0.1   (b) 0.2   (c) 0.3   (d) 0.4
Rearranging the addition theorem:
4. If P(A) = 3/7 and P(B) = 2/7 and A, B are mutually exclusive, then P(A′ ∩ B′) is
(a) 2/7   (b) 5/7   (c) 3/7   (d) 1/7
Since A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B). By De Morgan's law, P(A′ ∩ B′) = P((A ∪ B)′) = 1 − P(A ∪ B).
5. A card is drawn at random from a pack of 52 cards. The probability that it is a king or a heart is
(a) 4/13   (b) 1/52   (c) 17/52   (d) 4/52
Let A = king, B = heart. The king of hearts is in both events.
6. If P(A) = 2/5, P(B) = 1/3 and P(A ∩ B) = 1/15, then P(A|B) is
(a) 1/5   (b) 2/5   (c) 1/3   (d) 1/15
7. Three factories F1, F2, F3 produce 50%, 30%, 20% of items. Defect rates are 2%, 4%, 5%. A randomly chosen item is found defective. The probability it came from F2 is
(a) 12/31   (b) 10/31   (c) 9/31   (d) 6/31
Using Bayes' theorem:
II. Fill in the Blanks
1. The probability of an impossible event is ______.
An impossible event has no favourable outcomes, so P(∅) = 0.
2. If P(A) = 0.4, then P(A′) = ______.
By the complementary rule: P(A′) = 1 − P(A) = 1 − 0.4 = 0.6.
3. If A and B are mutually exclusive, then P(A ∪ B) = ______.
For mutually exclusive events, A ∩ B = ∅, so P(A ∩ B) = 0 and the addition theorem simplifies.
4. The probability of drawing a face card from a pack of 52 cards is ______.
5. If A and B are independent with P(A) = 0.5 and P(B) = 0.4, then P(A ∪ B) = ______.
6. For any event A, P(A) lies between ______ and ______.
Probability is always between 0 (impossible) and 1 (certain), inclusive.
7. If P(A) = 2/3 and P(B|A) = 1/4, then P(A ∩ B) = ______.
8. The sum of probabilities of all mutually exclusive and exhaustive events is ______.
If E1, E2, …, En are mutually exclusive and exhaustive, then P(E1) + P(E2) + … + P(En) = P(S) = 1.
9. When two dice are thrown, the number of elements in the sample space is ______.
Each die has 6 faces. n(S) = 6 × 6 = 36.
10. The probability of getting a sum of 7 when two dice are rolled is ______.
11. If P(A ∩ B) = P(A) · P(B), then A and B are ______ events.
This is the defining condition for statistical independence of two events.
12. Bayes' theorem gives the ______ probability of a cause given the observed effect.
Bayes' theorem updates prior probabilities using observed evidence to produce posterior probabilities.
III. True or False
1. If A and B are mutually exclusive, then P(A ∩ B) = P(A) × P(B).
False. Correction: Mutually exclusive means P(A ∩ B) = 0. The product P(A) × P(B) is non-zero unless one of the events has probability zero. The product formula applies to independent events, not mutually exclusive events.
2. The probability of any event always lies between 0 and 1 (inclusive).
True. By the axioms of probability, for any event A, 0 ≤ P(A) ≤ 1.
3. If P(A|B) > P(A), then the occurrence of B increases the probability of A.
| True. When P(A | B) > P(A), the conditional probability exceeds the unconditional probability, meaning B positively influences A. |
|---|
4. Two independent events are always mutually exclusive.
False. Correction: Independent events can occur together. In fact, if A and B are independent and both have non-zero probability, then P(A ∩ B) = P(A) × P(B) > 0, so they are not mutually exclusive. Mutual exclusivity and independence are opposite concepts (except in the trivial case where one event has probability zero).
5. Bayes' theorem is useful for revising probability estimates based on new information.
True. Bayes' theorem updates prior probabilities with observed evidence (likelihood) to produce posterior probabilities — a core idea in statistical inference and decision-making.
IV. Match the Following
| Column A | Column B | |
|---|---|---|
| 1. P(A) + P(A′) | (a) P(A) + P(B) | |
| 2. P(A ∪ B) when A, B are mutually exclusive | (b) 1 | |
| 3. P(A ∩ B) when A, B are independent | (c) P(A ∩ B) / P(B) | |
| 4. P(A \ | B) | (d) P(A) · P(B) |
| 5. P(S) | (e) 1 | |
| 6. P(A ∩ B) when A, B are mutually exclusive | (f) 0 |
- 1 → (b) 1 — A and its complement partition the sample space.
- 2 → (a) P(A) + P(B) — No overlap for mutually exclusive events.
- 3 → (d) P(A) · P(B) — Defining property of independence.
- 4 → (c) P(A ∩ B) / P(B) — Definition of conditional probability.
- 5 → (e) 1 — Probability of the entire sample space.
- 6 → (f) 0 — Mutually exclusive events have empty intersection.
V. Answer Briefly
1. Define sample space and give an example.
A sample space is the set of all possible outcomes of a random experiment, denoted by S.
Example: When two coins are tossed simultaneously:
Each element (HH, HT, TH, TT) is called a sample point.
2. State the addition theorem of probability for two events.
The addition theorem states that for any two events A and B:
Special cases:
- If A and B are mutually exclusive: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
- For three events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C).
