13_p5

Making use only of Kolmogorov definition of probability and the rules of algebra of sets we will prove the following theorem.

Theorem 13.5.1

Let Ω be a sample space and A and B be any events. Then

P( ∅ ) = 0,

if A ⊆ B, then P(A) ≤ P(B),

for each A ⊆ Ω, P(A) ≤ 1,

P(A') =1 - P(A),

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Ad. (a) As the probability of a certain event is 1 and as an empty event and a certain event are mutually exclusive then P( ∅ ) + P( Ω) = P( ∅ ∪ Ω) = P(Ω) = 1. That gives us P( ∅ ) = 0.

Ad. (b) If A ⊆ B, then B = (B\A) ∪ A and (B\A) ∩ A = ∅. As a consequence of the probability definition we have P(B) = P((B\A) ∪ A ) = P(B\A) + P(A). As P(A) ≥ 0 and P(B\A) ≥ 0, hence P(A) ≤ P(B).

Ad. (c) It is an immediate consequence of the point (b) and the probability definition.

Ad. (e) According to the rules of the set theory, for any sets A and B we have A ∪ B = A ∪ B\A = B\A ∪ A\B and A ∩ B\A = ∅ , B\A ∩ A\B= ∅. From here, according to the probability definition,

The applications of the theorems 13.4.1 and 13.5.1 will be shown by the following examples. We assume that all the elementary events in the considered sample spaces are equally probable.

We roll a die 3 times. What is the probability of the event A = "6 received at least once"?

Solution. The sample space Ω= {(x,y,z) : x,y,z ∈ {1,2,...6}}, hence | Ω| = 6³. There are many events favorable to the event A. The complement of the event A, A' = "6 never received" is a little bit simplier to determine. We have A' = {(x,y,z): x,y,z ∈ {1,2,3,4,5}} and P(A') = 5³/6³. Making use of the proved property 13.5.1(d), we obtain P(A) = 1- 5³/6³.

We roll a pair of dice (each die has a different colour) and we consider two events A = "the sum of pips is 8" and B = "odd number received on each die". What is the probability of the event A ∪ B?

Solution. As A = {(2,6), (3,5), (4,4), (5,3), (6,2)}, B ={(x,y): x, y = 1,3,5} and A ∩ B = {(3,5),(5,3)}, then, according to the property 13.5.1(e), P(A ∪ B) = 5/36 + 9/36 - 2/36 = 1/3.

Question 13.5.1 We toss a coin 10 times. What is the probability of getting tails at least once? Clue: Find the probability of a complement of this event first.

Example 13.5.3 We roll a die 3 times. What is the probability of receiving 5 or 6 at least once.

Solution. The sample space in this trial is a set of triples (x,y,z) which represent the number of pips from the first, second and third roll. There are 6³ of them. Let's cosider the complement of this event, i.e. the event "5 nor 6 never received". Each elementary event in a form of (x,y,z), where x, y and z have the values 1,2,3,4 is favorable to the considered event. There are 4*4*4 such elementary events. Hence, according to the property 13.5.1(d), the probability in question is 1- (4/6)³, i.e. 19/27.

Question 13.5.2 What was incorrect in Mr Ilatort's argumentation: "Because in the Lottery choosing 7 numbers I can either win or lose, then the probability of a win is 1/2. So there is quite a big chance that I'll win."?

Question 13.5.3 We have a deck of 6 cards, numbered from 0 to 5. Two cards are drawn (without replacement) from the deck. What is the probability that the sum of cards'numbers is 3?