Chapter 2. Probability
1.Sample space
Sample space
In the study of statistics we are concerned with the presentation and interpretation of chance outcomes that occur in a planned study or scientific investigation.
For example, we may record the number of accidents that occur monthly at the intersection of Driftwood Lane and Royal Oak Drive, hoping to justify the installation of a traffic light; we might classify items coming of an assembly line as "defective" or "non defective"; or we may be interested in the volume of gas released in a chemical reaction when the concentration of an acid is varied.
The statistician is often dealing with either experimental data, representing counts or measurements, or perhaps with categorical data that can be classified according to some criterion.
We shall refer to any recoding of information, whether it be numerical or categorical, as an observation.
The number 2, 0, 1 and 2, representing the number of accidents that occurred for each month from January through April during the past year at the intersection of Driftwood Lane and Royal Oak Drive, constitute a set of observations.
The categorical data N, D, N, D and D,representing the items found to be defective or non defective when five items are inspected, are recorded as observations.
experiment
Experiment：any process that generates a set of data.
tossing of a coin is a statistical experiment .There are only two possible outcomes, heads or tails.
Another experiment might be the launching of a missile and observing its velocity at specified times.
The opinions of voters concerning a new sales tax can also be considered as observations of an experiment.
particularly interested in the observations obtained by repeating the experiment several times.
example: a coin is tossed repeatedly
? cannot predict the result of a given toss
(i.e. the outcome depend on chance)
? know the entire set of possibilities for each toss.
Sample space
Definition 2.1 The set of all possible outcomes of a statistical experiment is called the sample space and is represented by S. Each outcome in a sample space is called an element or a sample point.
What's the possible outcomes when a coin is tossed?
The sample space may be written
S =｛H, T｝
H and T corresponds to 'heads' and 'tails', respectively.
Example 2.1 Consider the experiment of tossing a die. interested in the number that shows on the top face
S1 =｛｝
S=｛｝
S1provides more information than S2 If we know which elements in S1occurs, we can tell which outcome in S2 occurs; however, a knowledge of what happens in S2 is of little help in determining which element in S1 occurs.
Remarks:
? more than one sample space can be used to describe the outcomes of an experiment.
? It is desirable to use a sample space that gives the most information.
Example 2.3Suppose that three items are selected at random from a manufacturing process. Each item is inspected and classified defective, D, or non defective, N. List the elements of the sample space providing the most information.
The sample space is
S =｛DDD, DDN, DND, DNN, NDD, NDN, NND, NNN｝.
sample space with a large or infinite number of sample points ，How to describe?
by a statement or rule
examples
? S =｛x | ｝
? all points (x,y) on the boundary or the interior of a circle of radius 2 with center at the origin
sample space with a large or infinite number of sample points ，How to describe?
by a statement or rule
examples
? S =｛x | x is a city with a population over 1 million｝
? all points (x,y) on the boundary or the interior of a circle of
radius 2 with center at the origin
｛| x 2+ y 2≤｝
2.Events
Event
For any given experiment we may be interested in the occurrence of certain events rather than in the outcome of a specific element in the sample space.
event A: the outcome (when a die is tossed) is divisible by 3.
Event A will occur if the outcome is an element of the set A =｛3, 6｝
A =｛3, 6｝is the subset of the sample space
S1 =｛｝in Example 2.1.
event B: the number of defectives is greater than 1 in Example 2.3; this will occur if the outcome is an element of the subset
B =｛｝
Example 2.4.Given the sample space S =｛t|t≥ 0｝, where t is the life in years of a certain electronic component. Event A is that the component fails before the end of the fifth year.
subset A =｛t|0≤t < 5}
Two special subset
a subset that includes the entire sample space S
a subset contains no elements at all, denoted by ?, called null set.
｛x|x is an even factor of 7｝
then B must be the null set, since the only possible factors of 7 are odd numbers 1 and 7.
complement
Definition 2.3The complement of an event A with respect to S is the subset of all elements of S that are not in A. We denote the complement of A by the symbol A’
Example 2.5 Let R be the event that a red card is selected from an ordinary deck of 52 playing cards and let S be the entire deck. What is R’?
R’ is the event that the card selected from the deck is not red but a black card.
intersection
operations with events result in the formation of new events
Definition 2.4The intersection of two events A and B, denoted by the symbol A∩B, is the event containing all elements that are common to A and B.
Example 2.7
Let P be the event that a person selected at random while dining at a popular cafeteria is a taxpayer, and let Q be the event that the person is over 65 years of age. Then the event P∩Q is the set of all taxpayers in the cafeteria who are over 65 years of age.
Example 2.8 Let M =｛｝and N =｛r, s, t｝; then it follows that M∩?
That is, M and N have no elements in common and, therefore, cannot both occur simultaneously.
Definition 2.5 Two events A and B are mutually exclusive, or disjoint if A∩?, that is, if A and B have no elements in common.
Union
Definition 2.6 The union of the two events A and B, denote by symbol A∪B, is the event containing all the elements that belong to A or B or both.
Example 2.10 Let A =｛a, b, c｝｛｝; then
A∪｛a, b, c, d, e｝.
Example 2.11 Let P be the event that an employee selected at random from an oil drilling company smokes cigarettes. Let Q be the event that the employee selected drinks alcoholic beverages. Then the event P∪Q is the set of all employees who either drink or smoke, or do both.
Example 2.12If M =｛x|3 < x < 9｝｛x|5 < x < 12｝
then M∪N =｛x|3 < x < 12｝
Several results
The relationship between events and the corresponding sample space can be illustrated graphically by means of Venn diagrams.
A∩??，A∪?= A
A∩ A’= ?，A∪A’= S
S’=??’=S
（A’）’=A
（A∩B）’=A’∪B’，（A∪B）’=A’∩B’
3.Counting sample points
One the problems that the statistician must consider and attempt to evaluate is the element of chance associated with the occurrence of certain events when an experiment is performed.
In many cases we shall be able to solve a probability problem by counting the number of points in the sample space without actually listing each element.
Multiplication rule
Theorem 2.1 If an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, then the two operations can be performed together in n1n2ways.
Example 2.13How many sample points are in the sample space when a pair of dice is thrown once?
Solution The first dice can land in any one of n1= 6 ways. For each of these 6 ways the second die can also land in n2 = 6 ways. Therefore, the pair of dice can land in n1n2= 6× 6 = 36 possible ways.
The multiplication rule of Theorem 2.1 may be extended to cover any number of operations.
Theorem 2.2 If an operation can be performed in n1ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed i n 3 ways, and so forth, then the sequence of k operations can be performed in n1n2 . . n k ways.
Example 2.15 Sam is going to assemble a computer by himself. He has the choice of ordering chips from two brands, a hard drive from four, memory from three and an accessory bundle from five local stores. How many different ways can Sam order the parts?
Solution Since n1= 2,n2 = 4, n3 = 3, and n4 = 5, there are
n1×n2×n3×n4 = 2×4×3×5 = 120
different ways to order the parts.
Permutation
How many different arrangements are possible for sitting 6 people around a table?
How many different orders are possible for drawing 2 lottery tickets from a total of 20?
The different arrangements are called permutation.
Consider the three letters a, b, and c.
The possible permutations are abc, acb, bac, bca, cab, and cba. There are 6 different arrangement.
Using Theorem 2.2 we could arrive at the answer 6 without listing the diferent orders.
There are n1 = 3 choices for the first position, then n2 = 2 for
the second, leaving only n3 = 1 choice for the last position, giving a total of
n1×n2×n3= 3×2×1 = 6 permutations.
Theorem 2.3 The number of permutations of n distinct objects is n!.
n ! is read 'n factorial'
n! = n(n-1)(n-2)…(3)(2)(1)
The number of permutations of the four letters a, b, c, and d will be 4! = 24.
What's the number of permutations that are possible by taking the four letters two at a time?
