Class 11 — Mathematics (NCERT)

Probability: Axiomatic Approach and Events

Class 11

  • ✓By the end of this lesson students will be able to define random experiments, outcomes, and sample space.
  • ✓By the end of this lesson students will be able to identify and classify different types of events.
  • ✓By the end of this lesson students will be able to understand and apply the axiomatic approach to probability.
  • ✓By the end of this lesson students will be able to calculate probabilities of events using the axiomatic approach and related theorems.
  • ✓By the end of this lesson students will be able to solve problems involving mutually exclusive and exhaustive events.

Key concepts

Random Experiment

An experiment whose outcome cannot be predicted with certainty, but all possible outcomes are known in advance. For example, tossing a fair coin or rolling a die.

Outcome

A possible result of a random experiment. For example, when tossing a coin, 'Head' is an outcome. When rolling a die, '3' is an outcome.

Sample Space (S)

The set of all possible outcomes of a random experiment. It is denoted by 'S'. Each element of the sample space is called a sample point. For example, if a coin is tossed, S = {H, T}. If a die is rolled, S = {1, 2, 3, 4, 5, 6}.

Event (E)

An event is a subset of the sample space. It is a collection of some outcomes of the experiment. For example, in rolling a die, the event 'getting an even number' is E = {2, 4, 6}.

Impossible Event

An event which has no outcomes. It is the empty set (∅) and cannot occur. For example, in rolling a die, the event 'getting a number greater than 6' is an impossible event.

P(∅) = 0
Sure Event (Certain Event)

An event which contains all outcomes of the sample space. It is the sample space (S) itself and is certain to occur. For example, in rolling a die, the event 'getting a number less than 7' is a sure event.

P(S) = 1
Simple Event (Elementary Event)

An event consisting of a single sample point of the sample space. For example, in rolling a die, getting '3' is a simple event.

Compound Event

An event consisting of more than one sample point of the sample space. For example, in rolling a die, getting an 'even number' (E = {2, 4, 6}) is a compound event.

Mutually Exclusive Events

Two events A and B are said to be mutually exclusive if the occurrence of one precludes the occurrence of the other. In set theory terms, their intersection is an empty set (A ∩ B = ∅). They cannot occur simultaneously. For example, in rolling a die, 'getting an even number' (E={2,4,6}) and 'getting an odd number' (O={1,3,5}) are mutually exclusive events.

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

A set of events E1, E2, ..., En is said to be exhaustive if their union is the entire sample space (E1 ∪ E2 ∪ ... ∪ En = S). This means that at least one of these events must occur. If events are both mutually exclusive and exhaustive, then P(E1) + P(E2) + ... + P(En) = P(S) = 1.

If E1, E2, ..., En are exhaustive, then E1 ∪ E2 ∪ ... ∪ En = S.
Axiomatic Approach to Probability

This approach defines probability based on a set of axioms (postulates) or rules. For a sample space S and an event E, the probability P(E) must satisfy the following three axioms:

Axiom 1: For any event E, 0 ≤ P(E) ≤ 1.\nAxiom 2: P(S) = 1.\nAxiom 3: If E1, E2, E3, ... are a sequence of mutually exclusive events, then P(E1 ∪ E2 ∪ E3 ∪ ...) = P(E1) + P(E2) + P(E3) + ...
Probability of Complementary Event

The complement of an event A, denoted by A' or Aᶜ, consists of all outcomes in the sample space S that are not in A. The sum of the probability of an event and its complement is 1.

P(A') = 1 - P(A)
Probability of Union of Two Events

For any two events A and B, the probability of their union (A or B occurring) is given by the sum of their individual probabilities minus the probability of their intersection (both A and B occurring).

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Key facts to remember

  • 1A random experiment has outcomes that cannot be predicted with certainty but all possibilities are known.
  • 2The sample space (S) is the set of all possible outcomes of a random experiment.
  • 3An event (E) is any subset of the sample space.
  • 4The probability of any event E, P(E), must satisfy 0 ≤ P(E) ≤ 1.
  • 5The probability of the sure event S is P(S) = 1, and the probability of the impossible event ∅ is P(∅) = 0.
  • 6For mutually exclusive events E1, E2, ..., En, P(E1 ∪ E2 ∪ ... ∪ En) = P(E1) + P(E2) + ... + P(En).
  • 7The probability of the complement of an event A is P(A') = 1 - P(A).
  • 8For any two events A and B, P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Worked examples

Example 1

A coin is tossed three times. Write the sample space. Also, identify the following events: (i) A: 'Exactly two heads appear', (ii) B: 'At least two heads appear', (iii) C: 'No heads appear'. Are events A and C mutually exclusive?

IStep 1: Determine the sample space (S) for tossing a coin three times.
IIThe possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
IIISo, S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
IVStep 2: Identify event A: 'Exactly two heads appear'.
VA = {HHT, HTH, THH}.
VIStep 3: Identify event B: 'At least two heads appear'.
VIIThis means two or three heads. B = {HHT, HTH, THH, HHH}.
VIIIStep 4: Identify event C: 'No heads appear'.
9C = {TTT}.
10Step 5: Check if events A and C are mutually exclusive.
11A ∩ C = {HHT, HTH, THH} ∩ {TTT} = ∅.
12Since their intersection is the empty set, A and C are mutually exclusive events.

Answer

Sample Space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.\nEvent A = {HHT, HTH, THH}.\nEvent B = {HHT, HTH, THH, HHH}.\nEvent C = {TTT}.\nYes, events A and C are mutually exclusive.

