Words you need first
- Random experiment: an action whose result we cannot know in advance, like tossing a coin.
- Outcome: one possible result.
- Sample space S: the set of all outcomes. For two coins, S = {HH, HT, TH, TT}.
- Event: any subset of S. It "happens" if the outcome we get is inside it.
Types of events
- Impossible event: the empty set ∅ (getting 7 on a die).
- Sure event: S itself (getting a number from 1 to 6).
- Simple (elementary) event: exactly one outcome, like {4}.
- Compound event: more than one outcome, like {2, 4, 6}.
Algebra of events: not, and, or
- Not A (complement A′): all outcomes of S not in A. A′ = S − A.
- A and B (A ∩ B): outcomes in both A and B.
- A or B (A ∪ B): outcomes in A, or in B, or in both.
- A but not B (A − B = A ∩ B′): outcomes in A that are not in B.
Think of the die cubes: A lifts some cubes, B lifts some cubes. "And" is the cubes lifted by both (purple). "Or" is every lifted cube.
Mutually exclusive and exhaustive events
Mutually exclusive: A ∩ B = ∅. Both cannot happen together. Example: "even" and "odd" on one die.
Exhaustive: A ∪ B ∪ … = S. At least one of them must happen.
If events are both mutually exclusive and exhaustive, their probabilities add to exactly 1. The simple events {1}, {2}, …, {6} of a die are an example.
Axiomatic approach to probability
An axiom is a basic rule we accept without proof. Probability is a rule P that gives each event E a real number P(E) such that:
- P(E) ≥ 0 for every event E.
- P(S) = 1.
- If E and F are mutually exclusive, P(E ∪ F) = P(E) + P(F).
So for a finite S = {ω₁, …, ωₙ}, we just need numbers P(ωᵢ) ≥ 0 that add to 1. Then P(E) = sum of P(ωᵢ) for the outcomes in E. This works even when outcomes are not equally likely. When they are equally likely, it gives the old formula P(E) = n(E) ÷ n(S).
Results that follow from the axioms
- P(∅) = 0.
- 0 ≤ P(E) ≤ 1.
- P(not A) = 1 − P(A), because A and A′ are mutually exclusive and together make S.
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The overlap was counted twice, so subtract it once.
- P(A but not B) = P(A) − P(A ∩ B).
- If A ⊆ B then P(A) ≤ P(B).
Try it: a paper-slip experiment
Write 1 to 6 on six slips. Circle the even numbers in blue and the numbers above 3 in orange. Count the slips with both colours (A and B) and with at least one colour (A or B). Check that 3 + 3 − 2 = 4. Then press the buttons in the last 3D step to build the same A and B and see the same numbers.
Board exam pattern
The Statistics and Probability unit has 12 marks in CBSE Class 11. Common questions: write the sample space, list A ∪ B or A ∩ B, check if events are mutually exclusive or exhaustive, test whether a given assignment is a valid probability, and use P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Key formulas and definitions
- Event E ⊆ S; P(E) = n(E) ÷ n(S) when outcomes are equally likely
- Axioms: P(E) ≥ 0, P(S) = 1, P(E ∪ F) = P(E) + P(F) if E ∩ F = ∅
- P(∅) = 0 and 0 ≤ P(E) ≤ 1
- P(A′) = 1 − P(A)
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- P(A ∩ B′) = P(A) − P(A ∩ B)
- Mutually exclusive: A ∩ B = ∅; exhaustive: A ∪ B = S
Worked examples
1. Two coins are tossed. Write S and find P(at least one head).
S = {HH, HT, TH, TT}, n(S) = 4. At least one head = {HH, HT, TH}, 3 outcomes. P = 3/4.
2. A die is rolled. A = even, B = prime. Find A ∩ B, A ∪ B and P(A ∪ B).
A = {2, 4, 6}, B = {2, 3, 5}. A ∩ B = {2}. A ∪ B = {2, 3, 4, 5, 6}. P(A ∪ B) = 3/6 + 3/6 − 1/6 = 5/6. Check by counting: 5 outcomes out of 6.
3. S = {ω₁, ω₂, ω₃, ω₄}. Which is a valid assignment? (a) 0.3, 0.2, 0.4, 0.1 (b) 0.5, 0.6, −0.1, 0 (c) 0.4, 0.3, 0.2, 0.2
(a) all ≥ 0 and sum = 1, valid. (b) has −0.1, breaks axiom 1, not valid. (c) sum = 1.1, breaks P(S) = 1, not valid.
4. P(A) = 0.42, P(B) = 0.48 and P(A ∩ B) = 0.16. Find P(A ∪ B), P(A′) and P(A but not B).
P(A ∪ B) = 0.42 + 0.48 − 0.16 = 0.74. P(A′) = 1 − 0.42 = 0.58. P(A ∩ B′) = 0.42 − 0.16 = 0.26.
5. One card is drawn from 52 cards. Find P(king or heart).
P(king) = 4/52, P(heart) = 13/52, P(king of hearts) = 1/52. P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.
6. A and B are mutually exclusive with P(A) = 0.35 and P(B) = 0.45. Find P(A or B) and P(neither).
No overlap, so P(A ∪ B) = 0.35 + 0.45 = 0.8. P(neither) = 1 − 0.8 = 0.2.
7. Three coins are tossed. A = exactly two heads, B = no head, C = at least two heads. Which pairs are mutually exclusive? Find P(C).
S has 8 outcomes. A = {HHT, HTH, THH}, B = {TTT}, C = {HHT, HTH, THH, HHH}. A ∩ B = ∅ and B ∩ C = ∅, so A, B and B, C are mutually exclusive. A ∩ C = A, not empty. P(C) = 4/8 = 1/2.
8. From 4 boys and 3 girls, 2 students are chosen at random. Find P(both girls) and P(at least one boy).
n(S) = ⁷C₂ = 21. Both girls: ³C₂ = 3, so P = 3/21 = 1/7. "At least one boy" = not (both girls), so P = 1 − 1/7 = 6/7.
9. P(A) = 1/2, P(B) = 1/3 and P(A ∩ B) = 1/6. Find P(neither A nor B).
P(A ∪ B) = 1/2 + 1/3 − 1/6 = 3/6 + 2/6 − 1/6 = 4/6 = 2/3. Neither = not (A or B), so P = 1 − 2/3 = 1/3.
Common mistakes
- Adding P(A) + P(B) for "A or B" when the events overlap. Subtract P(A ∩ B).
- Thinking mutually exclusive means exhaustive. {1, 2} and {5, 6} share nothing but do not cover S.
- Accepting an assignment with a negative number or a total other than 1 as a valid probability.
- Writing A′ as a probability instead of an event. A′ is a set; P(A′) is its probability.