Year 10 Extended · Unit 4 Sets, Venn diagrams and conditional probability · Lesson 3 of 6

Which probability rules do we have to accept, and which can be proved?

Three statements are accepted. Every other rule of probability, such as the complement rule and the addition rule, follows from them by logic alone, and each step has to be justified.

Statistics and ProbabilityNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 8: the chance of rolling a 5 on a fair die is 1/6. What is the chance of not rolling a 5?
    Answer5/6. Together they make the whole: 1/6 + 5/6 = 1.
  2. From lesson 2: n(A) = 14, n(B) = 9 and n(A ∩ B) = 4. Find n(A ∪ B).
    Answer19. 14 + 9 − 4: the 4 in both were counted twice.
  3. From Year 9: solve 4x + 0.6 = 2.2.
    Answerx = 0.4. Take off 0.6 to get 1.6, then divide by 4.
Words used in this lesson
axiom
a statement accepted without proof. For example: P(A) ≥ 0.
theorem
a statement proved from the axioms and earlier theorems. For example: the addition rule for any two events.
corollary
a result that follows from a theorem in a step or two. For example: P(∅) = 0, from the complement rule.
mutually exclusive
events that cannot happen at the same time. For example: rolling a 1 and rolling a 6 on one throw.
proof
a chain of steps, each justified, from the axioms to the result. For example: the proof of the complement rule.
sample space
the set of all possible outcomes, written U. For example: {H, T} for a coin.

A forecast of one hundred and twenty per cent

A radio forecaster in Cape Town says: “There is a 70% chance of rain on Saturday and a 50% chance on Sunday, so the chance of rain at some time this weekend is 120%.” A listener says that cannot be right.

Predict first: what rule is the forecaster using, and what extra piece of information is needed before it could be used? What is the least that the chance of some rain this weekend could be? Test your ideas on the figure by setting P(A) and P(B), then moving P(A ∩ B).

40
30
10
60%P(A or B)
70%P(A) + P(B)
40%P(neither)

P(A) = 40%, P(B) = 30%, P(A ∩ B) = 10%: P(A ∪ B) = 60%, P(neither) = 40%. The sum P(A) + P(B) = 70% counts the overlap twice.

Show me the steps
  1. Start at P(A) = 40%, P(B) = 30% and P(A ∩ B) = 10%: the parts are 30%, 10%, 20% and 40%, which make 100%. P(A ∪ B) = 40 + 30 − 10 = 60%, and the plain sum 70% has counted the 10% twice.
  2. Set P(A ∩ B) to 0%: the events cannot happen together, the parts are 40%, 0%, 30% and 30%, and now P(A ∪ B) = 70% is exactly the plain sum. That is the third axiom.
  3. Set P(A) = 70%, P(B) = 60% and P(A ∩ B) = 0%: neither would be 100 − 70 − 60 + 0 = −30%, so no such model exists. Raise P(A ∩ B) to 30% and the model comes back, with neither at 0%.
  4. Set P(A) = 40%, P(B) = 30% and P(A ∩ B) = 35%: B only would be 30 − 35 = −5%, so no such model exists. P(A ∩ B) can never be larger than P(B).

Three axioms. An axiom is a statement we accept without proof.
1. P(A) ≥ 0 for every event A.
2. P(U) = 1, where U is the whole sample space.
3. If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B).

Everything else is proved. A theorem is a statement proved from the axioms and earlier theorems; a corollary follows from a theorem in a step or two. Each line of a proof must be an axiom, a definition, or something already proved.

So we can only add probabilities for mutually exclusive events. For events that can overlap, the addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

P(A) + P(B) is not P(A ∪ B) unless the events are mutually exclusive. Roll a fair die with A = {2, 4, 6} and B = {4, 5, 6}. Then P(A) + P(B) = 3/6 + 3/6 = 1, but A ∪ B = {2, 4, 5, 6}, which has probability 4/6. The outcomes 4 and 6 are in both events and were counted twice.