Always list all possible outcomes systematically to ensure the sample space is complete.

Example 2

A die is rolled. Let E be the event 'getting an even number' and F be the event 'getting a number greater than 4'. Find P(E), P(F), P(E ∩ F), and P(E ∪ F). Verify P(E ∪ F) = P(E) + P(F) - P(E ∩ F).

IStep 1: Determine the sample space (S) for rolling a die.
IIS = {1, 2, 3, 4, 5, 6}. The total number of outcomes is n(S) = 6.
IIIStep 2: Identify event E: 'getting an even number' and calculate P(E).
IVE = {2, 4, 6}. The number of outcomes in E is n(E) = 3.
VP(E) = n(E)/n(S) = 3/6 = 1/2.
VIStep 3: Identify event F: 'getting a number greater than 4' and calculate P(F).
VIIF = {5, 6}. The number of outcomes in F is n(F) = 2.
VIIIP(F) = n(F)/n(S) = 2/6 = 1/3.
9Step 4: Identify the intersection event E ∩ F: 'getting an even number AND a number greater than 4' and calculate P(E ∩ F).
10E ∩ F = {2, 4, 6} ∩ {5, 6} = {6}. The number of outcomes in E ∩ F is n(E ∩ F) = 1.
11P(E ∩ F) = n(E ∩ F)/n(S) = 1/6.
12Step 5: Identify the union event E ∪ F: 'getting an even number OR a number greater than 4' and calculate P(E ∪ F) directly.
13E ∪ F = {2, 4, 6} ∪ {5, 6} = {2, 4, 5, 6}. The number of outcomes in E ∪ F is n(E ∪ F) = 4.
14P(E ∪ F) = n(E ∪ F)/n(S) = 4/6 = 2/3.
15Step 6: Verify the formula P(E ∪ F) = P(E) + P(F) - P(E ∩ F).
16R.H.S. = P(E) + P(F) - P(E ∩ F) = 1/2 + 1/3 - 1/6
17R.H.S. = 3/6 + 2/6 - 1/6 = (3 + 2 - 1)/6 = 4/6 = 2/3.
18L.H.S. = P(E ∪ F) = 2/3.
19Since L.H.S. = R.H.S., the formula is verified.

Answer

P(E) = 1/2, P(F) = 1/3, P(E ∩ F) = 1/6, P(E ∪ F) = 2/3. The formula P(E ∪ F) = P(E) + P(F) - P(E ∩ F) is verified.

Remember that for equally likely outcomes, P(E) = (Number of favourable outcomes) / (Total number of outcomes).

Example 3

In a class of 30 students, 18 opted for Mathematics, 15 opted for Physics, and 7 opted for both. If a student is selected at random, what is the probability that the student opted for (i) Mathematics or Physics? (ii) Neither Mathematics nor Physics?

IStep 1: Define events and given probabilities.
IILet M be the event that a student opted for Mathematics.
IIILet P be the event that a student opted for Physics.
IVTotal number of students, n(S) = 30.
VNumber of students opted for Mathematics, n(M) = 18. So, P(M) = 18/30 = 3/5.
VINumber of students opted for Physics, n(P) = 15. So, P(P) = 15/30 = 1/2.
VIINumber of students opted for both Mathematics and Physics, n(M ∩ P) = 7. So, P(M ∩ P) = 7/30.
VIIIStep 2: Calculate the probability that the student opted for Mathematics or Physics (M ∪ P).
9Using the formula P(M ∪ P) = P(M) + P(P) - P(M ∩ P):
10P(M ∪ P) = 18/30 + 15/30 - 7/30
11P(M ∪ P) = (18 + 15 - 7)/30 = 26/30 = 13/15.
12Step 3: Calculate the probability that the student opted for neither Mathematics nor Physics.
13This is the complement of the event (M ∪ P). Let E be the event 'neither Mathematics nor Physics'.
14E = (M ∪ P)'
15P(E) = P((M ∪ P)') = 1 - P(M ∪ P)
16P(E) = 1 - 13/15 = (15 - 13)/15 = 2/15.

Answer

(i) The probability that the student opted for Mathematics or Physics is 13/15.\n(ii) The probability that the student opted for neither Mathematics nor Physics is 2/15.

Visualise with Venn diagrams if you find it helpful to understand the union and intersection of events.

Common mistakes

  • ✗Incorrectly identifying the sample space, leading to incorrect calculations of total outcomes.
  • ✗Confusing mutually exclusive events with independent events. Mutually exclusive means they cannot occur together (A ∩ B = ∅), while independent means the occurrence of one does not affect the other.
  • ✗Forgetting to subtract P(A ∩ B) when calculating P(A ∪ B) for non-mutually exclusive events.
  • ✗Misinterpreting 'at least', 'at most', 'exactly' in defining events.
  • ✗Not understanding that the sum of probabilities of all simple events in a sample space must be 1.

Exam tips

  • ★Always clearly define the sample space (S) and the events (A, B, C, etc.) before attempting to calculate probabilities.
  • ★Use set notation (∪, ∩, ∅, S) correctly to represent unions, intersections, impossible events, and sure events.
  • ★For problems involving multiple events, draw a Venn diagram to visualise the relationships between them, especially for union and intersection.
  • ★Show all steps of your calculations, including the formulas used, to ensure partial marks even if the final answer is incorrect.

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