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Probability

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Chapter Notes

Overview
This chapter formalizes the study of chance and uncertainty by defining events as subsets of a sample space and introducing an axiomatic framework for probability. It builds on intuitive ideas of likelihood to provide a rigorous mathematical foundation, covering event types, algebra of events, and probability calculations using axioms. The one big idea a student must walk away with is that probability is a real-valued function on subsets of a sample space satisfying three axioms, and all results—including the addition theorem and complement rule—follow from these axioms.
Key Concepts (17)
Random experiment
An experiment whose outcome cannot be predicted in advance, but all possible outcomes are known.
e.g. Tossing a coin, rolling a die, drawing a card from a shuffled deck.
Sample space
The set of all possible outcomes of a random experiment, denoted by S.
e.g. For tossing a coin twice, S = {HH, HT, TH, TT}.
Event
Any subset E of a sample space S is called an event.
e.g. In the experiment of throwing a die, the event 'a number less than 4 appears' corresponds to E = {1, 2, 3}.
Occurrence of an event
An event E is said to have occurred if the outcome ω of the experiment is such that ω ∈ E.
e.g. If a die shows 1, then the event 'a number less than 4 appears' has occurred.
Impossible event
The empty set φ is called an impossible event because no outcome satisfies its condition.
e.g. Getting a multiple of 7 on a die is an impossible event.
Sure event
The whole sample space S is called a sure event because every outcome ensures its occurrence.
e.g. Getting a number that is odd or even when rolling a die is a sure event.
Simple event
An event that has only one sample point of a sample space is called a simple (or elementary) event.
e.g. In tossing two coins, {HH} is a simple event.
Compound event
An event that has more than one sample point is called a compound event.
e.g. In tossing a coin thrice, the event 'exactly one head appeared' = {HTT, THT, TTH} is a compound event.
Complementary event
For every event A, the complementary event A' (or 'not A') is the set of all outcomes in S that are not in A.
e.g. If A = {HTH, HHT, THH} in three coin tosses, then A' = {HHH, HTT, THT, TTH, TTT}.
Event 'A or B'
The event 'A or B' is the union A ∪ B, consisting of outcomes that are in A or in B or in both.
e.g. If A = {2, 3, 5} and B = {1, 3, 5} from a die roll, then A or B = {1, 2, 3, 5}.
Event 'A and B'
The event 'A and B' is the intersection A ∩ B, consisting of outcomes common to both A and B.
e.g. If A = {2, 3, 5} and B = {1, 3, 5}, then A and B = {3, 5}.
Event 'A but not B'
The event 'A but not B' is the set difference A – B = A ∩ B'.
e.g. If A = {2, 3, 5} and B = {1, 3, 5}, then A but not B = {2}.
Mutually exclusive events
Two events A and B are called mutually exclusive if they cannot occur simultaneously, i.e., A ∩ B = φ.
e.g. In rolling a die, 'odd number' and 'even number' are mutually exclusive.
Exhaustive events
Events E1, E2, ..., En are called exhaustive if their union is the entire sample space S, i.e., at least one of them necessarily occurs.
e.g. In rolling a die, A = {1,2,3}, B = {3,4}, C = {5,6} are exhaustive because A ∪ B ∪ C = S.
Mutually exclusive and exhaustive events
Events that are pairwise disjoint (mutually exclusive) and whose union is the sample space (exhaustive).
e.g. In tossing a coin three times, A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} are mutually exclusive and exhaustive.
Axiomatic approach to probability
Probability P is a real-valued function defined on the power set of the sample space S, mapping to [0,1], satisfying three axioms: (i) P(E) ≥ 0 for any event E, (ii) P(S) = 1, (iii) if E and F are mutually exclusive, then P(E ∪ F) = P(E) + P(F).