Using Theorem 2.1, we have n1 = 4 choices for the first position and n2 = 3 for the second.
A total of n1n2 = 12 permutations.
In general, n distinct objects taken r at a time can be arranged in n(n-1)(n-- r + 1) ways.
Theorem 2.4 The number of permutations of n distinct objects taken r at a time is
Example 2.17 Three awards (research, teaching and service) will be given one year for a class of 25 graduate students in a statistic department. If each student can receive at most one award, how many possible selections are there?
Solution Since the awards are distinguishable, it is a permutation problem. The total number of sample points is
circular permutation:permutation that occur by arranging objects in a circle
Example. 4 people are playing bridge, we do not have a new permutation if they all move one position in a clockwise direction.
By considering one person in a fixed position and arranging the other three in 3! ways.
Theorem 2.5 The number of permutations of n distinct objects arranged in a circle is (n-1)!
combinations
combinations:the number of ways of selecting r objects from n without regard order. The number of such combinations is denoted by C r,n
Consider the four letters a, b, c, and d.
? the ways of selecting one letter from four? a , b, c, and d that is
? the ways of selecting two letters from four? ab, ac, ad, bc, bd, cd; that is
? the ways of selecting three letters from four?
abc, abd, acd, bcd; that is
Theorem 2.8 The number of combinations of n distinct objects taken r at a time is
Question 1: What's the number of distinct permutations of n things of which n1 of one kind,n2 of a second kind,..., n k of a k th kind?
hint: n objects, n positions
Example 2.19In a college football training session, the defensive coordinator needs to have 10 player standing in a row. Among these 10 players, there are 1 freshman, 2 sophomore, 4 juniors and 3 seniors, respectively. How many different ways can they be arranged in a row if only their class level will be distinguished?
Solution The total number of arrangements is
Question 2: What's the number of arrangements of a set of n objects into r cells with n1 elements in the first cell, n2elements in the second, and so forth?
Example 2.20In how many ways can 7 scientists be assigned to one triple and two double hotel rooms?
Solution: The total number of possible partitions would be
4. Probability of an Event
Perhaps it was man's unquenchable thirst for gambling that led to the early development of probability theory. In an effort to increase their winnings, gamblers called upon mathematicians to provide optimum strategies for various games of chance. Some of the mathematicians providing these strategies were Pascal, Leibniz, Fermat, and James Bernoulli.
Probability theory has branched out far beyond games of chance to encompass many other fields associated with chance occurrences, such as politics, business, weather forecasting, and scientific research.
probability
The remainder of chapter 2 only consider sample space contains a finite number of elements
To every point in the sample space we assign a probability such that the sum of all probabilities is 1.
The probability of an event A is denoted by P (A).
Definition 2.8 The probability of an event A is the sum of the weights of all sample points in A. Therefore,
0≤P (A)≤1, P (?) = 0 P (S) = 1,
Furthermore, if A1, A2, A3,…is a sequence of mutually exclusive events, then
Example 2.22 A coin is tossed twice. What is the probability that at least one head occurs?
Solution The sample space is S =｛HH, HT, T H, T T｝
The coin is balanced, each of these outcomes would be equally likely to occur. That is, the probability of each sample point is 1/4.
event A: at least one head occurring. A =｛HH, HT, T H｝
Example 2.23 A dice is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number less than 4 occurs on a single toss of the die, find P (E).
Solution The sample space is S =｛1, 2, 3, 4, 5, 6｝
assign a probability of w to each odd number, and 2 to each even number, we have 9 w= 1.
1/ 9 , and 2/9 are assigned to each odd and even number, respectively.
Since E =｛1, 2, 3｝
Example 2.24In Example 2.23 let A be the event that an even number turns up and let B be the event that a number divisible by 3 occurs. Find P (A∪B) and P (A∩B)
Solution We can get A =｛2, 4, 6｝｛3, 6｝, therefore,
A∪B =｛2, 3，4, 6｝and A∩ B =｛6｝
assign: 1/ 9→each odd number, 2/9→ each even number,
we have
If the sample space contains N elements, all of which are equally likely to occur, We assign 1/N to each element.
Event A containing n of these N sample points. P (A)?
Theorem 2.9 If an experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes correspond to event A, then the probability of event A is
Example 2.25 A statistic class for engineers consists of 25 industrial, 10 mechanical, 10 electrical, and 8 civil engineering students. If a person is randomly selected by the instructor to answer a question, find the probability that the student chosen is (a) an industrial engineering major, (b) a civil engineering or an electrical engineering major.
Solution
Denote by I, M, E, and C the students majoring in industrial, mechanical, electrical and civil engineering, respectively. students in the class: equally to be selected;
the total number: 53.
(a) Since 25 of the 53 students are majoring in industrial engineering, the probability of the event I is
(b) Since 18 of the 53 students are civil and electrical engineering majors, it follows that
Example 2.26In a poker hand consisting of 5 cards, find the probability of holding 2 aces and 3 jacks.
Solution
The number of ways of being dealt 2 aces from 4 is
The number of ways of being dealt 3 jacks from 4 is
By the multiplication rule of Theorem 2.1, there are n = 6× 4 = 24 hands with 2 aces and 3 jacks.
The total number of 5-card poker hands is
Therefore, the probability of event C of getting 2 aces and 3 jacks in a 5-card poker hand is
If event A occurs n A times in N repeated experiments under a certain conditions, then frequency of A occurring in N experiments is defined as:
When N is large enough, the frequency turns out to have a kind of stability.
i.e., the values of F N (A) show fluctuations which become progressively weaker as N increases, until ultimately F N (A) stabilizes to a constant.
Tossing a fair coin, let A =｛head comes up｝
Frequency stabilizes to 1/2
The statistical definition of probability:
The constant to which the frequency of the event A stabilizes is called the probability of the occurrence of event A (P (A).
5. Additive Rules
Often it is easier to calculate the probability of some event from other events.
the event in question can be represented as the union of two other events, or as the complement of some event.
Lay out several important laws that simplify the computation of probabilities.
Additive Rules
additive rule
Theorem 2.10If A and B are any two events, then
P (A∪B) = P (A) + P (B)-P (A∩B).
PROOF. Use the V enn diagram.
Corollary 2 If A1, A2, . . , A n are mutually exclusive, then
P (A1∪A2∪. . , ∪A n) = P (A1) + P (A2) + . . + P (A n ).
a partition of S
1.｛A1, A2, . . , A n ｝
2. A1, A2, . . , A n are mutually exclusive and A1∪A2∪. . , ∪A n= S.
Corollary 3If｛A1, A2, . . , A n ｝is a partition of S, then
P (A1∪A2∪. . , ∪A n) =P (A1) + P (A2) + . . + P (A n ) = P (S) = 1
Example 2.27John is going to graduate from an industrial engineering department in a university by the end of the semester. After being interviewed at two companies he likes, he assesses that his probability of getting an offer from company A is 0.8, and the probability he gets an offer from company B is 0.6. If, on the other hand, he believes that the probability that he will get offers from both companies is 0.5.
What is the probability that he will get at least one offer from these two companies?
Solution Using the additive rule we have
P (A∪B) = P (A) + P (B)-P (A∩B) = 0.8 + 0.6-0.5 = 0.9.
Example 2.28What is the probability of getting a total of 7 or 11 when a pair of fair dice are tossed?
Solution
A: the event that 7 occurs; B: the event that 11 occurs.
● a total of 7 occurs for 6 of the 36 sample points and a total of 11 occurs for only 2 of the sample points.
●all sample points are equally likely, we have P (A) = 1/6 and P (B) = 1/18.
The events A and B are mutually exclusive, therefore
（此时两件事件不能同时发生！）
How about the union of 3 events?
Theorem 2.11 For three events A, B, and C,
P (A∪B∪C) = P (A) + P (B) + P (C)-P (A∩B) -P (A∩ C)-P (B∩C) + P (A∩ B∩C). How to prove?
