Course: 13.7 相互排斥的活动 | The Way To Learn

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相互排斥活动
Mutually Exclusive Events
::相互排斥活动

Imagine you are going to see a movie. Your friend has just bought the tickets, and you are not sure which movie you are seeing. There are 4 movies playing. Harry Potter (which you have already seen, but your friend has not) is one of them.
::想象一下你会去看电影。你的朋友刚刚买了票,你不知道你正在看哪部电影。有四部电影在播放。哈利·波特(你已经看过了,但你朋友没有看过)就是其中之一。

What are the chances you will be seeing Harry Potter?
::你见到哈利波特的机会有多大?

What are the chances you will NOT be seeing Harry Potter?
::你见不到哈利波特的机会有多大?

This is an easy example of a mutually exclusive event : you will either see Harry Potter, or you will not. You cannot do both!
::这是一个互相排斥的事件的简单例子:你们要么见哈利波特,要么不见。你们不能两者兼而有之。

Finding the probability of mutually exclusive events is easy; what’s not as easy is finding the probability of events that can overlap depend on each other. In this lesson, you’ll learn how to find the probability of any two events that can be related to each other.
::找到相互排斥事件的概率是很容易的;最不容易的是找到可能重叠事件的概率取决于彼此。在这一教训中,你将学会如何找到任何两个可能相互联系的事件的概率。

Find the Probability of Mutually Exclusive Events
::查明相互排斥活动的可能性

In probability, when two events are mutually exclusive, the probability of both happening together is zero.
::很可能,当两个事件相互排斥时,两者同时发生的可能性是零。

Examples of mutually exclusive events in probability include:
::可能发生相互排斥事件的例子包括:

Flipping a coin and:

getting heads
::获取头部

getting tails
::获取尾尾巴

::抛硬币和:让头部得到尾巴

Picking a single card from a deck and:

getting an ace
::获得 AA 时

getting a 7
::获得 7

getting a queen
::获得 Q

etc...
::等...

::从甲板上挑一张单张卡片,然后从甲板上取一张A 得到一个七张A 得到一个皇后等...

Picking a single colored marble from a bag and:

getting a red marble
::得到红大理石

getting a blue marble
::得到蓝色大理石

getting a green marble
::获得绿色大理石

etc..
::等

::从一个袋子里摘出一个单色大理石, 和:得到一个红色大理石, 得到一个蓝色大理石, 得到一个绿色大理石等等。

What this means mathematically is two-fold. If our two mutually exclusive events are $\begin{aligned} A \end{aligned}$ and $\begin{aligned} B \end{aligned}$ :
::这在数学上是双重的。如果我们的两个相互排斥的事件是A和B:

$\begin{aligned} P (A a n d B) = 0. \end{aligned}$ There is no possibility of both events happening.
::P(A和B)=0. 这两种事件不可能同时发生。

$\begin{aligned} P (A o r B) = P (A) + P (B) . \end{aligned}$ To find the probability of either event happening, sum the individual probabilities.
::P(A 或 B) = P(A)+P(B) 。要找到发生任一事件的概率, 请将个别概率相加。

Calculating Probability
::计算概率

There are 7 marbles in a bag: 4 green, 2 blue and 1 red. Peter reaches into the bag and blindly picks a single marble. The following letters refer to these events:
::包里有7个大理石:4个绿色,2个蓝色,1个红色。彼得伸手到袋子里,盲目地挑出一个大理石。

$\begin{aligned} A \end{aligned}$ – the marble is red
::A - 大理石是红色的

::A - 大理石是红色的

$\begin{aligned} B \end{aligned}$ – the marble is blue
::B - 大理石是蓝色的

$\begin{aligned} C \end{aligned}$ – the marble is green
::C - 大理石是绿色的

Find the following:
::发现下列情况:

1. Look at the 3 events $\begin{aligned} A, B \end{aligned}$ and $\begin{aligned} C \end{aligned}$ . They must be mutually exclusive: if we pick a single marble from the bag then it must be either red or blue or green. There is no possibility of it being both blue and green!
::1. 看看A、B和C三件事件,它们必须是相互排斥的:如果我们从包里挑一个大理石,那么它必须是红色或蓝色或绿色的,不可能是蓝色和绿色的。

a. $\begin{aligned} P (A) \end{aligned}$
::a. P(A)

There are 7 marbles and only one is red, so $\begin{aligned} P (A) = \frac{1}{7} \end{aligned}$ .
::有7颗大理石,只有一个是红色,所以P(A)=17。

b. $\begin{aligned} P (B) \end{aligned}$
::b. P(B)