e.g. For a fair coin, P(H) = 1/2, P(T) = 1/2 satisfies all axioms.
Equally likely outcomes
Outcomes are equally likely if each simple event has the same probability, i.e., P(ωi) = 1/n for a sample space with n outcomes.
e.g. In rolling a fair die, each face has probability 1/6.
Formulas & Equations (7)
P(E) = n(E) / n(S)
Reads as: Probability of event E equals number of outcomes favourable to E divided by total number of possible outcomes
Variables: n(E) = number of outcomes in event E (dimensionless), n(S) = total number of outcomes in sample space S (dimensionless)
Use when: When all outcomes in the sample space are equally likely
⚠ Students forget to check that outcomes are equally likely before applying this formula; they use it even when probabilities are not uniform.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Reads as: Probability of A or B equals probability of A plus probability of B minus probability of A and B
Variables: P(A) = probability of event A, P(B) = probability of event B, P(A ∩ B) = probability of both A and B occurring
Use when: When finding the probability of the union of two events that are not necessarily mutually exclusive
⚠ Students forget to subtract P(A ∩ B) when events are not mutually exclusive, leading to double-counting.
P(A ∪ B) = P(A) + P(B) (when A and B are mutually exclusive)
Reads as: For mutually exclusive events, probability of A or B equals probability of A plus probability of B
Variables: P(A) = probability of event A, P(B) = probability of event B
Use when: When events A and B are mutually exclusive (disjoint)
⚠ Students apply this formula without verifying that A ∩ B = φ.
P(not A) = P(A') = 1 – P(A)
Reads as: Probability of not A equals 1 minus probability of A
Variables: P(A) = probability of event A
Use when: When finding the probability of the complement of an event
⚠ Students sometimes write P(A') = 1 – P(A) but then incorrectly compute P(A) when A is compound.
P(φ) = 0
Reads as: Probability of an impossible event is zero
Variables: φ = empty set (impossible event)
Use when: Derived from axiom (iii) by taking F = φ; used in proofs and to verify consistency
⚠ Students think P(φ) = 0 is an axiom, but it is a consequence of the axioms.
P(ω1) + P(ω2) + ... + P(ωn) = 1
Reads as: Sum of probabilities of all elementary events in the sample space equals 1
Variables: ωi = elementary outcomes of the sample space
Use when: To verify a probability assignment or to find an unknown probability when all but one are known
⚠ Students forget that this sum must be exactly 1; they sometimes assign probabilities that sum to more or less.
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
Reads as: Probability of A or B or C equals sum of individual probabilities minus sum of pairwise intersections plus triple intersection
Variables: P(A), P(B), P(C) = probabilities of events A, B, C; P(A ∩ B), etc. = pairwise intersections; P(A ∩ B ∩ C) = triple intersection
Use when: When finding probability of union of three events (inclusion-exclusion principle)
⚠ Students forget to add back the triple intersection after subtracting pairwise intersections.
How It Works
Defining an event from a sample space
  1. Step 1: Identify the random experiment and list all possible outcomes to form the sample space S.
  2. Step 2: Describe the condition of interest (e.g., 'exactly one head').
  3. Step 3: Collect all outcomes from S that satisfy that condition into a set E.
  4. Step 4: This set E is the event associated with that condition.
💡 An event is always a subset of the sample space; the empty set and the whole sample space are also events.
Checking if an event has occurred
  1. Step 1: Perform the random experiment and note the actual outcome ω.
  2. Step 2: Check if ω belongs to the event E.
