Lektsii (1) / Lecture 8

ICEF, 2012/2013 STATISTICS 1 year LECTURES

Lecture 8

30.10.2012

	Compound events
Let A, B be some events. We shall consider the following events:
•	A B − the union of A and B, A or B. The set A B consists of the elements that are
	contained in A or B. This event occurs if A or B occurs.
•	A ∩B − the intersection of A and B, A and B. The set A ∩B consists of the elements that
	are contained in A and B. This event occurs if A and B occurs.
•	A \ B − the complement of B to A, A minus B. The set A \ B consists of the elements that
	are contained in A but do not belong to B. The event A \ B occurs if A occurs and B does
	not occur. The complement S \ A is denoted as		.
		A

Example. Suppose that a student randomly selected from the ICEF students. Let A = {a student is a male}, B = {a student is honours student}. Then A B means that selected student is either a male or honours student or both. The event A ∩B means that a male honours student was selected, while A \ B means that selected student is a male student but not honours student.

Definition. If A ∩B = then the events A, B are called disjoint.

Properties of probability of events

1.0 ≤ Pr(A) ≤1, Pr(S) =1.

2.If A B then Pr(A) ≤ Pr(B) .

3.Additivity: if A ∩B = then Pr(A B) = Pr(A) +Pr(B) .

4.Pr(A) =1−Pr(A) .

5.If A B then Pr(A \ B) = Pr(A) −Pr(B) .

6.Generalization of 3:

Pr(A B) = Pr(A) +Pr(B) −Pr(A ∩B) .

CONDITIONAL PROBABILITY (условная вероятность)

Example. In some University there is a group of 25 mathematics students and there is a group of 30 economics students. It s known that 8 mathematicians prefer football and 17 remaining mathematicians prefer basketball, while in economics group 20 students prefer football and 10 prefer basketball.

	Football	Basketball	Total
Mathem.	8	17	25
Econ.	20	10	30
Total	28	27	55

A student is randomly selected from these two groups. What is the probability that she/he prefers football? Obviously Pr( football) = 258 ++2030 = 5528 . Now assume that we know that a selected student is a mathem. How we should recalculate the corresponding probability? It is clear that

given the event B = {a student is a math.} the probability that she/he prefers football is Pr( football | math.) = 258 = 25/8/ 5555 .

But 558 = Pr( football ∩math.), 5525 = Pr(math.) . And we get in this example

Pr( football | math.) = Pr( football ∩math.) . Pr(math.)

This is the reason to give the general definition:

Definition. Conditional probability of an event A given an event B is

Pr(A | B) = Pr(A ∩B) .

Pr(B)

Example. Consider the families with two children. What is the probability that both children are boys if it is known that there is a boy in the family? Here A = {both children are boys}, B = {there is a boy in the family}. Clearly Pr(A) =1/ 4, Pr(A ∩B) = Pr(A), Pr(B) =3/ 4 and

Pr(A | B) =1/ 3 − slightly contradicts to our intuition because the expected answer is ½.

FULL PROBABILITY AND BAYES RULE

Example. There are two machines producing the same detail. The first machine produces 60% of the amount of details and gives 3% of defective details while the second machine produces 40% of the amount of details and gives 5% defective details. A detail is randomly selected from the common box. What is the probability that it is defective?

Let A = {a detail is defective}, H1 = {a detail is produced by the first machine}, H2 ={a detail is produced by the second machine}. Then

Pr(H1 ) =0.6, Pr(A | H1 ) =0.03, Pr(H1 ) =0.4, Pr(A | H2 ) =0.05 .

Clearly, A =(A ∩H1 ) (A ∩H2 ) and (A ∩H1 ) ∩(A ∩H2 ) = . By additivity of probability,

we get:

Pr(A) = Pr(A | H1) Pr(H1 ) +Pr(A | H2 ) Pr(H2 ) = 0.03 0.6 +0.05 0.4 = 0.038

Generally, let H1,..., Hn are events such that: 1) Hi ∩H j = , i ≠ j ;

2) n	Hi = S (the space of elementary outcomes).
i=1

Definition. The set of events H1,..., Hn with the properties 1), 2) is called a full group of events.

In other words, events H1,..., Hn may be considered as a partition of S. Then

Pr(A) = ∑n Pr(A | Hi ) Pr(Hi ) −

i=1

full probability (формула полной вероятности).

Example (continued). It is known that a selected detail is defective. What is the probability that it is produced by the first machine? We have

	Pr(H1	| A) =	Pr(H1	∩ A)	=	Pr(A | H1) Pr(H1 )		=	0.018	= 0.474 .
			Pr(A)			Pr(A | H1) Pr(H1 ) +Pr(A | H2 ) Pr(H	2 )		0.038

Generally, let H1,..., Hn be a full group of events and let A be some event. Then for any k

Pr(Hk | A) = nPr(A | Hk ) Pr(Hk ) − Bayes rule

∑Pr(A | Hi ) Pr(Hi )

i=1

Example (from Wonnacott book)

Pr(OK) =0.7, Pr(F) =0.3, Pr("F "| F) =0.9, Pr("OK "| F) =0.1

Pr("OK "| OK) =0.8, Pr("OK "| F) =0.2 How to find
	Pr(F |"F "),					Pr(F |"OK ") ?
Using Bayes rule we get
P(F |"F ") =						P(" F "| F) P(F)	=
		P("F "| F) P(F) +P("F "| OK ) P(OK)
=	0.9 0.3					= 0.659 .
	0.9 0.3 +0.2 0.7
Similarly
P(F | "OK ") =						P("OK "| F) P(F)		=
			P("OK "| F) P(F) +P("OK "| OK) P(OK )
=	0.1 0.3					= 0.051
	0.1 0.3 +0.8 0.7

INDEPENDENCE

An event A does not depend on an event B if Pr(A | B) = Pr(A) . This is equivalent to the equality

Pr(A ∩B) = Pr(A) Pr(B) .

(*)

It follows that if A does not depend on B then B does not depend on A. Definition. Two events A and B are called independent if the equality (*) holds.

Example (independence of value and color of a card). One card is randomly selected from a standard 52-cards deck. Consider two events:

A = {ace}, B = {red color}.

Then

Pr(A) = 524 , Pr(B) = 5226 . A ∩B = {red ace} amd

Pr(A ∩B) = 522 = 524 5226 = Pr(A) Pr(B) , i.e. events A, B are independent, as expected.

Let A1, A2 ,..., An be the collection of random events.

Definition. Random events A1, A2 ,..., An are called mutually independent if for any subcollection Ak1 ,..., Akm the equality

Pr(Ak1 ∩...∩Akm ) = Pr(Ak1 ) ... Pr( Akm )

holds.

For example, Pr(A2 ∩ A4 ∩A13 ) = Pr(A2 ) Pr(A4 ) Pr( A13 ) must hold.

Example (pairwise independence does not imply mutual independence). Consider a tetrahedron (triangle pyramid with equal sides). Three sides of it have the colors red, white and black, respectively, while the fourth side has all these three colors. The tetrahedron is rolled once. Let’s consider the events

R = {the bottom side has red color}, W = {the bottom side has white color}, B = {the bottom side has black color}.

Obviously,

Pr(R) = Pr(W ) = Pr(B) = 12

Pr(R ∩W ) = 14 = Pr(R) Pr(W ), Pr(R ∩B) = 14 = Pr(R) Pr(B) ,

Pr(B ∩W ) = 14 = Pr(B) Pr(W ) ,

i.e. the events R, W, B are pairwise independent. But Pr(R ∩W ∩B) = 14 ≠ Pr(R) Pr(W ) Pr(B)