概率论总结

北京邮电大学国际学院

高欢

Introduction to the Probability

1.1the fundamental knowledge about probability

Experiments : It is used in probability theory to describe virtually every progress whose outcome is not known in advance with certainty.

1. repeat again and again

2. Many outcomes

Experiments’ characteristic

3. Know all the outcomes, but do not know which one

will be obtained in some experiment

4. Simple events , multiple events

the sample space: the collection of all possible outcomes of an experiment.

For example : roll a six-sided dice, the sample space contains the six numbers 1,2,3,4,5,6.

It can be written as

S={1,2,3,4,5,6}

1.2 the definition of Probability

In a given experiment, it is necessary to assign to each event A in the sample space S a number Pr(A) which indicates the probability that A will occur

1. For every event A, Pr(A)?0

Whatever is worth doing is worth doing well

Axiom 2. Pr(S)=1

3. For every infinite sequence of disjoint events A1, A2, ……..

???? Pr??Ai???Pr(Ai)

?i?1?i?1

1. Pr(?)=0

2 . For every finite sequence of n disjoint events A1, …….,An,

?n?nTheorems Pr??Ai???Pr(Ai)

?i?1?i?1 3. For every event A, Pr(Ac)=1-Pr(A)

4. For every event A, 0?Pr(A)?1

5. For every two events A and B, Pr(A?B)=Pr(A)+Pr(B)-Pr(AB)

Corollaries from the Axioms:

If A?B, then Pr(BA) = Pr(B) ? Pr(A) If A?B, then Pr(A)? Pr(B)

Example 1

Consider two events A and B such that Pr(A)=1/3 and Pr(B)=1/2. Determine the value of Pr(BAc) for each of the following conditions: (a) A and B are disjoint; (b) A?B; (c) Pr(AB)=1/8 Solution:

(a) Pr(BAC)= 1/2 (b) Pr(BAC)= 1/2-1/3=1/6

(c) Pr(BAC)=Pr(B)-Pr(AB)=1/2-1/8=3/8

c

1.3 simple Sample Spaces and Combination Method

The definition of simple sample space:

A sample space S containing n outcomes s1, …….,sn is called a simple

sample space if the probability assigned to each of the outcomes s1,…,sn is 1/n. If an

Whatever is worth doing is worth doing well

event A in this simple sample space contains exactly m outcomes, then Pr(A)=

m. n Example1:

Consider an experiment in which a fair coin is tossed once and a balanced die is rolled once.(a) Describe the sample space for this experiment. (b) What is the probability that a head will be obtained on the coin and an odd number will be obtained on the die? Solution: (a) 1 2 3 4 5 6 head tail (head,1) (tail, 1) (head, 2) (head,3) (head, 4) (head,5) (head, 6) (tail ,2) (tail, 3) (tail, 4) (tail, 5) (tail, 6)

S={(head,1) (head, 2) (head,3) (head, 4) (head,5) (head, 6)

(tail, 1) (tail ,2) (tail, 3) (tail, 4) (tail, 5) (tail, 6)} (b)

3Pr(A)==1/4

2?6?n?n!Binomial cofficients ???

?k?k!(n?k)!For all numbers x and y and each positive integer n, (x+y)=

n

?n?kn?k???xy k?0?k?nExample 2

The United States Senate contains two senators from each of the 50 states.(a) If a committee of eight senators is selected at random, what is the probability that it will contain at least one of the two senators from a certain specified state?(b) what is the probability that a group of 50 senators selected at random will contain one senator from each state?

1726C2C98?C2C98 Solution: (a) Pr(A)= 8C100250 (b) Pr(B)=50

C100

Whatever is worth doing is worth doing well

1.4 The probability of a Union of Events

Theorem 1:

Pr(A1?A2?A3)= Pr(A1) +Pr(A2)+Pr(A3)-[Pr(A1A2)+Pr(A2A3)+Pr(A1A3)]+Pr(A1A2A3). Example 1:

Suppose that four guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guests will receive the proper hat.