There are 7 marbles and 2 are blue, so $\begin{aligned} P (B) = \frac{2}{7} . \end{aligned}$
::有7颗大理石和2颗蓝色所以P(B)=27

c. $\begin{aligned} P (C) \end{aligned}$
::c. P(C)

There are 7 marbles and 4 are green, so $\begin{aligned} P (C) = \frac{4}{7} \end{aligned}$ .
::有7颗大理石和4颗绿色所以P(C)=47

d. $\begin{aligned} P (B o r A) \end{aligned}$
::d. P(B或A)

The events are mutually exclusive, therefore $\begin{aligned} P (B o r A) = P (B) + P (A) = \frac{2}{7} + \frac{1}{7} = \frac{3}{7} \end{aligned}$ .
::这些事件是相互排斥的,因此P(B或A)=P(B)+P(A)=27+17=37。

e. $\begin{aligned} P (C o r A) \end{aligned}$
::e. P(C或A)

The events are mutually exclusive, therefore $\begin{aligned} P (C o r A) = P (C) + P (A) = \frac{4}{7} + \frac{1}{7} = \frac{5}{7} \end{aligned}$ .
::这些事件相互排斥,因此P(C或A)=P(C)+P(A)=47+17=57。

f. $\begin{aligned} P (C o r B o r A) \end{aligned}$
::f. P(C或B或A)

The events are mutually exclusive, therefore $\begin{aligned} P (C o r B o r A) = P (C) + P (B) + P (A) = \frac{4}{7} + \frac{2}{7} + \frac{1}{7} = \frac{7}{7} = 1. \end{aligned}$
::这些事件相互排斥,因此P(C或B或A)=P(C)+P(B)+P(A)=47+27+17=77=1。

g. $\begin{aligned} P (A a n d C) \end{aligned}$
::g. P(A和C)

The events are mutually exclusive, so $\begin{aligned} P (A a n d C) = 0 \end{aligned}$ .
::事件是相互排斥的,所以P(A和C)=0。

The last two results make sense: $\begin{aligned} P (C o r B o r A) \end{aligned}$ means the probability of the marble being green blue or red. It must be one of these. And $\begin{aligned} P (A a n d C) \end{aligned}$ is the probability of the marble being both red and green. There are no such marbles in the bag!
::最后两种结果是有道理的:P(C)或B或A(P(C)或B或A)意指大理石的概率是绿色的蓝色或红色。它必须是其中之一。P(A)和C是大理石的概率是红的和绿色的。包里没有这种大理石!

Earlier we learned about permutations and combinations. We often have to use these calculations when determining the probability of mutually exclusive events.
::早些时候,我们学会了变异和组合。我们常常在确定相互排斥事件的概率时使用这些计算方法。

2. There are 7 marbles in a bag: 4 green, 2 blue and 1 red. Peter reaches into the bag and blindly picks out 4 marbles. Find the probability that he removes at least 3 green marbles.
::2. 袋内有7个大理石:4个绿色,2个蓝色,1个红色。彼得伸手到袋子里,盲目地挑出4个大理石。找出至少3个绿色大理石被移走的可能性。

There are 2 distinct ways this could occur:
::出现这种情况有两种不同的方式:

a) Peter picks 3 green marbles and 1 other.
:a) 彼得选取3颗绿色弹珠和1颗其他弹珠。

b) Peter picks 4 green marbles.
:b) 彼得选取4颗绿色弹珠。

These events are mutually exclusive – he cannot remove (three green and one other) and (four green) at once. If we find $\begin{aligned} P (A) \end{aligned}$ and $\begin{aligned} P (B) \end{aligned}$ we know that the total probability $\begin{aligned} P (A o r B) = P (A) + P (B) \end{aligned}$ .
::这些事件是相互排斥的 — — 他不能同时删除(三个绿色和一个绿色)和(四个绿色 ) 。如果我们发现P(A)和P(B),我们知道总概率P(A或B)=P(A)+P(B ) 。

We are choosing 4 marbles from a bag containing 7 marbles. The total number of marble combinations there are is $\begin{aligned} _{7} C_{4} = \frac{7!}{4! 3!} = 35 \end{aligned}$ possible combinations.
::我们从一个装有7个大理石的袋子中选择了4个大理石。大理石组合的总数是 7C4=7! 4! 3! =35 可能的组合。

a) The number of combinations that contain 3 green marbles +1 other:
:a) 含有3个绿色大理石+1其他绿色大理石的组合数目:

