Course: 12.7 基本概率 | The Way To Learn

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基本概率

Most are familiar with how flipping a coin or rolling dice works and yet probability remains one of the most counterintuitive branches of mathematics for many people. The idea that flipping a coin and getting 10 heads in a row is just as unlikely as getting the following of heads and tails is hard to comprehend.
::大多数人都熟悉如何翻硬币或滚动骰子起作用,但概率仍是许多人最反直觉数学分支之一。翻硬币和一排获得10个头的想法与获得以下头和尾一样难以理解。

$\begin{aligned} H H T H T T T H T H \end{aligned}$
::氢氢

Assume a plane crashes on average once every 100 days (extremely inaccurate). Given a plane crashed today, what day in the next 100 days is the plane most likely to crash next?
::假设飞机平均每100天坠毁一次(极不准确 ) 。鉴于今天的飞机坠毁,接下来的100天中哪一天最有可能下飞机坠毁?

Introduction to Probability
::概率介绍导言

Probability is the chance of an event occurring. Simple probability is defined as the number of outcomes you are looking for (also called successes) divided by the total number of outcomes. The notation $\begin{aligned} P (E) \end{aligned}$ is read “the probability of event $\begin{aligned} E \end{aligned}$ ”.
::概率是事件发生的概率。简单概率的定义是,您所寻找的结果(也称为成功)除以结果总数。P(E)符号为“事件E的概率”。

$\begin{aligned} P (E) = \frac{# s u c c e s s e s}{# p o s s i b l e o u t c o m e s} \end{aligned}$
::P(E) 成功 # 可能的成果

Probabilities can be represented with fractions, decimals, or percents. Since the number of possible outcomes is in the denominator, the probability is always between zero and one. A probability of 0 means the event will definitely not happen, while a probability of 1 means the event will definitely happen.
::概率可以用分数、小数或百分比表示。由于分母中含有可能的结果, 概率总是介于零和一之间。 0 的概率意味着事件绝对不会发生, 而 1 的概率则意味着事件绝对会发生。

$\begin{aligned} 0 \leq P (E) \leq 1 \end{aligned}$
::0P(E)1

The probability of something not happening is called the complement and is found by subtracting the probability from one.
::不发生事情的概率被称为补充,通过从一个中减去概率来发现。

$\begin{aligned} P (E^{C}) = 1 - P (E) \end{aligned}$
::P(EC)=1-P(E)

You will often be looking at probabilities of two or more independent experiments. Experiments are independent when the outcome of one experiment has no effect on the outcome of the other experiment. If there are two experiments, one with outcome $\begin{aligned} A \end{aligned}$ and the other with outcome $\begin{aligned} B \end{aligned}$ , then the probability of $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ is:
::您通常会看到两个或两个以上独立实验的概率。当一个实验的结果对另一个实验的结果没有影响时, 实验是独立的。如果有两个实验, 一个是结果A, 另一个是结果B, 那么A和B的概率是:

$\begin{aligned} P (A a n d B) = P (A) \cdot P (B) \end{aligned}$
::P(A和B) = P(A) P(B)

The probability of $\begin{aligned} A \end{aligned}$ or $\begin{aligned} B \end{aligned}$ is:
::A或B的概率是:

$\begin{aligned} P (A o r B) = P (A) + P (B) - P (A a n d B) \end{aligned}$
::P(A或B)=P(A)+P(B)-P(A和B)

Take the common probability example of a deck of cards. If you are dealt one card from a 52 card deck, what is the probability that you are dealt a heart? What is the probability that you are dealt a 3? What is the probability that you are dealt the three of hearts?
::以牌牌牌甲板的常见概率示例为例。如果从52张牌牌牌上划出一张牌,您被划入心脏的概率是多少?被划入心脏的概率是多少?被划入心脏3的概率是多少?被划入心脏3的概率是多少?

There are 13 hearts in a deck of 52 cards. %3D%5Cfrac%7B13%7D%7B52%7D%3D%5Cfrac%7B1%7D%7B4%7D"> $\begin{aligned} P (h e a r t) = \frac{13}{52} = \frac{1}{4} \end{aligned}$
::52张牌牌牌牌牌牌牌上有13个红心。 P( 心脏)=1352=14

There are 4 threes in the deck of 52. $\begin{aligned} P (t h r e e) = \frac{4}{52} = \frac{1}{13} \end{aligned}$
::52. P(3)=452=113的甲板上有4个3个。

There is only one three of hearts in a deck of 52. $\begin{aligned} P (t h r e e a n d h e a r t) = \frac{1}{52} \end{aligned}$
::在52P3和心脏152的甲板上只有一个心