Sometimes, it is more difcult to calculate the probability that an event occurs than it is to calculate the probability that the event does not occur.
Theorem 2.12 If A and A ’ are complementary events, then P (A) + P (A’) = 1.
Example 2.30If the probabilities that an automobile mechanic will service 3,4,5,6,7, or 8 or more cars on any given workday are, respectively, 0.12, 0.19, 0.28, 0.24, 0.10 and 0.07, what is the probability that he will service at least 5 cars on his next day at work?
Solution. Let E be the event that at least 5 cars are serviced.
Now, P (E) = 1 - P (E’), where E’ is the event that fewer than 5 cars are serviced.
P (E’) = 0.12 + 0.19 = 0.31, it follows that P (E) = 1-0.31 = 0.69.
6. Conditional Probability
Conditional Probability
The probability of an event B occurring when it is known that some event A has occurred.
denoted by P (B|A), read "the probability of B, given A".
Example
A die is constructed such that the even numbers are twice as likely to occur as the odd numbers.
i.e. the sample space is S =｛1, 2, 3, 4, 5, 6｝
event B: get a perfect square when a die is tossed. the probability that B occurs: P (B) = 1/3.
If it is known that the toss of the die resulted in a number greater than 3. What's the probability that B occurs under this condition?
｛4, 5, 6｝, a subset of S.
to find the probability that B occurs, relative to the space A, denote this event by B|A,
B|A =｛4｝
Remark: Events may have different probabilities when considered relative to different sample spaces.
We can also write
P (A∩B) and P (A): from the original sample space S.
Definition 2.9The conditional probability of B, given A, denoted by P (B|A), is defined by
a conditional probability relative to a subspace A of S may be calculated directly from the probabilities assigned to the elements of the original sample space S.
Example 2.31The probability that a regularly scheduled flight departs on time is P (D) = 0.83; the probability that it arrives on time is P (A) = 0.82; and the probability that it departs and arrives on time is P (D∩A) = 0.78. Find the probability that a plane
(a) arrives on time given that it departed on time,
(b) departed on time given that it has arrived on time.
Solution
a.
b.
It is important to know the probability that the flight arrives on time. One is given the information that the flight did not depart on time. Armed with this additional information, the more pertinent probability is P
(A|D’).
（注：A∩D’∪A∩D=A）
The probability of an on-time arrival is diminished severely in the presence of the additional information. The probability P (A|D’) is an "updating" of P (A) based on the knowledge that event D’has occurred.
Independent Events
previous example: the die-tossing experiment.
we have P（B|A) = 2/5, P (B) = 1/3.，P (B|A)= P (B), indicating that B depends on A.
Example 2 cards are drawn in succession from an ordinary deck, with replacement.
A: the first card is an ace,
B: the second card is a spade.
Since the first card is replaced, our sample space for both the first and second draws is the same.
P (B|A) = 13/52 and P (B) = 13/52.
We have P (B|A) = P (B), the events A and B are said to be independent.
Definition 2.10Two events A and B are independent if and only if
P (B|A) = P (B) or P (A|B) = P (A). Otherwise, A and B are dependent.
Remark The condition P(B|A) = P (B) implies that P (A|B) = P (A), and conversely.
7. Multiplicative Rules
Multiplicative Rules
Multiplying the formula of Definition 2.9 by P (A), we obtain the multiplicative rule
Theorem 2.13 If in an experiment the events A and B can both occur, then
P (A∩B) = P (A)P (B|A).
We can also write P (A∩B)= P (B)P (A|B).
Example 2.32Suppose that we have a fuse box containing 20 fuses, of which 5 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first, what is the probability that both fuses are defective?
Solution
event A: the first fuse is defective
event B: the second fuse is defective
A∩B: A occurs, and then B occurs after A has occurred
Example 2.33 One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black?
Solution B1: a black ball from bag 1;
B2: a black ball from 2;
W1: a white ball from bag 1.
If the first fuse is replaced in Example 2.32, then the probability of a defective fuse on the second selection is still 1/4.
i.e. P (B|A) = P (B) = 1/4. The events A and B are independent.
Theorem 2.14 Two events A and B are independent if and only if P (A∩B) = P (A)P (B).
Example 2.34 A small town has one fire engine and one ambulance available for emergencies. The probability that the fire engine is available when needed is 0.98, and the probability that the ambulance ia available when called is 0.92. In the event of an injury resulting from a burning building, find the probability that both the ambulance and the fire engine will be available. Solution Let A and B represent the respective events that the fire engine and the ambulance are available. Then
P (A∩B) = P (A)P (B) = 0.98×0.92 = 0.9016
Theorem 2.15 If the events A1, A2, A3, …Ak can occur,
then P (A1∩A2∩A3∩ …∩Ak) = P (A1)P (A2|A1)P (A3|A1∩A2）… P (A k|A1∩A2∩A3∩ …∩Ak-1）
Example 2.36 Three cards are draw in succession, without replacement, from an ordinary deck of playing cards. Find the probability that the event A1∩A2∩A3occurs, where A1 is the event that the first card is a red ace, A2 the second card is a 10 or a jack, and A3is the event that is the event that the third card is greater than 3 but less than 7.
Solution We have P (A1) = 2/52, P (A2|A1) = 8/51, P （A3|A1∩A2) = 12/50,
by Theorem 2.15 : P (A1∩A2∩A3 )= P (A1) P (A2|A1)P（A3|A1∩A2) = 8/5525.
Example 2.37A coin is biased so that a head is twice likely to occur as a tail. If the coin is tossed 3 times, what is the probability of getting 2 tails and 1 head?
Solution the sample space
｛｝
P (H) = 2/3, P (T ) = 1/3
event A: get 2 tails and 1 head; A =｛T T H, T HT, HT T｝
P (T T H) = P (T )P (T )P (H) = ( 1/3)(1/3 )(2/3 ) = 2/27.
Similarly, P (T HT ) = P (HT T ) = 2/27,
P (A) = 2/27 + 2/27 + 2/27 = 2/9.
8. Bayes' Rule
P (A) = P [(E∩A)∪(E∩A)] = P (E∩A) + P (E∩A)
= P (E)P (A|E) + P (E’)P (A|E’) draw the Venn diagram
Bayes' Rule
A generalization,theorem of total probability
Theorem 2.16 If the events B
1, B
2
, . . . , B
k
constitute a partition of the sample space S such that P (B
i
)≠0 for
i = 1 ，2 ..,,.,k, then for any event A of S
Example 2.38In a certain assembly plant, three machines, B1, B2,B3make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3% and 2% of the products made by each machine, respectively, are defective. Suppose that a finished products is randomly selected. What is the probability that it is defective?
Solution Applying Theorem 2.16, i.e. the theorem of total probability, we can write
P (A) = P (B1)P (A|B1) + P (B2)P (A|B2) + P (B3)P (A|B3).
On the other hand,
P (B1)P (A|B1) = 0.3×0.02 = 0.006,
P (B2)P (A|B2) = 0.45×0.03 = 0.0135,
P P (B3)P (A|B3).= 0.25×0.02 = 0.005.
Therefore,
P (A) = 0.006 + 0.0135 + 0.005 = 0.0245
Example 2.39With reference to Example 2.38, if a product were chosen randomly and found to be defective, what is the probability that it was made by machine B3?
Solution
Theorem 2.17 (Bayes' Rule)
If the events B
1, B
2
, . . . , B
k
constitute a partition of the sample space S, where P (B
i
)≠0 for i = 1, 2, . . . , k, then for any event
A in S such that P (A)≠0,
Exercises for Chapter 2
1. A truth serum has the property that 90% of the guilty suspects are properly judged while, of course, 10% of guilty suspects are improperly found innocent. On the other hand, innocent suspects are misjudged 1% of the time. If the suspect was selected from a group of suspects of which only 5% have ever committed a crime, and the scrum indicates that he is guilty, what is the probability that he is innocent?