  3. Step 3: If ω ∈ E, the event E has occurred; if ω ∉ E, the event E has not occurred.
💡 Occurrence is determined by the actual outcome, not by the event's description alone.
Constructing complementary, union, intersection, and difference events
  1. Step 1: For complementary event A', take all elements of S that are not in A.
  2. Step 2: For 'A or B', take the union A ∪ B.
  3. Step 3: For 'A and B', take the intersection A ∩ B.
  4. Step 4: For 'A but not B', take the set difference A – B = A ∩ B'.
💡 These operations follow the same rules as set operations; De Morgan's laws apply: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'.
Determining if events are mutually exclusive
  1. Step 1: Find the intersection A ∩ B of the two events.
  2. Step 2: If A ∩ B = φ, the events are mutually exclusive; otherwise, they are not.
💡 Simple events of a sample space are always mutually exclusive.
Determining if events are exhaustive
  1. Step 1: Find the union of all given events: E1 ∪ E2 ∪ ... ∪ En.
  2. Step 2: If this union equals the sample space S, the events are exhaustive; otherwise, they are not.
💡 Exhaustive events cover all possible outcomes; at least one of them must occur.
Assigning probabilities using the axiomatic approach
  1. Step 1: Define the sample space S = {ω1, ω2, ..., ωn}.
  2. Step 2: Assign a real number P(ωi) to each outcome such that 0 ≤ P(ωi) ≤ 1 for each i.
  3. Step 3: Ensure that the sum of all P(ωi) equals 1.
  4. Step 4: For any event A, define P(A) as the sum of probabilities of outcomes in A.
💡 There are infinitely many valid probability assignments; the assignment must satisfy both non-negativity and total sum = 1.
Calculating probability for equally likely outcomes
  1. Step 1: Count the total number of outcomes in the sample space, n(S).
  2. Step 2: Count the number of outcomes favourable to the event E, n(E).
  3. Step 3: Apply the formula P(E) = n(E) / n(S).
💡 This formula is valid only when all outcomes are equally likely; otherwise, use the axiomatic sum of probabilities.
Using the addition theorem for two events
  1. Step 1: Compute P(A) and P(B) individually.
  2. Step 2: Compute P(A ∩ B).
  3. Step 3: Apply P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
💡 If A and B are mutually exclusive, P(A ∩ B) = 0, so the formula reduces to P(A ∪ B) = P(A) + P(B).
Using the complement rule
  1. Step 1: Identify the event A for which you need the complement.
  2. Step 2: Compute P(A).
  3. Step 3: Apply P(A') = 1 – P(A).
💡 A and A' are mutually exclusive and exhaustive, so P(A) + P(A') = 1.
Exam Traps — Don't Make These Mistakes
⚠ Confusing 'mutually exclusive' with 'exhaustive'
✗ Students often write that two events are mutually exclusive if they cover all outcomes, or that exhaustive events cannot occur together.
→ Mutually exclusive means A ∩ B = φ (no common outcome), while exhaustive means A ∪ B = S (cover all outcomes). They are independent properties.
✓ Mutually exclusive events cannot occur simultaneously; exhaustive events together cover the entire sample space.
⚠ Using P(E) = n(E)/n(S) without checking equally likely condition
✗ Students directly write P(E) = number of favourable outcomes / total outcomes even when outcomes are not equally likely.
→ The formula P(E) = n(E)/n(S) is valid only when all outcomes are equally likely. If not, probabilities must be assigned via the axiomatic sum.
✓ First state that outcomes are equally likely, then use P(E) = n(E)/n(S). Otherwise, use P(E) = sum of probabilities of outcomes in E.
⚠ Forgetting to subtract P(A ∩ B) in the addition theorem
✗ Students write P(A ∪ B) = P(A) + P(B) even when A and B are not mutually exclusive.
→ When A and B are not disjoint, the outcomes in A ∩ B are counted twice in P(A) + P(B), so they must be subtracted once.
✓ Always use P(A ∪ B) = P(A) + P(B) – P(A ∩ B). Only if A and B are mutually exclusive does P(A ∩ B) = 0.
