What this chapter is about
Probability is the branch of mathematics that measures how likely an event is to occur. In Class 11, you study the axiomatic approach to probability, which builds the subject on a small set of clearly stated rules (axioms) rather than on intuitive ideas alone. This framework lets you handle experiments whose outcomes are equally likely, as well as those that are not.
You already met probability informally in earlier classes through coin tosses and dice rolls. Now you formalise those ideas with precise definitions of sample space, events, and the probability function. You learn how to combine events using set operations (union, intersection, complement) and how to compute probabilities when events overlap.
After this chapter you should be able to list the sample space of a random experiment, identify different types of events, apply the addition rule for two or more events, and solve problems involving complementary and mutually exclusive events.
Key ideas
- A random experiment is a process whose outcome cannot be predicted with certainty but whose set of all possible outcomes is known. Examples: rolling a die, drawing a card from a shuffled pack.
- The sample space S is the set of all possible outcomes of a random experiment. Each element of S is called a sample point.
- An event is any subset of the sample space. The event occurs if the actual outcome belongs to that subset.
- Two events are mutually exclusive (disjoint) if they cannot occur together; their intersection is empty.
- The complement of an event A, written A′ (or Aᶜ), contains every outcome in S that is not in A.
- Axiomatic definition: For every event A in sample space S, the probability P(A) satisfies three axioms:
1. P(A) ≥ 0. 2. P(S) = 1. 3. If A and B are mutually exclusive, then P(A ∪ B) = P(A) + P(B).
- Equally likely outcomes: When every sample point has the same chance, P(A) = (number of outcomes favourable to A) / (total number of outcomes in S).
- The addition rule for any two events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Formulas and facts to remember
- Formula / Fact: 0 ≤ P(A) ≤ 1 · Meaning: A probability is always between 0 and 1 inclusive.
- Formula / Fact: P(S) = 1 · Meaning: The certain event has probability 1.
- Formula / Fact: P(∅) = 0 · Meaning: The impossible event has probability 0.
- Formula / Fact: P(A′) = 1 − P(A) · Meaning: Probability of complement equals 1 minus probability of the event.
- Formula / Fact: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) · Meaning: General addition rule for two events.
- Formula / Fact: If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B) · Meaning: Special case; the intersection is empty.
- Formula / Fact: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C) · Meaning: Inclusion-exclusion for three events.
Worked examples
### Example 1 – Listing sample space and finding a simple probability
A fair six-sided die is rolled once. Let A be the event that the number shown is a multiple of 3.
Solution
Sample space S = {1, 2, 3, 4, 5, 6}; so n(S) = 6.
Multiples of 3 in S: 3 and 6. Hence A = {3, 6} and n(A) = 2.
Since all outcomes are equally likely,
P(A) = n(A) / n(S) = 2 / 6 = 1/3.
---
### Example 2 – Using the addition rule
In a class of 40 students, 18 play cricket, 12 play football, and 5 play both. A student is chosen at random. Find the probability that the student plays at least one of the two games.
Solution
Let C = event the student plays cricket, F = event the student plays football.
P(C) = 18/40, P(F) = 12/40, P(C ∩ F) = 5/40.
By the addition rule,
P(C ∪ F) = P(C) + P(F) − P(C ∩ F) = 18/40 + 12/40 − 5/40 = 25/40 = 5/8.
So the probability that a randomly chosen student plays cricket or football (or both) is 5/8.
---
### Example 3 – Using the complement
A bag contains 4 red, 3 green, and 2 blue marbles. Two marbles are drawn at random without replacement. Find the probability that at least one marble is red.
Solution
Total marbles = 4 + 3 + 2 = 9.
Number of ways to draw 2 marbles from 9 = C(9, 2) = 36.
Let A = event that at least one marble is red.
It is easier to find A′ (no red marble drawn). Non-red marbles = 3 + 2 = 5.
Ways to choose 2 from these 5 = C(5, 2) = 10.
P(A′) = 10/36 = 5/18.
Therefore P(A) = 1 − P(A′) = 1 − 5/18 = 13/18.
Common mistakes
- Forgetting to subtract the intersection when using the addition rule → Always check whether A ∩ B is empty; if not, subtract P(A ∩ B).
- Treating events that can happen together as mutually exclusive → Events are mutually exclusive only when they share no common outcome.
- Writing probability greater than 1 or less than 0 → Recheck your counts; a valid probability lies between 0 and 1.
- Confusing the complement of an event with just one alternative outcome → The complement includes every outcome in S that is not in the event.
- Mixing up ordered and unordered counting → Use permutations when order matters, combinations when it does not.
Quick revision
1. Sample space = set of all possible outcomes; event = any subset of the sample space.
2. P(A) always lies between 0 and 1; P(S) = 1 and P(∅) = 0.
3. P(A′) = 1 − P(A) — often the quickest route to "at least one" problems.
4. Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
5. Mutually exclusive events have no common outcome, so P(A ∩ B) = 0.
6. For equally likely outcomes, P(A) = favourable outcomes / total outcomes.