3. By comparing appropriate regions of Venn diagrams
(a) (A∩B）∪(A∩B’) =?
5. How many bridge hands are possible containing 4 spades, 6 diamonds, 1 club, and 2 hearts?
7. A large industrial firm uses 3 local motels to provide overnight accommodations for its clients. From past experience it is known that 20% of the clients are assigned rooms at the Ramada Inn, 50% at the Sheraton, and 30% at the Lakeview Motor Lodge. If the plumbing is faulty in 5% of the rooms at the Ramada Inn, in 4% of the rooms at the Sheraton, and in 8% of the rooms at the Lakeview Motor Lodge, what is the probability that
(a) a client will be assigned a room with faulty plumbing?
(b) a person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge?
9. The probability that a patient recovers from a delicate heart operation is 0.8. What is the probability that
(a) exactly 2 of the next 3 patients who have this operation survive?
(b) all of the next 3 patients who have this operation survive?
11. From 4 red, 5 green, and 6 yellow apples, how many selections of 9 apples are possible if 3 of each color are to be selected?
13. A shipment of 12 television sets contains 3 defective sets. In how many ways can a hotel purchase 5 of these sets and receive at least 2 of the defective sets?
15. A certain federal agency employs three consulting firms (A, B, and C) with probabilities 0.40, 0.35, and 0.25, respectively. From past experience it is known that the probability of cost overruns for the firms are 0.05, 0.03, and 0 .15, respectively. Suppose a cost overrun is experienced by the agency.
(a) What is the probability that the consulting firm involved is company C?
(b) What is the probability that it is company A?
概率论与数理统计 第二章习题 1 考虑为期一年的一张保险单，若投保人在投保一年内意外死亡，则公司赔付20万元，若投保人因其它原因死亡，则公司赔付5万元，若投保人在投保期末自下而上，则公司无需传给任何费用。若投保人在一年内因意外死亡的概率为0.0002，因其它原因死亡的概率为0.0010，求公司赔付金额的分崣上。 解 设赔付金额为X ，则X 是一个随机变量，取值为20万，5万，0，其相应的概率为0.0002；0.0010； 2.（1）一袋中装有5只球，编号为1,2，3,4，5。在袋中同时取3只，以X 表示取出的3只球中的最大号码，写出随机变量X 的分布律 （2）将一颗骰子抛掷两次，以X 表示两次中得到的小的点数，试求X 的分布律。 解 （1）在袋中同时取3个球，最大的号码是3,4，5。每次取3个球，其总取法： 3554 1021 C ?= =?，若最大号码是3，则有取法只有取到球的编号为1,2，3这一种取法。因而其概率为 2 2335511 {3}10 C P X C C ==== 若最大号码为4，则号码为有1,2，4；1，3,4； 2，3，4共3种取法， 其概率为23335533 {4}10 C P X C C ==== 若最大号码为5，则1，2,5；1,3，5；1，4,5；2,3，5；2,4，5；3,4，5共6种取法 其概率为 25335566 {5}10 C P X C C ==== 一般地 3 5 21 )(C C x X p x -==，其中21-x C 为最大号码是x 的取法种类数，则随机变量X 的分布律为
（2）将一颗骰子抛掷两次，以X表示两次中得到的小的点数，则样本点为S＝｛（1,1），（1,2），（1,3），…，（6,6）｝，共有36个基本事件， X的取值为1,2，3,4，5,6， 最小点数为1，的共有11种，即（1,1，），（1,2），（2,1）…，（1,6），（6,1），11 {1} 36 P X==； 最小点数为2的共有9种，即（2,2），（2,3），（3,2），…，（3,6），（6,3）， 9 {2} 36 P X==； 最小点数为3的共有7种， 7 {3} 36 P X==； 最小点数为4的共有5种， 5 {4} 36 P X==； 最小点数为5的共有3种， 3 {5} 36 P X==； 最小点数为6的共有1种， 1 {6} 36 P X== 于是其分布律为 3 设在15只同类型的产品中有2只次品，在其中取3次，每次任取1只，作不放回抽样，以X表示取出的次品的次数， （1）求X的分布律； （2）画出分布律的图形。 解从15只产品中取3次每次任取1只，取到次品的次数为0，1，2。在不放回的情形下， 从15只产品中每次任取一只取3次，其总的取法为：3 15151413 P=??，其概率为 若取到的次品数为0，即3次取到的都是正品，其取法为3 13131211 P=?? 其概率为 13121122 {0} 15141335 p X ?? === ??
《概率论》第二章练习答案 一、填空题： ”2x c S 1 1.设随机变量X的密度函数为f(x)= 则用丫表示对X的3次独立重复的 0 其匕 '- 观察中事件(X< -)出现的次数，则P (丫= 2)= ___________________ 。 2 2.设连续型随机变量的概率密度函数为： ax+b 0 4. 设为随机变量，E =3, E 2=11,则 E (4 10) = 4E TO =22 5. 已知X的密度为(x)二ax?"b Y 01 0 . x :: 1 1 1 (x ) =P(X?),则 3 3 6. 7. 1 1 (X〈一)= P ( X〉一)一 1 (ax b)dxjQx b) 联立解得: dx 若f(x)为连续型随机变量X的分布密度，则J［f(x)dx= ________ 1 ——'J 设连续型随机变量汕分布函数F(x)=x2/：, 丨1, x ：： 0 0 岂 x ：：： 1，则 P ( E =0.8 ) = _0_; P(0.2 ：：：： 6) = 0.99 8. 