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概率与统计入门教程:理论与实战应用
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"《概率与统计入门》是一本深入浅出的教材,专为Spring2014年春季开设的18.05课程设计。本书旨在引导学生预习课程内容,尽管在开始时可能会遇到许多陌生术语,但随着课程的进展,这些概念将逐步解释清楚。 本书的核心主题是概率论与统计学之间的紧密联系。概率论是一个逻辑自洽的学科,基于一套明确的规则,通过逻辑推理得出答案,尽管计算可能涉及复杂的技巧。它主要关注的是确定性问题,如抛一枚公平硬币100次出现60个或更多头的概率,这个问题有一个确切的答案(大约为0.028444),学生们会学习如何计算这类概率。 相比之下,统计学则将概率应用于数据分析,旨在从数据中提取信息并做出结论。这通常是一个实践性更强、科学与艺术相结合的过程。例如,在实际场景中,假设你有一枚未知公正性的硬币,你需要通过掷100次来检验其是否公平。如果你观察到60次正面朝上,作为统计学家,你的任务是: 1. 基于观察到的数据(即60个头)进行推断。 2. 利用概率理论,可能需要构建假设检验,来判断这是否与硬币是公平的(即预期50%的正面朝上)相符。 3. 分析数据分布,如二项分布,来估计硬币公平性的置信度。 4. 可能还会涉及到样本大小的选择、误差分析以及决策制定,这些都是统计学中的关键环节。 总结来说,《概率与统计入门》涵盖了基础概率理论的概念,如随机事件、概率分布,以及如何通过实验设计和数据分析来应用这些理论。它强调了理论与实践的结合,让学生在理解基本原理的同时,掌握实际问题解决的能力。通过深入研究这本书,学生将不仅能掌握数学工具,还能培养批判性思维和数据驱动的决策能力。"
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18.05 class 2, Probability: Terminology and Examples, Spring 2014
2.4 The probability function
So far we’ve been using a casual definition of the probability function. Let’s give a more
precise one.
Careful definition of the probability function.
For a discrete sample space S a probability function P assigns to each outcome ω anumber
P (ω) called the probability of ω. P must satisfy two rules:
Rule 1. 0 ≤ P (ω) ≤ 1 (probabilities are between 0 and 1).
Rule 2. The sum of the probabilities of all possible outcomes is 1 (something must occur)
In symbols Rule 2 says: if S = {ω
1
,ω
2
,...,ω
n
} then P (ω
1
)+ P (ω
2
)+ ...+ P (ω
n
)=1.
n
Using summation notation: P (ω
j
)=1.
j=1
The probability of an event E is the sum of the probabilities of all the outcomes in E.That
is,
P (E)= P (ω).
ω∈E
Think: Check Rules 1 and 2 on Examples 1 and 2 above.
Example 7. Flip until heads (A classic example)
Suppose we have a coin with probability p of heads.
Experiment: Toss the coin until the first heads. Report the number of tosses.
Sample space: Ω = {1, 2, 3, . . . }.
Probability function: P (n)=(1 − p)
n−1
p.
Challenge 1: show the sum of all the probabilities equals 1 (hint: geometric series).
Challenge 2: justify the formula for P (n) (we will do this soon).
Stopping problems. The previous toy example is an uncluttered version of a general
class of problems called stopping rule problems. A stopping rule is a rule that tells you
when to end a certain process. In the toy example above the process was flipping a coin
and we stopped after the first heads. A more practical example is a rule for ending a series
of medical treatments. Such a rule could depend on how well the treatments are working,
how the patient is tolerating them and the probability that the treatments would continue
to be effective. One could ask about the probability of stopping within a certain number of
treatments or the average number of treatments you should expect before stopping.
Some rules of probability
For events A, L and R contained in a sample space Ω.
Rule 1. P (A
c
)=1 − P(A).
Rule 2. If L and R are disjoint then P (L ∪ R)= P (L)+ P (R).
Rule 3. If L and R are not disjoint, we have the inclusion-exclusion principle:
P (L ∪ R)= P (L)+ P (R) − P (L ∩ R)
![](https://csdnimg.cn/release/download_crawler_static/33278833/bg11.jpg)
c
L R L R
5 18.05 class 2, Probability: Terminology and Examples, Spring 2014
We visualize these rules using Venn diagrams.
Ω= A ∪ A ,nooverlap L ∪ R,n
ooverlap L ∪ R, overlap = L ∩ R
We can also justify them logically.
Rule 1: A and A
c
split Ω into two non-overlapping regions. Since the total probability
P (Ω) = 1 this rule says that the probabiity of A and the probability of ’not A’are comple-
mentary, i.e. sum to 1.
Rule 2: L and R split L ∪ R into two non-overlapping regions. So the probability of L ∪ R
is is split between P (L)and P (R)
Rule 3: In the sum P (L)+ P (R)the overlap P (L ∩ R) gets counted twice. So P (L)+
P (R) − P (L ∩ R) counts everything in the union exactly once.
Think: Rule 2 is a special case of Rule 3.
For the following examples suppose we have an experiment that produces a random integer
between 1 and 20. The probabilities are not necessarily uniform, i.e., not necessarily the
same for each outcome.
Example 8. If the probability of an even number is .6 what is the probability of an odd
number?
answer: Since being odd is complementary to being even, the probability of being odd is
1-.6 = .4.