First, the 3 greens. We are choosing 3 green marbles out of a total of 4: $\begin{aligned} _{4} C_{3} = \frac{4!}{3! 1!} = 4 \end{aligned}$
::首先,3个绿球。我们选择了3个绿球在总共4: 4C3=4! 3! 1!=4中选择3个绿球。

Now to choose 1 other – we’re picking 1 marble out of the 3 non-green ones: $\begin{aligned} _{3} C_{1} = \frac{3!}{2! 1!} = 3 \end{aligned}$
::现在选择另外一种 — — 我们选取3个非绿色的大理石中的1大理石: 3C1=3! 2! 1! =3

The total number of combinations of 3 greens and 1 other = $\begin{aligned} _{4} C_{3} \times_{3} C_{1} = 4 \times 3 = 12 \end{aligned}$
::3个绿和1个其他 = 4C3x3C1= 4x3=12的组合总数

So $\begin{aligned} P (A) = \frac{12}{35} \end{aligned}$ .
::所以P(A)=1235。

b) The number of combinations that contain 4 green balls:
:b) 含有4个绿球的组合数:

We are choosing 4 from 4 possible green balls: $\begin{aligned} _{4} C_{4} = 1 \end{aligned}$ possible
::我们从4个可能的绿球中选择4个:4C4=1

So $\begin{aligned} P (B) = \frac{1}{35} \end{aligned}$
::因此P(B)=135

$\begin{aligned} P (A o r B) = P (A) + P (B) \end{aligned}$ , so the probability of getting at least 3 green balls is $\begin{aligned} \frac{12}{35} + \frac{1}{35} = \frac{13}{35} \end{aligned}$ or slightly better than 1 in 3 .
::P(A或B)=P(A)+P(B),因此获得至少3个绿球的概率为1235+135=1335,或略高于三分之一的1。

Finding the Probability of Overlapping Events
::查找重叠活动的概率

Sometimes we wish to look at overlapping events . In essence, this means two events that are NOT mutually exclusive. For instance, if you pick a card at random from a standard deck, what is the probability that you’ll get a card which is either a seven or a diamond? Let’s use this as our next example:
::有时候,我们想看看重叠事件。本质上,这意味着两个事件并不相互排斥。比如,如果你随机从标准甲板抽取一张牌,那么你拿到一张牌的概率是7张还是1张钻石?让我们用这个作为下一个例子:

If you pick a card at random from a standard 52-card deck, what is the probability that you get a card which is either a seven or a diamond?
::如果你随机从标准的52张牌牌牌上选一张牌,那么获得一张牌的概率是7张还是1张钻石?

One thing we can say for certain is that these two events are not mutually exclusive (it is possible to get a card which both a seven and a diamond). First of all, let’s look at the information we have:
::有一点我们可以肯定地说,这两个事件不是相互排斥的(有可能得到一张牌,牌牌是7和1。 )首先,让我们看看我们掌握的信息:

There are 52 cards – the chances of picking any particular one is $\begin{aligned} \frac{1}{52} \end{aligned}$ .
::有52张贺卡, 选择特定贺卡的可能性是152张。

There are 4 sevens (diamond, heart, club, spade) – the chances of picking a seven is $\begin{aligned} \frac{4}{52} = \frac{1}{13} \end{aligned}$ .
::选择七人的机会是452=113。

There are 13 diamonds (ace through king) – the chances of picking a diamond is $\begin{aligned} \frac{13}{52} = \frac{1}{4} \end{aligned}$ .
::钻石采摘机会为1352=14。

The chances of picking the seven of diamonds is $\begin{aligned} \frac{1}{52} \end{aligned}$ .
::采摘钻石7的机会是152个。

So there are 4 sevens, 13 diamonds and 1 card that is both (the seven of diamonds). That means we can’t just add up the number of sevens and the number of diamonds, because if we did that, we’d be counting the seven of diamonds twice. Instead, we have to add up the number of sevens and the number of diamonds, and then subtract the number of cards that fit in both categories. That means there are $\begin{aligned} (4 + 13 - 1) = 16 \end{aligned}$ cards that are seven or diamond. The probability of getting a seven or a diamond is therefore $\begin{aligned} \frac{16}{52} = \frac{4}{13} \end{aligned}$ .
::因此,有4个七、13个钻石和1张卡,两者都是(钻石的7个),这意味着我们不能将7个和钻石的数量相加,因为如果我们这样做了,我们就会把7个钻石计算成2倍。相反,我们必须把7个和钻石的数量相加,然后减去两个类别都适用的卡片数量。这意味着有4+13-1=16卡片是7个或钻石。因此,获得7个或1个钻石的可能性是1652=413。