某型号电子管，其寿命(以小时记)为一随机变量，概率密度：(x)二 x _100 x2，某一个电子设备内配有3个这样的电子管，则电子管使用150小时都不0(其他) 需要更换的概率为_____ 厂100 8/27 _________ x> 100 第二章 随机变量及其分布 1、解： 设公司赔付金额为X ，则X 的可能值为； 投保一年内因意外死亡：20万，概率为0.0002 投保一年内因其他原因死亡：5万，概率为0.0010 投保一年内没有死亡：0，概率为1-0.0002-0.0010=0.9988 所以X 2、一袋中有5X 表示取出的三只球中的最大号码，写出随机变量X 的分布律 解：X 可以取值3，4，5，分布律为 10 61)4,3,2,1,5()5(1031)3,2,1,4()4(10 11)2,1,3()3(35 2 435 2 335 2 2=?= === ?==== ?= ==C C P X P C C P X P C C P X P 中任取两球再在号一球为中任取两球再在号一球为号两球为号一球为 也可列为下表 X ： 3， 4，5 P ：10 6, 103,101 3、设在15只同类型零件中有2只是次品，在其中取三次，每次任取一只，作不放回抽样，以X 表示取出次品的只数，（1）求X 的分布律，（2）画出分布律的图形。 解：任取三只，其中新含次品个数X 可能为0，1，2个。 35 22 )0(315313= ==C C X P 3512)1(3 15213 12=?==C C C X P 35 1)2(3 15 113 22= ?= =C C C X P 再列为下表 X ： 0， 1， 2 P ： 35 1, 3512,3522 4、进行重复独立实验，设每次成功的概率为p ，失败的概率为q =1－p (0 第二章 随机变量及其分布 I 教学基本要求 1、了解随机变量的概念以及它与事件的联系； 2、理解随机变量的分布函数的概念与性质；理解离散型随机变量的分布列、连续型随机变量的密度函数及它们的性质； 3、掌握几种常用的重要分布：两点分布、二项分布、泊松分布、均匀分布、指数分布、正态分布，且能熟练运用； 4、会求简单随机变量函数的分布. II 习题解答 A 组 1、检查两个产品，用T 表示合格品，F 表示不合格品，则样本空间中的四个样本点为 1(,)F F ω=、2(,)T F ω=、3(,)F T ω=、4(,)T T ω= 以X 表示两个产品中的合格品数. (1) 写出X 与样本点之间的对应关系； (2) 若此产品的合格品率为p ，求(1)p X =？ 解：(1) 10ω→、21ω→、31ω→、42ω→； (2) 1 2(1)(1)2(1)p X C p p p p ==-=-. 2、下列函数是否是某个随机变量的分布函数？ (1) 021()2021 x F x x x <-??? =-≤ 求常数A 及(13)p X <≤？ 解：由()1F +∞=和lim (1)x x A e A -→+∞ -=得 1A =； (13)(3)(1)(3)(1)p X p X p X F F <≤=≤-≤=- 3113(1)(1)e e e e ----=---=-. 4、设随机变量X 的分布函数为 2 00()0111 x F x Ax x x ≤??=<≤??>? 求常数A 及(0.50.8)p X <≤？ 解：由(10)(1)F F +=得 1A =； (0.50.8)(0.8)(0.5)(0.8)(0.5)p X p X p X F F <≤=≤-≤=- 220.80.50.39=-=. 5、设随机变量X 的分布列为 ()a p X k N == (1,2,,)k N =L 求常数a ？ 解：由 1 1i i p +∞ ==∑得 1 1N k a N ==∑ 1a ?=. 6、一批产品共有100个，其中有10个次品，求任意取出的5个产品中次品数的分布列？ 解：设X 表示5个产品中的次品数，则X 是离散型随机变量，其所有可能取值为0、1、…、 5，且 0510905100(0)C C p X C ==、1410905100(1)C C p X C ==、2310905100(2)C C p X C ==、321090 5100 (3)C C p X C ==、 4110905100(4)C C p X C ==、50 1090 5100 (5)C C p X C == 于是X 的分布列为 两人各投中两次的概率为： P(A ^ A 2B 1B 2^0.0784O 所以: 作业题解： 2.1掷一颗匀称的骰子两次，以X 表示前后两次出现的点数之和 ，求X 的概率分布，并验 证其满足(222) 式. 解: Q Q Q Q 根据 v P(X = k) =1，得 k =0 故 a 二 e 「1 2.3 甲、乙两人投篮时，命中率分别为0.7和0.4 ,今甲、乙各投篮两次，求下列事件的 概率： (1)两人投中的次数相同；(2) 甲比乙投中的次数多. 解：分别用A ，B j (i =1,2)表示甲乙第一、二次投中，则 P(A) = P(A 2)=0.7,P(A) = P(A 2)=0.3,P(B 1)= P(B 2)=0.4,P(B 1)= P(D) =0.6, 两人两次都未投中的概率为： P(A A 2 B^! B 2) = 0.3 0.3 0.6 0.6二0.0324， 两人各投中一次的概率为： 并且，P(X P(X P(X P(X = 12) = 1 36 =10) 煤 =8) 嗥； =k)=( =2) =P(X =4) =P(X =6) =P(X 2.2 2 P(X =3) =P(X =11)= ； 36 4 P(X =5) =P(X =9)= p (X =7)」。 36 k =2,3,4,5,6,7,8,9,10,11,12) P{X =k}二ae°,k =1,2…，试确定常数 解: k ae ae = 1 ，即 1=1。 k -0 1 - e P(AA2BB2)P(AA2B2B1)P(A2AB1B2)P(AA2B2B1)= 4 0.7 0.3 0.4 0.6 = 0.2016两人各投中两次的概率为：P(A^ A2B1B2^0.0784O所以: 第二章练习题(答案) 一、单项选择题 1. 已知连续型随机变量X 的分布函数为 3.若函数f(x)是某随机变量X 的概率密度函数，则一定成立的是(C ) A. f(x)的定义域是[0, 1] B. f(x)的值域为[0,1] 4.设X - N(l,l),密度函数为f(x),则有(C ) 5.设随机变量X ~ N (/M6), Y ?N 仏25)，记 P1 = P (X “ + 5), 则正确的是 (A)对任意“，均有Pi = p 2 (B)对任意“，均有Pi v p? (c)对任意〃，均有Pl > Pi (D )只对“的个别值有P1 = P2 6.设随机变量x ?N(10^s 2) 9 则随着s 的增加 P{|X- 10|< s} ( C ) F(x) = o, kx+b 、 x<0 0 < x< x> 则常数&和〃分别为 (A) k = —b = 0 龙， (B) k = 0,b 丄 (C) k = —,b = 0 (D) k = 0,b= 1 n In In 2.下列函数哪个是某随机变量的分布函数 (A ) z 7 fl -cosx ； 2 0, f sinx, A. f(x)』沁,xnO C. f (x)= a (a>0); B. f (x) 1, x < 0 [cosx, — - < X < - 1 2 2 D. f (x) 其他 0, 0 < X < 7T 其他 —-< x < - 2 2 其他 C- f(x)非负 D. f (x)在(-叫+00)内连续 A. P {X 《概率论》第二章 练习答案 一、填空题： 1．设随机变量X 的密度函数为f(x)=?? ?0 2x 其它1???o 则用Y 表示对X 的3次独立重复 的观察中事件(X≤ 2 1 )出现的次数，则P （Y ＝2）＝ 。 ?==≤4120 21)21(xdx X P 64 9 )43()41()2(1223===C Y p 2. 设连续型随机变量的概率密度函数为： ax+b 0 6．若f(x)为连续型随机变量X 的分布密度，则 ? +∞ ∞ -=dx x f )(＿＿1＿＿＿＿。 7. 设连续型随机变量ξ的分布函数?? ???≥<≤<=2,110, 4/0, 0)(2 x x x x x F ，则 P （ξ=0.8）= 0 ；)62.0(<<ξP = 0.99 。 8. 某型号电子管，其寿命（以小时记）为一随机变量，概率密度)(x ?＝ ()?????≥) (0100100 2其他x x ，某一个电子设备内配有3个这样的电子管，则电子管使用150小时都不需要更换的概率为＿＿＿8/27＿＿＿＿＿。 