Examples
::实例

Example 1
::例1

Earlier, you were asked about the probability of a plane crashing. Whether or not a plane crashes today does not matter. The probability that a plane crashes tomorrow is $\begin{aligned} p = 0.01 \end{aligned}$ . The probability that it crashes any day in the next 100 days is equally $\begin{aligned} p = 0.01 \end{aligned}$ . The key part of the question is the word “next”.
::早些时候,有人问你飞机坠毁的概率。今天飞机坠毁是否无关紧要。明天飞机坠毁的概率是p=0.01。在接下来的100天内,飞机坠毁的概率等于p=0.01。问题的关键部分是“下一个”一词。

The probability that a plane does not crash on the first day and does crash on the second day is a compound probability, which means you multiply the probability of each event.
::飞机第一天不坠毁,第二天不坠毁的概率是复数概率,这意味着每个事件的概率乘以。

$\begin{aligned} P (D a y 1 n o c r a s h A N D D a y 2 c r a s h) = 0.99 \cdot 0.01 = 0.0099 \end{aligned}$
::P(第1天无失事,第2天失事)=0.990.01=0.0099

Notice that this probability is slightly smaller than 0.01. Each successive day has a slightly smaller probability of being the next day that a plane crashes. Therefore, the day with the highest probability of a plane crashing next is tomorrow.
::请注意此概率略小于 0.01 。每连续一天的飞机坠毁概率稍小一点。因此, 下一班飞机坠毁概率最高的一天是明天。

Example 2
::例2

Dean and his friend Randy like to play a special poker game with their friends. Dean goes home a winner 60% of the time and Randy goes home a winner 75% of the time.
::Dean和他的朋友Randy喜欢和他们的朋友玩一场特别的扑克游戏。Dean60%的时间回家,兰迪75%的时间回家。

What is the probability that they both win in the same night?
::他们在同一晚获胜的概率有多大?

What is the probability that Randy wins and Dean loses?
::兰迪和迪恩输赢的可能性有多大?

What is the probability that they both lose?
::两者失去的可能性有多大?

First represent the information with probability symbols.
::首先代表带有概率符号的信息。

Let $\begin{aligned} D \end{aligned}$ be the event that Dean wins. Let $\begin{aligned} R \end{aligned}$ be the event that Randy wins. The complement of each probability is when Dean or Randy loses instead.
::让D成为Dean获胜的事件。让R成为Randy获胜的事件。每种可能性的补充是当Dean或Randy输的时候。

$\begin{aligned} P (D) = 0.60, P (D^{C}) = 0.40 \end{aligned}$
::P(D)=0.60,P(DC)=0.40

$\begin{aligned} P (R) = 0.75, P (R^{C}) = 0.25 \end{aligned}$
::P(R)=0.75,P(RC)=0.25

$\begin{aligned} P (D a n d R) = P (D) \cdot P (R) = 0.60 \cdot 0.75 = 0.45 \end{aligned}$
::P(D和R)=P(D)P(R)=0.600.75=0.45

$\begin{aligned} P (R a n d D^{C}) = P (R) \cdot P (D^{C}) = 0.75 \cdot 0.40 = 0.30 \end{aligned}$
::P(R和DC)=P(R)P(DC)=0.750.40=0.30

$\begin{aligned} P (D^{C} a n d R^{C}) = P (D^{C}) \cdot P (R^{C}) = 0.40 \cdot 0.25 = 0.10 \end{aligned}$
::P(DC和RC)=P(DC)_P(RC)=0.40_0.25=0.10。

Example 3
::例3

If a plane crashes on average once every hundred days, what is the probability that the plane will crash in the next 100 days?
::如果每100天平均发生一次飞机失事,飞机在今后100天中坠毁的可能性有多大?

The naïve and incorrect approach would be to interpret the question as “what is the sum of the probabilities for each of the days?” Since there are 100 days and each day has a probability of 0.01 for a plane crash, then by this logic, there is a 100% chance that a plane crashes. This isn’t true because if on average the plane crashes once every hundred days, some stretches of 100 days there will be more crashes and some stretches there will be no crashes. The 100% solution does not hold.
::最天真和不正确的做法是将问题解释为“每一天概率的总和是多少? ” 因为有100天,每天的飞机坠毁概率是0.01,然后按照这一逻辑,飞机坠毁概率是100%。这不是真的,因为如果飞机每100天平均坠毁一次,100天的几段距离就会有更多的飞机坠毁,而一些波段将不会有坠毁。 100%的解决方案将无法维持。

In order to solve this question, you need to rephrase the question and ask a slightly different one that will help as an intermediate step. What is the probability that a plane does not crash in the next 100 days?
::为了解决这个问题, 您需要重写这个问题, 并询问一个稍有不同的问题, 作为中间步骤。飞机在未来100天内不会坠毁的概率是多少 ?

