Introduction to Probability

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Introduction to Probability Probability Examples c-1

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Leif Mejlbro

Probability Examples c-1 Introduction to Probability

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Probability Examples c-1 – Introduction to Probability

ISBN 978-87-7681-515-8

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I ntroduction to Probability

4

Contents

1 Some theoretical background 6

3 Sampling with and without replacement 12

6 Binomial distribution 35

8 Huyghens’ exercise 39

9 Balls in boxes 41

10 Conditional probabilities, Bayes’s formula 42

11 Stochastic independency/dependency 48

12 Probabilities of events by set theory 51

13 The rencontre problem and similar examples 53

14 Strategy in games 57

15 Bertrand’s paradox 59

Contents

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I ntroduction

Introduction

This is the ﬁrst book of examples from the Theory of Probability This topic is not my favourite,

however, thanks to my former colleague, Ole Jørsboe, I somehow managed to get an idea of what it is

all about The way I have treated the topic will often diverge from the more professional treatment

On the other hand, it will probably also be closer to the way of thinking which is more common among

many readers, because I also had to start from scratch

Unfortunately errors cannot be avoided in a ﬁrst edition of a work of this type However, the author

has tried to put them on a minimum, hoping that the reader will meet with sympathy the errors

which do occur in the text

Leif Mejlbro 25th October 2009

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1 Some theoretical beckground

1 Some theoretical background

It is not the purpose here to produce a full introduction into the theory, so we shall be content just

to mention the most important concepts and theorems

The topic probability is relying on the concept σ-algebra A σ-algebra is deﬁned as a collection F of

subsets from a given set Ω, for which

1) The empty set belongs to the σ-algebra, ∅ ∈ F

2) If a set A ∈ F, then also its complementary set lies in F, thus ∁A ∈ F

3) If the elements of a ﬁnite or countable sequence of subsets of Ω all lie in F, i.e An ∈ F for e.g

n ∈ N, then the union of them will also belong to F, i.e

+∞

n=1

An∈ F

The sets of F are called events

We next introduce a probability measure on (Ω, F) as a set function P : F → R, for which

1) Whenever A ∈ F, then 0 ≤ P (A) ≤ 1

2) P (∅) = 0 and P (Ω) = 1

3) If (An) is a ﬁnite or countable family of mutually disjoint events, e.g Ai ∩ Aj = ∅, if i = j, then

P

+∞

n=1

An

=

+∞

n=1

P (An)

All these concepts are united in the Probability field, which is a triple (Ω, F, P ), where Ω is a

(non-empty) set, F is a σ-algebra of subsets of Ω, and P is a probability measure on (Ω, F)

We mention the following simple rules of calculations:

If (Ω, F, P ) is a probability ﬁeld, and A, B ∈ F, then

1) P (B) = P (A) + P (B\) ≥ P (A), if A ⊆ B

2) P (A ∪ B) = P (A) + P (B) − P (A ∩ B)

3) P (∁A) = 1 − P (A)

4) If A1⊆ A2⊆ · · · ⊆ An ⊆ · · · and A =

+∞

n=1

An, then P (A) = lim

n→+∞P (An)

5) If A1⊇ A2⊇ · · · ⊇ An ⊇ · · · and A =

+∞

n=1

An, then P (A) = lim

n→+∞P (An)

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Let (Ω, F, P ) be a probability ﬁeld, and let A and B ∈ F be events where we assume that P (B) > 0

We deﬁne the conditional probability of A, for given B by

P (A | B) := P (A ∩ B)

P (B) .

In this case, Q, given by

Q(A) := P (A | B), A ∈ F,

is also a probability measure on (Ω, F)

The multiplication theorem of probability,

P (A ∩ B) = P (B) · P (A | B)

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Two events A and B are called independent, if P (A | B) = P (A), i.e if

P (A ∩ B) = P (A) · P (B)

We expand this by saying that n events Aj, j = 1, , n, are independent, if we for any subset

J ⊆ {1, , n} have that

P

⎛

⎝

j∈J

Aj

⎞

j∈J

P (Aj)

We ﬁnally mention two results, which will become useful in the examples to come:

Given (Ω, F, P ) a probability ﬁeld We assume that we have a splitting (Aj)+∞j=1 of Ω into events

Aj∈ F, which means that the Aj are mutually disjoint and their union is all of Ω, thus

+∞

j=1

Aj= Ω, and Ai ∩ Aj = ∅, for every pair of indices (i, j), where i = j

If A ∈ F is an event, for which P (A) > 0, then

The law of total probability,

P (A) =

+∞

j=1

P (Aj) · P (A | Aj) ,

and

Bayes’s formula,

P (Ai| A) = +∞P (Ai) · P (A | Ai)

j=1P (Aj) · P (A | Aj).