2100 x x≥100 ∴ ?(x)= 0 其它 P （ξ≥150）=1－F(150)=1－??=-+=+=150 10015010023 2 132********x dx x [P(ξ≥150)]3=(32)3=27 8 9. 设随机变量X 服从B （n, p ）分布，已知EX ＝1.6，DX ＝1.28，则参数n ＝___________， P ＝_________________。 EX = np = 1.6 DX = npq = 1.28 ，解之得：n = 8 ，p = 0.2 10. 设随机变量x 服从参数为（2，p ）的二项分布，Y 服从参数为（4，p ）的二项分布，若P （X ≥1）＝9 5 ，则P （Y ≥1）＝＿65/81＿＿＿＿＿＿。 解： 11. 随机变量X ～N （2， σ2） ，且P （2＜X ＜4）=0.3，则P （X ＜0）=＿＿0.2＿＿＿ % 2.8081 65 811614014==-=-=q p C o ) 0(1)1(=-=≥Y P Y p 31,3294)0(94 )1(95)1(2 = =?=∴===??= ≥p q q X p X p X p 概率论与数理统计作业 班级 姓名 学号 任课教师 第二章 随机变量及其分布 教学要求： 一、理解随机变量的概念；理解离散型随机变量及其分布律的定义，理解分布律的性质；掌 握(0-1)分布、二项分布、Poisson 分布的概念、性质；会计算随机变量的分布律. 二、理解分布函数的概念及其性质；理解连续型随机变量的定义、概率密度函数的基本性质， 并熟练掌握有关的计算；会由分布律计算分布函数,会由分布函数计算密度函数,由密度函数计算分布函数. 三、掌握均匀分布、正态分布和指数分布的概念、性质. 一、掌握一维随机变量函数的分布． 重点:二项分布、正态分布，随机变量的概率分布． 难点:正态分布，随机变量函数的分布． 练习一 随机变量、离散型随机变量及其分布律 1．填空、选择 (1)抛一枚质地均匀的硬币，设随机变量?? ?=，，出现正面 ，，出现反面H T X 10 则随机变量X 在区间 ]22 1 ，（上取值的概率为21. (2)一射击运动员对同一目标独立地进行4次射击,以X 表示命中的次数,如果 {}81 80 1= ≥X P ,则{}==1X P 8. (3)设离散型随机变量X 的概率分布为{},,2,1, ===i cp i X P i 其中0>c 是常数， 则（ B ） （A ）11-=c p ； （B ）1 1 +=c p ； （C ）1+=c p ； （D ）0>p 为任意常数 2．一袋中装有5只球，编号为1,2,3,4,5.在袋中同时取出3只球,以X 表示取出的3只球中的最大号码,写出随机变量X 的分布律. 解：从1～5中随机取3个共有103 5=C 种取法. 以X 表示3个中的最大值.X 的所有可能取值为;5,4,3 {}3=X 表示取出的3个数以3为最大值,其余两个数是1,2,仅有这一种情况,则 第二章 事件与概率 1、字母M ，A ，X ，A ，M 分别写在一张卡片上，充分混合后重新排列，问正好得到顺序MAAM 的概率是多少？ 解：这五个字母自左往右数，排第i 个字母的事件为A i ，则 42)(,52)(121== A A P A P ，2 1)(,31)(1234123==A A A A P A A A P 1)(12345=A A A A A P 。 利用乘法公式，所求的概率为 ()()()() 12345123412312154321)()(A A A A A P A A A A P A A A P A A P A P A A A A A P =30 1 121314252=????= 2、有三个孩子的家庭中，已知有一个是女孩，求至少有一个男孩的概率。 解：有三个孩子的家庭总共有23=8个类型。设A={三个孩子中有一女}，B={三个孩子中至少有一男}，A 的有利场合数为7，AB 的有利场合为6，依题意所求概率为P （B|A ），则 ()7 6 8/78/6)()(=== A P A B P A B P . 3、若M 件产品中包含m 件废品，今在其中任取两件，求：（1）已知取出的两件中有一件是废品的条件下，另一件也是废品的条件概率；（2）已知两件中有一件不是废品的条件下，另一件是废品的条件概率；（3）取出的两件中至少有一件是废品的概率。 3、解：（1）M 件产品中有m 件废品，m M -件正品。设A={两件有一件是废品}，B={两件都是废 品}，显然B A ?，则 () 1122()/m M m m M P A C C C C -=+ 2 2/)(M m C C B P =， 题中欲求的概率为 )(/)()(/)()|(A P B P A P AB P A B P ==1 21 /)(/2 2112 2---=+=-m M m C C C C C C M m m M m M m . （2）设A={两件中有一件不是废品}，B={两件中恰有一件废品}，显然A B ?，则 () ,/)(2112M m M m m M C C C C A P --+= 2 11/)(M m M m C C C B P -=. 题中欲求的概率为 )(/)()(/)()|(A P B P A P AB P A B P ==1 2/)(/2 1122 11-+=+=---m M m C C C C C C C M m M m m M M m M m . （3）P{取出的两件中至少有一件废品}=( ) ) 1() 12(/2 2 11---= +-M M m M m C C C C M m m M m . 概率论与数理统计课后习题答案 第二章 1.一袋中有5 只乒乓球，编号为1，2，3，4，5，在其中同时取3只，以X 表示取出的3只 球中的最 大号码，写出随机变量X 的分布律. 【解】 35 35 24 35 3,4,51 (3)0.1C 3(4)0.3C C (5)0.6 C X P X P X P X ====== ==== 2.设在15只同 类型零件中有2只为次品，在其中取3次，每次任取1只，作不放回抽样，以X 表示取出 的次品个数，求： （1） X 的分 布律； （2） X 的分 布函数并作图； (3) — 133{},{1},{1},{12}222 P X P X P X P X ≤<≤≤≤<<. 【解】 31331512213 3151133 150,1,2. C 22 (0). C 35 C C 12(1). C 35 C 1 (2).C 35 X P X P X P X ========== 故X 的分布律为 （2） 当x <0时， F （x ）=P （X ≤x ）=0 当0≤x <1时 ，F （x ）=P （X ≤x ）=P (X =0)= 2235 当1≤x <2时 ，F （x ）=P （X ≤x ）=P (X =0)+P (X =1)=3435 当x ≥2时， F （x ）=P （X ≤x ）=1 故X 的分布函 数 0, 022 ,0135 ()34,12351,2x x F x x x 1、考虑为期一年的一张保险单，若投保人在投保一年后因意外死亡，则公司赔付20万元， 若投保人因其他原因死亡，则公司赔付5万元，若投保人在投保期末生存，则公司无需付给任何费用。若投保人在一年内因意外死亡的概率为0.0002，因其他愿意死亡的概率为0.0010，求公司赔付金额的分布律。 解：设X为公司的赔付金额，X=0,5,20 P（X=0）=1-0.0002-0.0010=0.9988 P（X=5）=0.0010 P（X=20）=0.0002 X 0 5 20 P 0.9988 0.0010 0.0002 2.(1) 一袋中装有5只球，编号为1,2,3,4,5.在袋中同时取3只球，以X表示取出的三只中的最大号码，写出随机变量的分布律. 解：方法一: 考虑到5个球取3个一共有=10种取法，数量不多可以枚举来解此题。 设样本空间为S S=｛123,124,125,134,135,145,234,235,245,345 ｝ 易得，P｛X=3｝=；P｛X=4｝=；P｛X=5｝=； X 3 4 5 1/10 3/10 6/10 方法二：X的取值为3,4,5 当X=3时，1与2必然存在，P｛X=3｝= =; 当X=4时，1,2,3中必然存在2个，P｛X=4｝= =； 当X=5时，1,2,3,4中必然存在2个，P｛X=5｝= =； X 3 4 5 1/10 3/10 6/10 (2)将一颗骰子抛掷两次，以X表示两次中得到的小的点数，试求X的分布律. 