3. Distinguish between mutually exclusive events and independent events.
| Aspect | Mutually Exclusive | Independent |
|---|---|---|
| Definition | A ∩ B = ∅ (cannot occur together) | P(A ∩ B) = P(A) · P(B) |
| Joint probability | P(A ∩ B) = 0 | P(A ∩ B) = P(A) · P(B) > 0 |
| Occurrence | If A happens, B cannot happen | If A happens, B's probability is unchanged |
| Relationship | Occurrence of one prevents the other | Occurrence of one does not affect the other |
| Example | Drawing a king and drawing a queen in one draw | Drawing a king on the 1st draw and drawing a king on the 2nd draw (with replacement) |
4. What is conditional probability? Write its formula.
Conditional probability measures the probability of event A occurring, given that event B has already occurred.
Interpretation: We restrict our attention to the reduced sample space B, and find the proportion of B where A also occurs.
Example: In a class of 40 students, 25 like Maths and 15 like both Maths and Science. If a student likes Science, what is the probability they also like Maths?
VI. Answer in Detail
1. State and prove the addition theorem and multiplication theorem. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is (i) a king or a queen, (ii) a heart or a diamond.
Addition Theorem
Statement: For any two events A and B of a sample space S,
Proof:
Multiplication Theorem
Statement: For any two events A and B,
Proof: By definition of conditional probability:
Worked Numerical Example
A card is drawn from a well-shuffled pack of 52 cards.
(i) Probability of drawing a king or a queen:
(ii) Probability of drawing a heart or a diamond:
2. Explain conditional probability with a worked example. Two dice are rolled. Given that the sum is 7, find the probability that one of the dice shows a 3.
Definition and Formula
| Conditional probability P(A | B) is the probability of event A occurring given that event B has already occurred. |
|---|
Key properties:
Worked Example
Two dice are rolled. Given that the sum is 7, find the probability that one die shows 3.
Verification using the formula:
| Both methods give the same answer: **P(A | B) = 1/3**. |
|---|
3. State and prove Bayes' theorem. Factory F1 produces 40% of items, F2 produces 35%, and F3 produces 25%. The defect rates are 3%, 2%, and 1% respectively. If a randomly selected item is found defective, find the probability that it was produced by F1, F2, and F3.
Statement of Bayes' Theorem
If E1, E2, …, En are mutually exclusive and exhaustive events with P(Ei) > 0 for all i, and A is any event with P(A) > 0, then:
Proof of Bayes' Theorem
Worked Numerical Example
Three factories produce items with the following data:
| Factory | Production Share | Defect Rate | |
|---|---|---|---|
| F1 | 40% (P(F1) = 0.40) | 3% (P(D\ | F1) = 0.03) |
| F2 | 35% (P(F2) = 0.35) | 2% (P(D\ | F2) = 0.02) |
| F3 | 25% (P(F3) = 0.25) | 1% (P(D\ | F3) = 0.01) |
A randomly selected item is found defective. Find the probability it came from each factory.
Common Mistakes to Avoid
1. Confusing mutual exclusivity with independence. Mutually exclusive events cannot occur together (P(A ∩ B) = 0). Independent events can occur together (P(A ∩ B) = P(A) × P(B)). They are opposite concepts, not the same.
2. Forgetting to subtract P(A ∩ B) in the addition theorem. The formula is P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Adding P(A) and P(B) directly overcounts the intersection. Only skip the subtraction when events are mutually exclusive.
| 3. Dividing by P(A) instead of P(B) in conditional probability. P(A | B) = P(A ∩ B) / P(B). The condition always goes in the denominator. P(A | B) ≠ P(A ∩ B) / P(A). |
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4. Using n(S) = 6 for two dice. A single die has 6 outcomes. Two dice have 36 outcomes (6 × 6). Similarly, three dice have 216 outcomes.
5. Counting overlapping cases twice in card problems. A king of hearts belongs to both "kings" and "hearts." Always identify and subtract P(A ∩ B) to avoid double counting.
| 6. Applying multiplication theorem without checking independence. P(A ∩ B) = P(A) × P(B) is valid only when A and B are independent. For dependent events, use P(A ∩ B) = P(A) × P(B | A). |
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| 7. Forgetting the denominator in Bayes' theorem. The denominator Σ P(Ej) × P(A | Ej) is the total probability P(A) computed over all causes. Omitting any cause gives an incorrect result. |
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| 8. **Assuming P(A | B) = P(B | A).** These are generally not equal. P(A | B) = P(A ∩ B)/P(B) while P(B | A) = P(A ∩ B)/P(A). The denominators differ. Confusing the two is the "prosecutor's fallacy" in legal contexts. |
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Exam Checklist
Use this checklist before your exam. Tick every item you are confident about.
- [ ] I can define sample space, sample point, and n(S) for common experiments (coins, dice, cards).
- [ ] I can classify events: simple, compound, mutually exclusive, exhaustive, independent, and complementary.
- [ ] I know the classical definition: P(A) = m/n (favourable/total) and can apply it to basic problems.
- [ ] I can state and apply the addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), including the mutually exclusive special case.
- [ ] I can state and apply the multiplication theorem: P(A ∩ B) = P(A) × P(B|A) and the independent-events simplification.
- [ ] I can compute conditional probability P(A|B) = P(A ∩ B)/P(B) and interpret the result.
- [ ] I can state and apply Bayes' theorem, including computing total probability as the denominator.
- [ ] I can solve card-probability problems correctly, accounting for overlapping categories (e.g., king and heart).
- [ ] I can solve dice problems, remembering n(S) = 36 for two dice, and listing favourable outcomes systematically.
- [ ] I can distinguish between mutually exclusive and independent events with examples, and I will not confuse them on the exam.