In order for this to happen it must not crash on day 1 and not crash on day 2 and not crash on day 3 etc.
::为了做到这一点,它不得在第一天坠毁,不得在第2天坠毁,不得在第3天坠毁。

The probability of the plane not crashing on any day is $\begin{aligned} P (n o c r a s h) = 1 - P (c r a s h) = 1 - 0.01 = 0.99 \end{aligned}$ .
::飞机在任何一天不坠毁的概率为 P( 无坠毁) = 1 - P( 坠毁) = 1 - 0.01 = 0. 99。

The product of each of these probabilities for the 100 days is:
::每种100天概率的产物如下:

$\begin{aligned} {0.99}^{100} \approx 0.366 \end{aligned}$

Therefore, the probability that a plane does not crash in the next 100 days is about 36.6%. To answer the original question, the probability that a plane does crash in the next 100 days is $\begin{aligned} 1 - 0.366 = 0.634 \end{aligned}$ or about $\begin{aligned} 63.4 % \end{aligned}$ .
::因此,未来100天飞机不坠毁的概率约为36.6%。要回答最初的问题,未来100天飞机坠毁的概率为1-0.366=0.634或63.4%左右。

Example 4
::例4

Jack is a basketball player with a free throw average of 0.77. What is the probability that in a game where he has 8 shots that he makes all 8? What is the probability that he only makes 1?
::杰克是一个篮球运动员,平均免费投球0.77,在他有8发球的比赛中,他投出8发球的概率有多大?他只投出1的概率有多大?

Let $\begin{aligned} J \end{aligned}$ represent the event that Jack makes the free throw shot and $\begin{aligned} J^{C} \end{aligned}$ represent the event that Jack misses the shot.
::让J代表杰克自由投球和JC代表杰克错过射击

$\begin{aligned} P (J) = 0.77, P (J^{C}) = 0.23 \end{aligned}$
::P(J)=0.77,P(JC)=0.23

The probability that Jack makes all 8 shots is the same as Jack making one shot and making the second shot and making the third shot etc.
::杰克射出所有8发子弹的概率和杰克射出一枪第二次射出第二枪以及第三枪等等的概率一样

$\begin{aligned} P (J)^{8} = {0.77}^{8} \approx 12.36 % \end{aligned}$
::P(J)8=0.778=12.36%

There are 8 ways that Jack could make 1 shot and miss the rest. The probability of each of these cases occurring is:
::Jack有八种方法能打中一枪错过其余的每一个案例的概率是:

$\begin{aligned} P (J^{C})^{7} \cdot P (J) = {0.23}^{7} \cdot 0.77 \end{aligned}$
::P(JC)7P(J)=0.2370.77

Therefore, the overall probability of Jack making 1 shot and missing the rest is:
::因此,杰克打一枪并丢失其余的概率是:

$\begin{aligned} {0.23}^{7} \cdot 0.77 \cdot 8 = 0.0002097 = 0.02097 % \end{aligned}$

Example 5
::例5

If it has a 20% chance of raining on Tuesday, your phone has 30% chance of running out of batteries, and there is a 10% chance that you forget your wallet, what is the probability that you are in the rain without money or a phone? What is the probability that at least one of the three events does occur?
::如果周二有20%的下雨机率, 你的手机有30%的电池用完的可能性, 还有10%的机会你忘了你的钱包, 那么在雨中没有钱或手机的概率是多少? 三个事件中至少有一个发生的可能性是多少?

While a pessimist may believe that all the improbable negative events will occur at the same time, the actual probability being out in the rain without money or a phone is less than one percent:
::虽然悲观主义者可能认为所有不可能发生的负面事件都会同时发生,

$\begin{aligned} 0.20 \cdot 0.30 \cdot 0.1 = 0.006 = 0.6 % \end{aligned}$

The naïve approach to determining the probability that at least one of the events occurs would be to simply add the three probabilities together. This is incorrect. The better way to approach the problem is to ask the question: what is the probability that none of the events occur?
::确定至少发生其中一起事件的概率的天真方法就是简单地将三种可能性加在一起。这是不正确的。解决问题的最好办法是问一个问题:没有发生这些事件的概率是多少?