Let’s redo this example a bit more formally, so you see how it’s done. First, so we can refer
to it, let’s name the random integer X. Let’s also name the event ‘X is even’ as A.Then
the event ‘X is odd’ is A
c
. We are given that P (A)= .6. Therefore P (A
c
)=1 − .6= .4 .
Example 9. Consider the 2 events, A:‘X is a multiple of 2’; B:‘X is odd and less than
10’. Suppose P (A)= .6and P (B)= .25.
(i) What is A ∩ B?
(ii) What is the probability of A ∪ B?
answer: (i) Since all numbers in A are even and all numbers in B are odd, these events are
disjoint. That is, A ∩ B = ∅.
(ii) Since A and B are disjoint P (A ∪ B)= P (A)+ P (B)= .85.
Example 10. Let A, B and C be the events X is a multiple of 2, 3 and 6 respectively. If
P (A)= .6, P (B)= .3and P (C)= .2whatis P (A or B)?
answer: Note two things. First we used the word ‘or’ which means union: ‘A or B’=
A ∪ B. Second, an integer is divisible by 6 if and only if it is divisible by both 2 and 3.
This translates into C = A ∩ B. So the inclusion-exclusion principle says
P (A ∪ B)= P (A)+ P (B) − P (A ∩ B)= .6+ .3 − .2= .7 .
Endofclass2notes
![](https://csdnimg.cn/release/download_crawler_static/33278833/bg13.jpg)
Conditional Probability, Independence and Bayes’ Theorem
Class 3, 18.05, Spring 2014
Jeremy Orloff and Jonathan Bloom
1 Learning Goals
1. Know the definitions of conditional probability and independence of events.
2. Be able to compute conditional probability directly from the definition.
3. Be able to use the multiplication rule to compute the total probability of an event.
4. Be able to check if two events are independent.
5. Be able to use Bayes formula to ‘invert’ conditional probabilities.
6. Be able to organize the computation of conditional probabilities using trees and tables.
7. Understand the base rate fallacy thoroughly.
2 Conditional Probability
Conditional probability answers the question ‘how does the probability of an event change
if we have extra information’. We’ll illustrate with an example.
Example 1. Toss a fair coin 3 times.
a) What is the probability of 3 heads?
answer: Sample space Ω = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}.
All outcomes are equally likely, so P (3 heads) = 1/8.
b) Suppose we are told that the first toss was heads. Given this information how should we
compute the probability of 3 heads?
answer: We have a new (reduced) sample space: Ω
'
= {HHH,HHT,HTH,HTT}.
All outcomes are equally likely, so
P (3 heads given that the first toss is heads) = 1/4.
This is called conditional probability, since it takes into account additional conditions. To
develop the notation, we rephrase (b) in terms of events.
b) Let A be the event ‘all three tosses are heads’ = {HHH}.
Let B be the event ‘the first toss is heads’ = {HHH,HHT,HTH,HTT}.
The conditional probability of A knowing that B occurred is written
P (A|B)
This is read as ‘the conditional probability of A given B’ or ‘the probability of A condi-
tioned on B’ or simply ‘the probability of A given B’.
We can visualize conditional probability as follows. Think of P (A) as the proportion of the
area of the whole sample space taken up by A.For P (A|B) we restrict our attention to B.
That is, P(A|B) is the proportion of area of B taken up by A, i.e. P (A ∩ B)/P (B).
![](https://csdnimg.cn/release/download_crawler_static/33278833/bg14.jpg)
2
3
18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014
B
A
A ∩ B
Conditional probability: Abstract visualization and coin example
Note, A ⊂ B in the right-hand figure, so there are only two colors shown.
The formal definition of conditional probability catches the gist of the above example and
visualization.
Formal definition of conditional probability
Let A and B be events. We define the conditional probability of A given B as
P (A ∩ B)
P (A|B)= , provided P (B) (1)=0.
P (B)
Let’s redo the coin tossing example using this definition (1). Recall A =‘3heads’and B
= ‘first toss is heads’. We have P (A)=1/8and P (B)=1/2. Since A ∩ B = A,we also
1/8
have P (A ∩ B)=1/8. Now according to (1), P (A|B)= =1/4, which agrees with our
1/2
answer in example (1b).
Multiplication Rule
The following formula is called the multiplication rule.
P (A ∩ B)= P (A|B) · P (B). (2)
This is simply a rewriting of the definition (1) of conditional probability. We will see that
our use of the multiplication rule is very similar to our use of the rule of product. In fact,
the multiplication rule is just a souped up version of the rule of product.
We start with a simple example where we can check all the probabilities directly by counting.
Example 2. Draw two cards from a deck.
Let S
1
= ‘first card is a spade’.
Let S
2
= ‘second card is a spade’.
What is the P (S
2
|S
1
)?
answer: We can do this directly by counting: if the first card is a spade then of the 51
cards remaining, 12 are spades.
P (S
2
|S
1
)=12/51.
Now, let’s recompute this using formula (1). We have to compute P (S
1
), P (S
2
)and
P (S
1
∩ S
2
):
P (S
1
)=1/4 because there are 52 equally likely ways to draw the first card and 13 of them
are spades.
B
A
A ∩ B
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