⚠ Misidentifying the sample space for compound experiments
✗ For tossing two coins, students write S = {HH, HT, TT} or S = {H, T}.
→ The sample space must list all ordered pairs: {HH, HT, TH, TT}. Missing TH is a common error.
✓ For two coin tosses, the sample space is S = {HH, HT, TH, TT} with 4 equally likely outcomes.
⚠ Confusing 'at least one' with 'exactly one'
✗ Students treat 'at least one head' as the same as 'exactly one head' when counting favourable outcomes.
→ 'At least one head' includes outcomes with one head, two heads, etc. 'Exactly one head' includes only those with exactly one.
✓ For 'at least one head' in three tosses, favourable outcomes = {HTT, THT, TTH, HHT, HTH, THH, HHH} (7 outcomes). For 'exactly one head', favourable = {HTT, THT, TTH} (3 outcomes).
⚠ Incorrectly applying De Morgan's laws
✗ Students write (A ∪ B)' = A' ∪ B' or (A ∩ B)' = A' ∩ B'.
→ De Morgan's laws state that the complement of a union is the intersection of complements, and vice versa.
✓ (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'.
Compare & Contrast
Definition and set property
Mutually exclusive events
Exhaustive events
Mutually exclusive events have empty intersection (A ∩ B = φ), meaning they cannot occur together. Exhaustive events have union equal to the sample space (A ∪ B = S), meaning at least one must occur.
Number of sample points
Simple event
Compound event
A simple event has exactly one sample point (e.g., {HH} in two coin tosses). A compound event has more than one sample point (e.g., {HT, TH} for exactly one head).
Exam Focus
Marks pattern: Typically 1 MCQ (1 mark) on basic probability or event types, 1 short answer (2-3 marks) on addition theorem or complement, and 1 long answer (4-5 marks) on a word problem involving selection (committees, cards, dice) with multiple sub-parts.
Most asked: Probability of an event using equally likely outcomes (cards, dice, coins, balls from bag) · Addition theorem: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) · Complement rule: P(not A) = 1 – P(A) · Mutually exclusive and exhaustive events · Problems involving 'at least one' or 'at most' conditions
Don't skip: Word problems on drawing cards or selecting balls from a bag with conditions like 'at least one', 'neither', 'either-or'; Problems involving union and intersection of events with given probabilities (e.g., P(A)=0.5, P(B)=0.4, P(A∩B)=0.2, find P(A∪B)); Verification of mutually exclusive and exhaustive events from a given sample space
One-Line Revision (read aloud before the exam)
  1. A random experiment has a known set of possible outcomes but an unpredictable single outcome.
  2. The sample space S is the set of all possible outcomes of a random experiment.
  3. An event is any subset of the sample space.
  4. The empty set φ is an impossible event; the sample space S is a sure event.
  5. A simple event has exactly one sample point; a compound event has more than one.
  6. Two events A and B are mutually exclusive if A ∩ B = φ.
  7. Events E1, E2, ..., En are exhaustive if their union equals the sample space S.
  8. The complementary event A' consists of all outcomes in S that are not in A.
  9. Probability P is a function from the power set of S to [0,1] satisfying P(S)=1 and additivity for mutually exclusive events.
  10. For equally likely outcomes, P(E) = number of favourable outcomes / total number of outcomes.
  11. P(A ∪ B) = P(A) + P(B) – P(A ∩ B) for any two events A and B.
  12. P(not A) = 1 – P(A).
  13. De Morgan's laws: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'.
  14. If A and B are mutually exclusive, then P(A ∪ B) = P(A) + P(B).
  15. The sum of probabilities of all elementary events in a sample space is 1.
  16. For three events, P(A ∪ B ∪ C) = P(A)+P(B)+P(C) – P(A∩B) – P(A∩C) – P(B∩C) + P(A∩B∩C).