Worked examples

1. Prove the complement rule, P(A′) = 1 − P(A), and deduce that P(A) ≤ 1.

  1. A and A′ cannot both happen, so they are mutually exclusive, and A ∪ A′ = U.
  2. By axiom 3, P(A ∪ A′) = P(A) + P(A′). By axiom 2, P(A ∪ A′) = P(U) = 1.
  3. So P(A) + P(A′) = 1, which gives P(A′) = 1 − P(A).
  4. Corollary: P(A′) ≥ 0 by axiom 1, so 1 − P(A) ≥ 0 and P(A) ≤ 1.

2. Deduce P(∅) = 0.

  1. The empty event is the complement of U, so ∅ = U′.
  2. By the complement rule, P(∅) = 1 − P(U) = 1 − 1 = 0.

3. Prove the addition rule, P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

  1. Split A ∪ B into two parts that cannot overlap: A ∪ B = A ∪ (A′ ∩ B). By axiom 3, P(A ∪ B) = P(A) + P(A′ ∩ B).
  2. Split B the same way: B = (A ∩ B) ∪ (A′ ∩ B), so P(B) = P(A ∩ B) + P(A′ ∩ B), and P(A′ ∩ B) = P(B) − P(A ∩ B).
  3. Substitute: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Your go. The first step is given. Prove that if A ⊆ B then P(A) ≤ P(B).

  1. Since A ⊆ B, the event B is A together with the part of B outside A: B = A ∪ (A′ ∩ B), and the two parts cannot overlap.
  2. Now use axiom 3, then axiom 1.
Check the answer

By axiom 3, P(B) = P(A) + P(A′ ∩ B). By axiom 1, P(A′ ∩ B) ≥ 0, so P(B) ≥ P(A).

Check before you go on

1. A student proves that P(A ∪ B) ≤ P(A) + P(B) for any events. Line 1: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), by the addition rule. Line 2: P(A ∩ B) ≥ 0. Line 3: so P(A ∪ B) ≤ P(A) + P(B). Which statement justifies line 2?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. P(A) = 2x and P(A′) = 3x + 0.1. Find x.

3. P(A) = 0.45, P(B) = 0.38 and P(A ∩ B) = 0.17. Find P(A′ ∩ B′), the probability of neither.

Level 2 · Building

4. P(A ∪ B) = 0.9, P(A) = 0.6 and P(B) = 0.5. Find P(A′ ∩ B), the probability of B but not A.

5. A and B are mutually exclusive, P(A) = 2 × P(B) + 0.1 and P(A ∪ B) = 0.85. Find P(B) as a decimal to two decimal places.

Level 3 · Stretch

6. P(A) = 0.7 and P(B) = 0.8. What is the least value that P(A ∩ B) can take?

7. A spinner stops on red, blue or green, which are mutually exclusive and cover everything. P(red) = x, P(blue) = 2x² and P(green) = 0.72. Find the value of x that is possible.

Something to think about

Given only P(A) = 0.4 and P(B) = 0.8, what are the least and the greatest values P(A ∩ B) can take, and what are the least and the greatest values of P(A ∪ B)? Say which axiom each limit comes from.

One way to think about it

P(A ∩ B) cannot exceed the smaller of the two, 0.4, and P(A ∪ B) cannot exceed 1, so 0.4 + 0.8 − P(A ∩ B) ≤ 1 gives P(A ∩ B) ≥ 0.2. Then P(A ∩ B) runs from 0.2 to 0.4. The union is 1.2 − P(A ∩ B), so it runs from 0.8 (when A ∩ B is 0.4, the larger event contains the smaller) to 1 (when A ∩ B is 0.2). Every limit comes from a region that cannot go below 0 (axiom 1) inside a whole of 1 (axiom 2): the region outside both events being at least 0 gives P(A ∩ B) ≥ 0.2 and P(A ∪ B) ≤ 1, and A only being at least 0 gives P(A ∩ B) ≤ 0.4 and P(A ∪ B) ≥ 0.8.

Exit ticket

8. For which set of values is there NO event model, so the values cannot be P(A), P(B) and P(A ∩ B)?