Egypt Secondary 3 Statistics, Week 12: The three axioms and the corollaries P(∅) = 0, P(A′) = 1 − P(A), P(A − B) and P(A ∪ B)

The axioms of probability and their corollaries

Week 12 · Axioms of probability and their corollaries: 0 ≤ P ≤ 1, P(S) = 1, additivity, P(A′), P(A − B) and P(A ∪ B), and solving problems · 3 lessons

The three axioms and the corollaries P(∅) = 0, P(A′) = 1 − P(A), P(A − B) and P(A ∪ B)

Statistics Sec 3 Term 1, Unit 3, Lesson 1

GoalState the three axioms of probability and use P(∅) = 0, P(A′) = 1 − P(A), P(A − B) and P(A ∪ B).

New words

axiom · a rule accepted for every probabilitycorollary · a rule that follows from the axiomsaddition rule · P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Explanation

Probability obeys three rules, the axioms. (1) 0 ≤ P(A) ≤ 1 for every event A of S. (2) P(S) = 1. (3) If A and B are mutually exclusive then P(A ∪ B) = P(A) + P(B); the same holds for any finite number of events that are exclusive in pairs. Everything else is a corollary. S and ∅ are exclusive and S ∪ ∅ = S, so axiom 3 gives P(S) = P(S) + P(∅), hence P(∅) = 0. A and A′ are exclusive and their union is S, so P(A) + P(A′) = 1, hence P(A′) = 1 − P(A). The event A splits into the exclusive parts A − B and A ∩ B, so P(A) = P(A − B) + P(A ∩ B), hence P(A − B) = P(A) − P(A ∩ B). The event A ∪ B splits into the exclusive parts A − B and B, so P(A ∪ B) = P(A) − P(A ∩ B) + P(B). Finally, if A ⊂ B then P(A) ≤ P(B): a larger set cannot be less likely. The picture names the four regions of two events.

Worked examples

Let P(A) = 0.55, P(B) = 0.40 and P(A ∩ B) = 0.25. Then P(A ∪ B) = 0.55 + 0.40 − 0.25 = 0.70, P(A′) = 1 − 0.55 = 0.45, P(A − B) = 0.55 − 0.25 = 0.30 and P((A ∪ B)′) = 1 − 0.70 = 0.30. The four regions 0.30, 0.25, 0.15 and 0.30 add up to 1.

A spinner lands on red with probability 0.35 and on blue with probability 0.25; one sector is hit at a time, so red and blue are exclusive. By axiom 3, P(red or blue) = 0.35 + 0.25 = 0.60, and P(neither) = 1 − 0.60 = 0.40.

A subset. If A ⊂ B with P(A) = 0.2 and P(B) = 0.5, then A ∩ B = A, so P(B − A) = P(B) − P(A ∩ B) = 0.5 − 0.2 = 0.3, and P(A) = 0.2 ≤ 0.5 = P(B), as the rule for subsets says.

Class activity

On squared paper draw a 10 × 10 square: its 100 small squares stand for S, so each small square is 0.01. Shade 55 squares for A and 40 squares for B so that exactly 25 are shaded twice. Count the squares in A ∪ B, in A′ and in the part outside both, and check that the counts agree with the corollaries when divided by 100.

Practice

P(A) = 0.7. Find P(A′).

P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1. Find P(A ∪ B) and P(A − B).

A and B are exclusive with P(A) = 1/5 and P(B) = 3/10. Find P(A ∪ B) and P((A ∪ B)′).

Say why each statement is impossible: (a) P(A) = −0.1; (b) P(A) = 0.6 and P(A′) = 0.5.

P(A) = 0.45 and P(A ∩ B) = 0.15. Find P(A − B).