MCQs — 10 Questions with Answers

Q1. When a die is rolled, which of the following represents a simple event?

  • A. {2, 4, 6}
  • B. {5} ✓
  • C. {1, 3, 5}
  • D. {2, 3, 5}

Answer: B — A simple event contains exactly one sample point; {5} contains only the outcome 5, making it a simple event.

Q2. In tossing two coins, let A be the event 'at least one head' and B be the event 'exactly one tail'. Which statement is true?

  • A. A ∩ B = φ
  • B. A ∩ B = {HT, TH} ✓
  • C. A ∩ B = {HH}
  • D. A ∩ B = {TT}

Answer: B — A = {HH, HT, TH} and B = {HT, TH}; their intersection contains outcomes common to both, which are {HT, TH}.

Q3. If E is the event 'getting a multiple of 7 on rolling a die', then E is classified as:

  • A. A simple event
  • B. A compound event
  • C. An impossible event ✓
  • D. A sure event

Answer: C — No outcome on a die (1–6) is a multiple of 7, so E = φ, making it an impossible event.

Q4. Two events A and B are mutually exclusive if and only if:

  • A. A ∪ B = S
  • B. A ∩ B = φ ✓
  • C. A ⊂ B
  • D. A′ = B

Answer: B — Mutually exclusive events cannot occur simultaneously, which means they have no common outcomes, so A ∩ B = φ.

Q5. In rolling a die, let A = 'prime number' = {2, 3, 5} and B = 'number greater than 3' = {4, 5, 6}. What is A but not B?

  • A. {2, 3} ✓
  • B. {5}
  • C. {4, 6}
  • D. {2, 3, 4, 5, 6}

Answer: A — A but not B = A − B = A ∩ B′; outcomes in A that are NOT in B are 2 and 3, so the answer is {2, 3}.

Q6. Which of the following is NOT correct about complementary events A and A′?

  • A. A ∩ A′ = φ
  • B. A ∪ A′ = S
  • C. A′ = S − A
  • D. A ∩ A′ = S ✓

Answer: D — Complementary events are disjoint (A ∩ A′ = φ), not equal to the entire sample space; option D is incorrect.

Q7. In the experiment of tossing three coins, the event 'at most two heads' corresponds to which set?

  • A. {HHH}
  • B. {HTT, THT, TTH, HHT, HTH, THH}
  • C. {TTT}
  • D. {TTT, HTT, THT, TTH, HHT, HTH, THH} ✓

Answer: D — 'At most two heads' means 0, 1, or 2 heads; this includes all 8 outcomes except HHH, giving {TTT, HTT, THT, TTH, HHT, HTH, THH}.

Q8. Assertion (A): Simple events of a sample space are always mutually exclusive. Reason (R): Each simple event contains exactly one sample point, and no two simple events share the same sample point.

  • A. Both A and R are true, and R is the correct explanation of A ✓
  • B. Both A and R are true, but R is not the correct explanation of A
  • C. A is true but R is false
  • D. Both A and R are false

Answer: A — Both statements are true: simple events contain one point each and cannot share outcomes, making them mutually exclusive; R correctly explains why A is true.

Q9. In rolling a die twice, let A = 'first roll is 6' and B = 'sum is at least 11'. How many outcomes satisfy both A and B?

  • A. 1
  • B. 2 ✓
  • C. 3
  • D. 4

Answer: B — A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, B = {(5,6), (6,5), (6,6)}; A ∩ B = {(6,5), (6,6)}, which contains 2 outcomes.

Q10. If S = {1, 2, 3, 4, 5, 6, 7, 8}, and events P = {2, 4, 6, 8} and Q = {3, 6}, then which statement is true?

  • A. P′ ∪ Q = {1, 3, 5, 7}
  • B. (P ∩ Q)′ = {1, 2, 3, 4, 5, 7, 8}
  • C. P ∩ Q′ = {2, 4, 8}
  • D. All of the above ✓

Answer: D — P′ = {1, 3, 5, 7}, so P′ ∪ Q = {1, 3, 5, 7}; P ∩ Q = {6}, so (P ∩ Q)′ = {1, 2, 3, 4, 5, 7, 8}; Q′ = {1, 2, 4, 5, 7, 8}, so P ∩ Q′ = {2, 4, 8}; all three are correct.

Flashcards

What is an event in probability?

An event is any subset E of the sample space S associated with a random experiment.

When does an event E occur?

Event E occurs if the outcome ω of the experiment satisfies ω ∈ E.

What is a simple event?

A simple event is an event containing exactly one sample point from the sample space.

What is a compound event?

A compound event is an event containing more than one sample point from the sample space.

Define impossible and sure events.

Impossible event = φ (empty set, never occurs); Sure event = S (entire sample space, always occurs).

What does A′ (complement) represent?

A′ is the event 'not A', containing all outcomes in S that are not in A, where A′ = S − A.

When are two events mutually exclusive?

Events A and B are mutually exclusive if they cannot occur simultaneously, meaning A ∩ B = φ.

