7 min readConcept sheet1-mark MCQ or assertion-reason; a 2-mark proof

Properties of conditional probability

Inside the reduced sample space F, conditional probability obeys the usual rules: P(S|F) = P(F|F) = 1 and P(E′|F) = 1 − P(E|F).

Property 1P(S|F) = P(F|F) = 1

Property 2P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F)

If A and B are disjointP((A ∪ B)|F) = P(A|F) + P(B|F)

Property 3P(E′|F) = 1 − P(E|F)

Proofs of the three properties

Let F be an event with P(F) ≠ 0.

1
P(S|F) = P(S ∩ F)/P(F) = P(F)/P(F) = 1. In the same way, P(F|F) = P(F ∩ F)/P(F) = 1.
2
(A ∪ B) ∩ F = (A ∩ F) ∪ (B ∩ F), by the distributive law.
3
So P((A ∪ B)|F) = [P(A ∩ F) + P(B ∩ F) − P(A ∩ B ∩ F)]/P(F).
4
Split the fraction: P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F).
5
E and E′ are disjoint and E ∪ E′ = S, so P(E|F) + P(E′|F) = P(S|F) = 1.
6★
So, P(E′|F) = 1 − P(E|F).

★ key step

Example 1: the complement rule

Problem

A die is thrown. F: the number is greater than 2. E: the number is even. Find P(E|F) and P(E′|F).

Working

F = {3, 4, 5, 6}, so n(F) = 4.
E ∩ F = {4, 6}, so P(E|F) = 2/4 = 1/2.
P(E′|F) = 1 − 1/2 = 1/2
Check: E′ ∩ F = {3, 5}, so P(E′|F) = 2/4 = 1/2. It matches.

Answer

P(E|F) = 1/2 and P(E′|F) = 1/2.

Example 2: the union rule

Problem

A card is drawn from cards numbered 1 to 20. Given that the number is even, find the probability that it is a multiple of 3 or a multiple of 5.

Working

F = even numbers, so n(F) = 10. A = multiple of 3, B = multiple of 5.
A ∩ F = {6, 12, 18}, so P(A|F) = 3/10.
B ∩ F = {10, 20}, so P(B|F) = 2/10.
A ∩ B ∩ F = even multiples of 15 up to 20: none, so P((A ∩ B)|F) = 0.
P((A ∪ B)|F) = 3/10 + 2/10 − 0 = 5/10 = 1/2

Answer

The probability is 1/2. (Check: 6, 10, 12, 18 and 20 are 5 of the 10 even numbers.)

P(E′|F) vs P(E|F′)

P(E′|F)Complement of the event, same condition F. It always equals 1 − P(E|F).

P(E|F′)Same event, complement of the condition. It is not 1 − P(E|F) in general; work it out as P(E ∩ F′)/P(F′).

Example: a die is thrown, E: the number is 6, F: the number is even. P(E|F) = 1/3, but P(E|F′) = 0, not 2/3.

In real life

If a weather app says there is a 70% chance of rain given a cloudy sky, then the chance of no rain given a cloudy sky is 30%. That is the complement property at work.

Exam angle

1 marks: “If P(A|B) = 0.3, find P(A′|B).” P(A′|B) = 1 − P(A|B) = 1 − 0.3 = 0.7.

2 marks: “Prove that P(E′|F) = 1 − P(E|F).” E ∪ E′ = S with E and E′ disjoint. So P(E|F) + P(E′|F) = P(S|F) = P(S ∩ F)/P(F) = 1. Hence P(E′|F) = 1 − P(E|F).

2 marks: “P(B) = 0.5, P(A ∩ B) = 0.2 and P(C ∩ B) = 0.15, where A and C are mutually exclusive. Find P((A ∪ C)|B).” P(A|B) = 0.2/0.5 = 0.4 and P(C|B) = 0.15/0.5 = 0.3. A and C are disjoint, so P((A ∪ C)|B) = 0.4 + 0.3 = 0.7.

Remember it

Complement the event, never the condition: P(E′|F) = 1 − P(E|F).