Solution: Pr(A)=

3?33= 4A48Example 2:

Among a group of 200 students, 137 students are enrolled in a mathematics class, 50 students are enrolled in a history class, and 124 students are enrolled in a music class. Furthermore, the number of students enrolled in both the math and history classes is 33, the number of students enrolled in both history and music classes is 29, the number of students enrolled in both math and music is 92. Finally the number of students enrolled in all three classes is 18. We shall determine the probability that a student selected at random from the group of 200 students will be enrolled in at least one of the three classes. Solution :

let A denote the event that the selected student is enrolled in the math class. Let B

denote the event that the selected student is enrolled in the history class. Let C denote the event that the selected student is enrolled in the music class.

Pr(A)=137/200 Pr(B)=50/200 Pr(C)=124//200 Pr(AB)=33/200 Pr(BC)=29/200 Pr(AC)=92/200 Pr(ABC)=18/200 We can get

Pr(A?B?C) =7/8

2 conditional probability

2.1

the definition of conditional probability

Pr(AB) Pr(AB)=Pr(B)Pr(A|B) or Pr(AB)=Pr(A)Pr(B|A)

Pr(B) Pr(A|B)=

Suppose that A1,A2,….An are events such that Pr(A1A2..An-1)>0. Then Pr(A1A2..An)=Pr(A1)Pr(A2|A1)Pr(A3|A1A2)…Pr(An|A1A2…An-1)

Whatever is worth doing is worth doing well

Suppose that A1,A2,…An, B are events such that Pr(A1A2..An-1|B)>0. Then Pr(A1A2…An|B)=Pr(A1|B)Pr(A2|A1B)…Pr(An|A1A2..An-1B)

Example 1

Selecting four balls. Suppose that four balls are selected one at a time, without replacement, from a box containing r red balls and b blue balls(r>=2, b>=2). We shall determine the probability of obtaining the sequence of outcomes red, blue, red, blue. Solution:

If we let Rj denote the event that a red ball is obtained on the jth draw and let Bj denote the event that a blue ball is obtained on the jth draw, then

Pr(R1B2R3B4)=Pr(R1)Pr(B2|R1)Pr(R3|R1B2)Pr(B4|R1B2R3)

rbr?1b?1??? = r?br?b?1r?b?2r?b?32.2 Independent Events

Definition: A and B are independent if and only if Pr(A|B)=Pr(A) and Pr(B|A)=Pr(B). Pr(AB)=Pr(A)Pr(B)>=0

Theorem If two events A and B are independent, then the events A and BC are also independent Pr(ABC)=Pr(A)Pr(BC)

In particular, in order for three events A,B, and C to be independent, the following four relations must be satisfied:

Pr(AB)=Pr(A)Pr(B) Pr(AC)=Pr(A)Pr(C) Pr(BC)=Pr(B)Pr(C)

Pr(ABC)=Pr(A)Pr(B)Pr(C)

If A and B are independent event, then AC and BC are also independent Proof: Pr(ACBC)=1-Pr(A?B)

=1-Pr(A)-Pr(B)+Pr(A)Pr(B) =[1-Pr(B)]-Pr(A)[1-Pr(B)]

=[1-Pr(B)][1-Pr(A)]=Pr(AC)Pr(BC) Example 1

Two students A and B are both registered for a certain course. Assume that student A attends class 80 percent of the time, student B attends class 60 percent of the time, and the absences of two students are independent.

(a)what is the probability that at least one of the two students will be in class on a given day?

(b) if at least one of the two students is in class on a given day, what is the probability that A is in class that day.

Solution:

(a) Pr(A?B)=Pr(A)+Pr(B)-Pr(A?B) =0.8+0.6-0.8?0.6 =0.92

Whatever is worth doing is worth doing well

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