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2 Set theory

Example 2.1 Let A1,A2, , An be subsets of the sets Ω Prove that

∁

n

i=1

Ai

=

n

i=1

n

i=1

∁Ai

=

n

i=1

∁Ai

These formulæ are calledde Morgan’s formulæ

1a If x ∈ ∁ ( ni=1Ai), then x does not belong to any Ai, thus x ∈ ∁Ai for every i, and therefore also

in the intersection, so

∁

n

i=1

Ai

n

i=1

∁Ai

1b On the other hand, if x ∈ni=1∁Ai, then x lies in all complements ∁Ai, so x does not belong to

any Ai, and therefore not in the union either, so

n

i=1

∁Ai ∁

n

i=1 Ai

Summing up we conclude that we have equality

2 If we put Bi= ∁Ai, then ∁Bi= ∁∁Ai= Ai, and it follows from (1) that

∁

n

i=1

∁Bi

=

n

i=1 Bi

Then by taking the complements,

n

i=1

∁Bi= ∁

n

i=1 Bi

We see that (2) follows, when we replace Bi by Ai

Example 2.2 Let A and B be two subsets of the set Ω We define the symmetric set difference AΔB

by

AΔB = (A \ B) ∪ (B \ A)

Prove that

AΔB = (A ∪ B) \ (A ∩ B)

Then letA, B and C be three subsets of the set Ω Prove that

(AΔB)ΔC = AΔ(BΔC)

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2 Set theory

B minus A

A f lles B

A minus B

Figure 1: Venn diagram for two sets

The claim is easiest to prove by a Venn diagram Alternatively one may argue as follows:

1a If x ∈ (A \ B) ∪ (B \ A), then x either lies in A, and not in B, or in B and not in A This means

that x lies in one of the sets A and B, but not in both of them, hence

AΔB = (A \ B) ∪ (B \ A) (A ∪ B) \ (A ∩ B)

1b Conversely, if x ∈ (A ∪ B) \ (A ∩ B), and A = B, then x must lie in one of the sets, because

x ∈ A ∪ B and not in both of them, since x /∈ A ∩ B, hence

(A ∪ B) \ (A ∩ B) (A \ B) ∪ (B \ A) = AΔB

1c Finally, if A = B, then it is trivial that

AΔB = (A \ B) ∪ (B \ A) = ∅ = (A ∪ B) \ (A ∩ B)

Summing up we get

AΔB = (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)

2 If x ∈ AΔB, then x either lies in A or in B, and not in both of them Then we have to check two

possibilities:

(a) If x ∈ (AΔB)ΔC and x ∈ (AΔB), then x does not belong to C, and precisely to one of the

sets A and B, so we even have with equality that

{(AΔB)ΔC} ∩ (AΔB) = (A \ (B ∪ V )) ∪ (B \ (A ∪ C))

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2 Set theory

D C

B A

Figure 2: Venn diagram of three discs A, B, C The set (AΔB)ΔC is the union of the domains in

which we have put one of the letters A, B, C and D

(b) If instead x ∈ (AΔB)ΔC and x ∈ C, then x does not belong to AΔB, so either x does not

belong to any A, B, or x belongs to both sets, so we obtain with equality,

{(AΔB)ΔC} ∩ C = {C \ (A ∪ B)} ∪ {A ∪ B ∪ C}

Summing up we get

∪ (B \ (A ∪ C)) only contained in B,

∪ (C \ (A ∪ B)) only contained in C,

∪ (A ∩ B ∩ C) contained in all three sets

By interchanging the letters we get the same right hand side for AΔ(BΔC), hence

(AΔB)ΔC = AΔ(BΔC)

possibilities:

(a) If x ∈ (AΔB)ΔC and x ∈ (AΔB), then x does not belong to C, and precisely to one of the

sets A and B, so...

(b) If instead x ∈ (AΔB)ΔC and x ∈ C, then x does not belong to AΔB, so either x does not

belong to any A, B, or x belongs to both sets, so we obtain with equality,

{(AΔB)ΔC} ∩... class="text_page_counter">Trang 9

I ntroduction to Probability< /p>

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2 Set theory

Example 2.1 Let A1,A2,

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