解：P｛X=1｝= P (第一次为1点)+P（第二次为1点）- P（两次都为一点） = =； P｛X=2｝= P (第一次为2点，第二次大于1点)+P（第二次为2点，第一次大于1点）- P（两次都为2点） For personal use only in study and research; not for commercial use 《概率论》第二章 练习答案 螂 一、填空题: "2x 莁 1 .设随机变量X 的密度函数为f(x)=丿 1 的观察中事件(XW —)出现的次数，则 P (Y = 2)= ___________________ 2 P(X J)「￡xdx 二 2 0 2 1 2 3 1 9 袇 P —F (3)2 螃 2.设连续型随机变量的概率密度函数为: -ax+b 0 莇 DX= 12 4.设 为随机变量，E =3, E 2 =11，则E （4： 10） 羀 D (4 10)=16D # =16 E 2 (E )2 32 100 r x -100 、 X ，某一个电子设备内配有 3个这样的电子管，则电子管使用150小时都不 、0（其他） 需要更换的概率为 8/27 二 4E 10 =22 蒇 5.已知X 的密度为（X ）二 ax + b 广 0 c x < 1 其他，且 1 1 P ( X 二)=P(X>-) , r (x ) dx=1 1 ax b ) dx 二 /ax b ) 3 联立解得: dx 肇 6?若f （x ）为连续型随机变量 X 的分布密度，则 J 「f （x ）dx= _1 ~ |*"^0 羆 7.设连续型随机变量旳布函数F （X ）=X 2/； 丨1, x ：： 0 0 乞 x ：：： 1，则 蚄 P ( E =0.8 ) = _; P(0.2 ：： ：： 6) = 0.99 螄 8. 某型号电子管，其寿命（以小时记）为一随机变量，概率密度 （X ）= 《概率论》第二章 练习答案 一、填空题： 1．设随机变量X 的密度函数为f(x)=? ??02x 其它1???o 则用Y 表示对X 的3次独立重复的观察中事 件(X≤ 2 1 )出现的次数，则P （Y ＝2）＝ 。 2. 设连续型随机变量的概率密度函数为： ax+b 0 第二章 条件概率与统计独立性 1、字母M ，A ，X ，A ，M 分别写在一张卡片上，充分混合后重新排列，问正好得到顺序MAAM 的概率是多少？ 2、有三个孩子的家庭中，已知有一个是女孩，求至少有一个男孩的概率。 3、若M 件产品中包含m 件废品，今在其中任取两件，求：（1）已知取出的两件中有一件是废品的条件下，另一件也是废品的条件概率；（2）已知两件中有一件不是废品的条件下，另一件是废品的条件概率；（3）取出的两件中至少有一件是废品的概率。 4、袋中有a 只黑球，b 吸白球，甲乙丙三人依次从袋中取出一球（取后来放回），试分别求出三人各自取得白球的概率（3≥b ）。 5、从{0，1，2，…，9}中随机地取出两个数字，求其和大于10的概率。 6、甲袋中有a 只白球，b 只黑球，乙袋中有α吸白球，β吸黑球，某人从甲袋中任出两球投入乙袋， 然后在乙袋中任取两球，问最后取出的两球全为白球的概率是多少？ 7、设的N 个袋子，每个袋子中将有a 只黑球，b 只白球，从第一袋中取出一球放入第二袋中，然后从第 二袋中取出一球放入第三袋中，如此下去，问从最后一个袋子中取出黑球的概率是多少？ 8、投硬币n 回，第一回出正面的概率为c ，第二回后每次出现与前一次相同表面的概率为p ，求第n 回 时出正面的概率，并讨论当∞→n 时的情况。 9、甲乙两袋各将一只白球一只黑球，从两袋中各取出一球相交换放入另一袋中，这样进行了若干次。以 pn ，qn ，rn 分别记在第n 次交换后甲袋中将包含两只白球，一只白球一只黑球，两只黑球的概率。试导出pn+1，qn+1，rn+1用pn ，qn ，rn 表出的关系式，利用它们求pn+1，qn+1，rn+1，并讨论当∞→n 时的情况。 10、设一个家庭中有n 个小孩的概率为 ??? ??=--≥=,0,11, 1,n p ap n ap p n n 这里p p a p /)1(0,10-<<<<。若认为生一个小孩为男孩可女孩是等可能的，求证一个家庭有 )1(≥k k 个男孩的概率为1)2/(2+-k k p ap 。 11、在上题假设下：（1）已知家庭中至少有一个男孩，求此家庭至少有两个男孩的概率； （2）已知家庭中没有女孩，求正好有一个男孩的概率。 12、已知产品中96%是合格品，现有一种简化的检查方法，它把真正的合格品确认为合格品的概率为0.98， 而误认废品为合格品的概率为0.05，求在简化方法检查下，合格品的一个产品确实是合格品的概率。 13、设A ，B ，C 三事件相互独立，求证B A AB B A -,,Y 皆与C 独立。 第2章 条件概率与统计独立性 1、字母M ，A ，X ，A ，M 分别写在一张卡片上，充分混合后重新排列，问正好得到顺序MAAM 的概率是多少？ 2、有三个孩子的家庭中，已知有一个是女孩，求至少有一个男孩的概率。 3、若M 件产品中包含m 件废品，今在其中任取两件，求：（1）已知取出的两件中有一件是废品的条件下，另一件也是废品的条件概率；（2）已知两件中有一件不是废品的条件下，另一件是废品的条件概率；（3）取出的两件中至少有一件是废品的概率。 5、袋中有a 只黑球，b 吸白球，甲乙丙三人依次从袋中取出一球（取后来放回），试分别求出三人各自取得白球的概率（3≥b ）。 6、甲袋中有a 只白球，b 只黑球，乙袋中有α吸白球，β吸黑球，某人从甲袋中任出两球投入乙袋，然后在乙袋中任取两球，问最后取出的两球全为白球的概率是多少？ 7、设的N 个袋子，每个袋子中将有a 只黑球，b 只白球，从第一袋中取出一球放入第二袋中，然后从第二袋中取出一球放入第三袋中，如此下去，问从最后一个袋子中取出黑球的概率是多少？ 9、投硬币n 回，第一回出正面的概率为c ，第二回后每次出现与前一次相同表面的概率为p ，求第n 回时出正面的概率，并讨论当∞→n 时的情况。 10、甲乙两袋各将一只白球一只黑球，从两袋中各取出一球相交换放入另一袋中，这样进行了若干次。以pn ，qn ，rn 分别记在第n 次交换后甲袋中将包含两只白球，一只白球一只黑球，两只黑球的概率。试导出pn+1，qn+1，rn+1用pn ，qn ，rn 表出的关系式，利用它们求pn+1，qn+1，rn+1，并讨论当∞→n 时的情况。 11、设一个家庭中有n 个小孩的概率为 ?????=--≥=,0,11,1,n p ap n ap p n n 这里p p a p /)1(0,10-<<<<。若认为生一个小孩为男孩可女孩是等可能的，求证一个家庭有)1(≥k k 个男孩的概率为1)2/(2+-k k p ap 。 12、在上题假设下：（1）已知家庭中至少有一个男孩，求此家庭至少有两个男孩的概率； （2）已知家庭中没有女孩，求正好有一个男孩的概率。 一、教学目的与要求 1、掌握随机变量的概念，离散型随机变量的分布列，会用Ch1求事件概率的方法，求随机变量的分布列； 2、熟悉随机变量的数学期望，方差的概念，会应用分布列求数学期望、方差；掌握数学期望，方差的性质； 3、掌握二维随机变量的分布，边际分布的概念，会应用联合分布列求边际分布，会计算二维随机变量的数字特征，会判定随机变量的独立性与相关性。 4、掌握随机变量函数分布的求法，会求随机变量函数的数字特征。 二、教学重点与难点 重点是分布列的求法，期望与方差的计算。 难点是二维随机变量联合分布列的求法，期望与方差性质的应用。 §2.1一维随机变量及分布列 一．随机变量及其分类 1．概念 在Ch1里，我们研究了随机事件及其概率，细心的同学可能会注意到在某些例子中，随机事件与实数之间存在某种客观的联系。例如袋中有五个球（三白两黑）从中任取三球，则取到的黑球数可能为0，1，2本身就是数量且随着随机试验结果的变化而变化的。又如在“n重贝努里试验中，事件A出现k次”这一事件的概率，若记ξ=n重贝努里试验中A出现的次数，则上述“n重贝努里试验中，事件A出现k次”这一事件可以简记为（ξ=k）,从而有 P(ξ=k)= C p q q=1-p 并且ξ的所有可能取值就是事件A可能出现的次数0，1，2，……n，有些初看起来与数值无关的随机现象，也常常能联系数值来描述。 