What is the difference between 'A or B' and 'A and B'?

'A or B' = A ∪ B (at least one occurs); 'A and B' = A ∩ B (both occur together).

How is 'A but not B' expressed using set notation?

'A but not B' is represented as A − B = A ∩ B′, containing outcomes in A that are not in B.

Why are simple events always mutually exclusive?

Simple events are mutually exclusive because each contains a unique single sample point, so no two simple events can share an outcome.

Important Board Questions

Define a simple event and a compound event. Give one example of each from the experiment of tossing a coin twice. [2 marks]

Simple event has exactly 1 sample point; compound event has 2+ sample points. For tossing 2 coins, use sample space S = {HH, HT, TH, TT}.

In rolling a die, let A be the event 'getting a number less than or equal to 3' and B be the event 'getting an odd number'. Find (i) A ∪ B, (ii) A ∩ B, (iii) A − B, and (iv) A′ ∩ B. Show all working steps. [5 marks]

First write A = {1, 2, 3} and B = {1, 3, 5}. Use definitions: A ∪ B includes all elements in A or B; A ∩ B is common elements; A − B = A ∩ B′; A′ = S − A then intersect with B.

Explain the concepts of mutually exclusive events and exhaustive events. Prove that if A and B are mutually exclusive and exhaustive events, then A ∪ B = S and A ∩ B = φ. Give a real-world example from the experiment of rolling a die. [6 marks]

Define mutually exclusive as A ∩ B = φ (cannot both occur) and exhaustive as A ∪ B = S (together cover all outcomes). Show both conditions hold simultaneously using set properties. Use example: A = 'even number' and B = 'odd number' on a die where A ∩ B = φ and A ∪ B = {1, 2, 3, 4, 5, 6}.

Assertion-Reason Questions for Probability

Q1. Assertion (A):

In the experiment of tossing three coins, the event 'exactly one head appears' is a compound event.

Reason (R):

A compound event is any event that contains more than one sample point.

Answer: A — Both A and R are true, and R is the correct explanation of A — The event 'exactly one head' = {HTT, THT, TTH}, which has three sample points. By definition, an event with more than one sample point is compound. The reason correctly explains why it is compound.

Q2. Assertion (A):

Two events A and B are mutually exclusive if A ∩ B = φ.

Reason (R):

Mutually exclusive events cannot occur at the same time because they share at least one common outcome.

Answer: C — A is true but R is false — The assertion is true—mutually exclusive events have no common outcomes (empty intersection). However, the reason is false; it incorrectly states they share 'at least one common outcome,' which contradicts the definition. The reason reverses the logic.

Q3. Assertion (A):

The complement of event A, denoted A', contains all outcomes in the sample space that are not in A.

Reason (R):

A' = S – A, and A ∪ A' = S while A ∩ A' = φ.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both the assertion and reason are true. The assertion correctly defines a complementary event. The reason provides mathematical properties of complementary events, but it does not directly explain why A' is defined as S – A; it only states consequences. Thus, the reason is true but not the direct explanation.

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Case-Based Questions (CBQ) for Probability

Case Study:

A school conducted a survey on student participation in extracurricular activities. Out of 200 students, 120 participate in sports (S), 90 participate in cultural programs (C), and 40 participate in both. The school wants to analyze different categories of student participation. For a randomly selected student from the survey: Event A represents students who participate in sports, Event B represents students who participate in cultural programs.

Q1. What is the event 'students participating in sports but not cultural programs'?

  • A. A ∩ B
  • B. A – B ✓
  • C. B – A
  • D. A ∪ B

Answer: B

Q2. How many students participate in either sports or cultural programs?

  • A. 120
  • B. 90
  • C. 170 ✓
  • D. 210

Answer: C

Q3. Which set represents students participating in neither sports nor cultural programs?

  • A. (A ∪ B)' ✓
  • B. A ∩ B
  • C. A – B
  • D. B – A

Answer: A

Q4. How many students participate in neither activity?

  • A. 30 ✓
  • B. 40
  • C. 50
  • D. 60

Answer: A

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