$\begin{aligned} 0.8 \cdot 0.7 \cdot 0.9 = 0.504 \end{aligned}$

The probability that at least one occurs is the complement of none occurring.
::至少发生一次的概率是无发生次数的补充。

$\begin{aligned} 1 - 0.504 = 0.496 = 49.6 % \end{aligned}$

Summary

Probability is the chance of an event occurring
::概率是事件发生的可能性

Simple probability is defined as the number of successes divided by the total number of outcomes.
::简单概率的定义是成功次数除以结果总数。

The complement of a probability is the probability of an event not happening, found by subtracting the probability from 1.
::概率的补充是事件不发生的概率,通过从 1 中减去概率而发现。

Experiments are independent when the outcome of one experiment has no effect on the outcome of the other experiment.
::当一项试验的结果对另一项试验的结果没有影响时,实验是独立的。

For two independent experiments with outcomes A and B:
- $\begin{aligned} P (A and B) = P (A) * P (B) \end{aligned}$
  ::P(A和B)=P(A)*P(B)
- $\begin{aligned} P (A or B) = P (A) + P (B) - P (A and B) \end{aligned}$
  ::P(A或B)=P(A)+P(B)-P(A和B)
::用于结果A和结果B的两个独立实验:P(A和B)=P(A)*P(B)P(A或B)=P(A)+P(B)-P(A和B)

Review
::回顾

A card is chosen from a standard deck.
::从标准甲板中选择一张牌。

1. What’s the probability that the card is a queen?
::1. 该卡片是女王的概率有多大?

2. What’s the probability that the card is a queen or a spade?
::2. 该卡片是皇后或的概率是多少?

You toss a nickel, a penny, and a dime.
::你扔一个硬币,一个硬币,一个硬币,一个硬币。

3. List all the possible outcomes (the elements in the sample space).
::3. 列出所有可能的结果(样本空间的要素)。

4. What is the probability that the nickel comes up heads?
::4. 镍浮出水面的概率有多大?

5. What is the probability that none of the coins comes up heads?
::5. 没有一枚硬币抬头的概率有多大?

6. What is the probability that at least one of the coins comes up heads?
::6. 至少有一枚硬币浮现出来的可能性有多大?

A bag contains 7 red marbles, 9 blue marbles, and 10 green marbles. You reach in the bag and choose 4 marbles, one after the other, without replacement.
::包里装有7个红色大理石、9个蓝色大理石和10个绿色大理石。你伸手在包里,选择4个大理石,一个接一个,没有替换。

7. What is the probability that all 4 marbles are red?
::7. 所有4颗大理石都红色的可能性有多大?

8. What is the probability that you get a red marble, then a blue marble, then 2 green marbles?
::8. 获得红大理石、蓝大理石、两颗绿大理石的可能性有多大?

You take a 40 question multiple choice test and believe that for each question you have a 55% chance of getting it right.
::你接受一个40个问题多重选择测试并相信每个问题你有一个55%的机会得到正确的答案。

9. What is the probability that you get all the questions right?
::9. 你答对所有问题的可能性有多大?

10. What is the probability that you get all of the questions wrong?
::10. 你把所有问题都弄错的可能性有多大?

A player rolls a pair of standard dice. Find each probability.
::玩家滚动一对标准骰子。查找每个概率。

11. $\begin{aligned} P (s u m i s e v e n) \end{aligned}$
::11. P(总和相等)

12. $\begin{aligned} P (s u m i s 7) \end{aligned}$
::12.(总和为7)

13. $\begin{aligned} P (s u m i s a t l e a s t 3) \end{aligned}$
::13. (总和至少为3)

14. You want to construct a 3 digit number at random from the digits 4, 6, 8, 9 without repeating digits. What is the probability that you construct the number 684?
::14. 您想要从数字4、6、8、9中随机构建一个3位数数字,而不重复数字。您构建数字684的概率是多少?

15. In poker, a straight is 5 cards in a row (ex. 3, 4, 5, 6, 7), NOT all the same suit (if they are all the same suit it is considered a straight flush or a royal flush). A straight can start or end with an Ace. What’s the probability of a straight? For an even bigger challenge, see if you can calculate the probabilities for all of the poker hands.
::15. 在扑克牌中,直牌是5张牌一排(如3、4、5、6、7),不是全部相同的西装(如果它们都是相同的西装,则被视为直冲或王牌冲。直牌可以通过Ace开始或结束。直牌的概率是多少? 对于更大的挑战,请看看您能否计算出所有扑克手的概率。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

基本概率

Introduction to Probability ::概率介绍导言

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)