例如抛掷一枚均匀的硬币可能出现正面，也可能出现反面，约定 若试验结果出现正面, 令η=1, 从而{试验结果出现正面}=（η=1）； 若试验结果出现反面, 令η=0, 从而{试验结果出现反面}=（η=0）。 为了计算n次投掷中出现正面数就只需计算其中“1”出现的次数了。 一般地，若A为某个随机事件，则一定可以通过如下示性函数使它与数值发生联系 在上面的例子中，我们遇到了两个随机变量ξ，η，这两个变量取什么值，在每次试验之前是不确定的，因为它的取值依赖于试验的结果，也就是说它的取值是随机的，通常称这种量为随机变量。从上面例子可以发现，有了随机变量，至少使随机事件的表达在形式上简洁得多了。 在上述前两个例子中，对每一个随机试验的结果自然地对应着一个实数，而在后两个例子中，这种对应关系是人为地建立起来，由此可见，无论哪一种性质， 西南财经大学《 概率论与数理统计》第二章单元测试 满分100分 考试时间 120分钟 一、选择题（每题2分，共20分） 1．设F(x) 是随机变量X 的分布函数，则下列结论不正确的是 （A ）若F(a)=0，则对任意x ≤a 有F(x)=0 （B ）若F(a)=1，则对任意x ≥a 有F(x)=1 （C ）若F(a)=1/2，则 P(x ≤a)=1/2 （D ）若F(a)=1/2，则 P(x ≥a)=1/2 2．设随机变量X 的概率密度f(x) 是偶函数，分布函数为F(x)，则 （A ）F(x) 是偶函数 （B ）F(x)是奇函数 （C ）F(x)+F(-x)=1 （D ）2F(x)-F(-x)=1 4．设随机变量X 1, X 2是任意两个独立的连续型随机变量，它们的概率密度分别为f 1 (x)和f 2 (x)，分布函数分别为F 1 (x)和F 2 (x)，则 （A ）f 1 (x) +f 2 (x) 必为某一随机变量的概率密度 （B ）f 1 (x) f 2 (x) 必为某一随机变量的概率密度 （C ）F 1 (x)+F 2 (x) 必为某一随机变量的分布函数 （D ）F 1 (x)F 2 (x) 必为某一随机变量的分布函数 5．设随机变量X 服从正态分布),(211σμN ，Y 服从正态分布),(2 22σμN ，且 )1|(|)1|(|21<-><-μμY P X P ，则必有 （A ）21σσ< （B ）21σσ> （C ）21μμ< （D ）21μμ> 6．设随机变量X 服从正态分布),(2σμN ，则随σ的增大，概率)|(|σμ<-X P （A ）单调增大 （B ）单调减小 （C ）保持不变 （D ）增减不定 9．下列陈述正确的命题是 （A ）若),1()1(≥=≤X P X P 则2 1 )1(= ≤X P （B ）若X~b(n, p), 则P(X=k)=P(X=n-k), k=0,1,2,?,n （C ）若X 服从正态分布,则F(x)=1-F(-x) （D ）1)]()([lim =-++∞ →x F x F x 第2章练习题 1、二手数据的特点是（） A.采集数据的成本低，但搜集比较困难 B. 采集数据的成本低，但搜集比较容易 C.数据缺乏可靠性 D. 不适合自己研究的需要 2、从含有N个元素的总体中，抽取n个元素作为样本，使得总体中的每一个元素都有相同的机会（概率）被抽中，这样的抽样方式称为（） A.简单随机抽样 B.分层抽样 C.系统抽样 D. 整群抽样 3、从总体中抽取一个元素后，把这个元素放回到总体中再抽取第二个元素，直至抽取n个元素为止，这样的抽样方法称为（） A.重复抽样 B.不重复抽样 C.分层抽样 D.整群抽样 4、一个元素被抽中后不再放回总体，然后从所剩下的元素中抽取第二个元素，直至抽取n个元素为止，这样的抽样方法称为（） A.不重复抽样 B.重复抽样 C.系统抽样 D.多阶段抽样 5、在抽样之前先将总体的元素划分为若干类，然后从各个类中抽取一定数量的元素组成一个样本，这样的抽样方式称 为（） A.简单随机抽样 B.系统抽样 C.分层抽样D?整群抽样 6、先将总体各元素按某种顺序排列，并按某种规则确定一个随机起点，然后每隔一定的间隔抽取一个元素，直至抽取 n个元素形成一个样本。这样的抽样方式称为（） A.分层抽样 B.简单随机抽样 C.系统抽样D?整群抽样 7、先将总体划分为若干群，然后以群作为抽样单位从中抽取部分群，再对抽中的各个群中所包含的所有元素进行观察， 这样的抽样方式称为（） A.系统抽样 B.多阶段抽样 C.分层抽样 D.整群抽样 8为了调查某校学生的购书费用支出，从男生中抽取60名学生调查，从女生中抽取40名学生调查，这种调查方是（） A.简单随机抽样 B.整群抽样 C.系统抽样 D.分层抽样 9、为了调查某校学生的购书费用支出，从全校抽取4个班级的学生进行调查，这种调查方法是（） A.系统抽样 B.简单随机抽样 C.分层抽样D?整群抽样 10、为了调查某校学生的购书费用支出，将全校学生的名单按拼音顺序排列后，每隔50名学生抽取一名学生进行调查， 这种调查方法是？（） A.分层抽样 B.整群抽样 C.系统抽样 D.简单随机抽样 11、为了了解女性对某种化妆品的购买意愿，调查者在街头随意拦截部分女性进行调查。这种调查方式是（） A.简单随机抽样 B.分层抽样C?方便抽样D?自愿抽样 12、研究人员根据研究对象的了解有目的的选择一些单位作为样本，这种调查方式是（） A.判断抽样 B.分层抽样 C.方便抽样 D.自愿抽样 13、下面的那种调查方式不是随机选取的（） A.分层抽样 B.系统抽样C?整群抽样D?判断抽样 14、下面的那种抽样调查结果不能用于对总体有关参数进行估计（） A.分层抽样 B.系统抽样 C.整群抽样 D.判断抽样 15、调查时首先选择一组调查单位，对其实施调查之后，再请他们提供另外一些属于研究总体的调查对象，调查人员根据所提供的线索，进行此后的调查。这样的调查方式称为（） A.系统抽样 B.整群抽样 C.滚雪球抽样 D.判断抽样 16、如果要搜集某一特定群体的有关资料。适宜采用的调查方式是（） A.滚雪球抽样 B.系统抽样 C.判断抽样 D.整群抽样 17、下面的那种抽样方式不属于概率抽样（） A.系统抽样 B.整群抽样 C.分层抽样 D.滚雪球抽样 18、下面的那种抽样方式属于非概率抽样（） A.系统抽样 B.简单随机抽样 C.整群抽样 D.方便抽样 19、先将总体中的所有单位按一定的标志（变量）分为若干类，然后在每个类中采用方便抽样或判断抽样的方式选取样本单位。这种抽样方式称为（） A.分类抽样 B.配额抽样 C.系统抽样 D.整群抽样 第二章 条件概率与统计独立性 1、字母M ，A ，X ，A ，M 分别写在一张卡片上，充分混合后重新排列，问正好得到顺序MAAM 的概 率是多少？ 2、有三个孩子的家庭中，已知有一个是女孩，求至少有一个男孩的概率。 3、若M 件产品中包含m 件废品，今在其中任取两件，求：（1）已知取出的两件中有一件是废品的条件 下，另一件也是废品的条件概率；（2）已知两件中有一件不是废品的条件下，另一件是废品的条件概率；（3）取出的两件中至少有一件是废品的概率。 4、袋中有a 只黑球，b 吸白球，甲乙丙三人依次从袋中取出一球（取后来放回），试分别求出三人各自取得白球的概率（3≥b ）。 5、从{0，1，2，…，9}中随机地取出两个数字，求其和大于10的概率。 6、甲袋中有a 只白球，b 只黑球，乙袋中有α吸白球，β吸黑球，某人从甲袋中任出两球投入乙袋，然 后在乙袋中任取两球，问最后取出的两球全为白球的概率是多少？ 7、设的N 个袋子，每个袋子中将有a 只黑球，b 只白球，从第一袋中取出一球放入第二袋中，然后从第 二袋中取出一球放入第三袋中，如此下去，问从最后一个袋子中取出黑球的概率是多少？ 8、投硬币n 回，第一回出正面的概率为c ，第二回后每次出现与前一次相同表面的概率为p ，求第n 回 时出正面的概率，并讨论当∞→n 时的情况。 9、甲乙两袋各将一只白球一只黑球，从两袋中各取出一球相交换放入另一袋中，这样进行了若干次。以 pn ，qn ，rn 分别记在第n 次交换后甲袋中将包含两只白球，一只白球一只黑球，两只黑球的概率。试导出pn+1，qn+1，rn+1用pn ，qn ，rn 表出的关系式，利用它们求pn+1，qn+1，rn+1，并讨论当∞→n 时的情况。 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