We can check this by listing all the cards in the deck and highlighting those that are sevens or diamonds:
::我们可以通过列出甲板上所有卡片, 并突出显示7或钻石:

You can see that there are 16 cards that fit, not the 17 we’d get if we just added up the sevens and the diamonds.
::你可以看到,有16张卡片合适, 而不是17张, 如果我们把7张和钻石加起来,

Look again at the numbers: the number of cards that are seven or diamond is (number of sevens) plus (number of diamonds) minus (number of seven and diamond). In probability terms we can write this as:
::再看看数字:7张卡片或钻石牌数(7张数)加上(钻石数)减去(7张数和钻石数)。

$\begin{aligned} P (seven or diamonds) = P (seven) + P (diamonds) - P (seven and diamonds) \end{aligned}$

::P(七或钻石)=P(七)+P(钻石)-P(七和钻石)

This leads to a general formula:
::这导致形成一个一般公式:

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

::对于重叠事件:P(A或B)=P(A)+P(B)-P(A和B)

Example
::示例示例示例示例

Example 1
::例1

A cooler contains 6 cans of Sprite, 9 cans of Coke, 4 cans of Dr Pepper and 7 cans of Pepsi. If a can is selected at random, calculate the probability that it is either Pepsi or Coke.
::冷却器包含6罐Sprite, 9罐可乐, 4罐Pepper博士和7罐Pepsi。如果随机选择一个罐, 计算它是百事可乐或可乐的概率。

Selecting a Pepsi or selecting a Coke are mutually exclusive events. This means we can use the formula $\begin{aligned} P (A or B) = P (A) + P (B) . \end{aligned}$
::选择百事可乐或选择百事可乐是相互排斥的事件。这意味着我们可以使用公式P( A 或 B) = P( A)+P( B) 。

Finding the probability of each event separately:
::分别查找每个事件的概率 :

$\begin{aligned} P (Pepsi) = \frac{Cans of Pepsi}{Total cans of soda} = \frac{7}{26} \approx 0.269 \end{aligned}$

::P(Pepsi) = Pepsi Total罐体苏打汽水=7260.269

$\begin{aligned} P (Coke) = \frac{Cans of Coke}{Total cans of soda} = \frac{9}{26} \approx 0.346 \end{aligned}$

::P(Coke) = 可口可乐罐头苏打汽水=9260.346

This means that the probability of selecting a Pepsi or a Coke is:
::这意味着选择百事可乐或百事可乐的概率是:

$\begin{aligned} P (Pepsi or Coke) = P (Pepsi) + P (Coke) = 0.269 + 0.346 = 0.615. \end{aligned}$
::P(百事可乐或可乐)=P(百事可乐)+P(粉)=0.269+0.346=0.615。

There is a 61.5% probability of selecting a Pepsi or a Coke.
::选择百事可乐或可乐的概率为61.5%。

Review
::回顾

For 1-6, determine whether the following pairs of events are mutually exclusive or overlapping:
::对于1-6,确定以下两组事件是相互排斥的还是重叠的:

The next car you see being red; the next car you see being a Ford.
::下一辆车是红色的下一辆车是福特牌的

A train being on time; the train being full.
::火车准时,火车满载。

Flipping a coin and getting heads; flipping a coin and getting tails.
::抛硬币和获得头; 抛硬币和获得尾巴。

Selecting 3 cards and getting an ace; selecting 3 cards and getting a king.
::选择 3 张卡并获得 A; 选择 3 张卡并获得国王。

Selecting 3 cards and getting 2 aces; selecting 3 cards and getting 2 kings.
::选择 3 张卡并获得 2 张 A; 选择 3 张卡并获得 2 王。

A person’s age is an even number; a person’s age is a prime number.
::一个人的年龄是偶数;一个人的年龄是质数。

For 7-10, a card is selected at random from a standard 52 card deck. Calculate the probability that:
::对于 7- 10, 从标准 52 牌牌上随机选择一张牌。计算概率 :

The card is either a red card or an even number (2, 4, 6, 8 or 10).
::该卡要么是红色卡,要么是偶数(2、4、6、8或10)。

The card is both a red card and an even number.
::该卡既是红卡,也是偶数。

The card is red or even but not both .
::该卡片是红色的,甚至不是两种红色的。

The card is black or red but not an ace .
::牌是黑色或红色的,不是王牌。

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。

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