课程： 6.8 通过寻找补充物寻找概率 | The Way To Learn

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选择节通过寻找补充来发现概率

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通过寻找补充来发现概率
Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?
::假设你被要求计算滚动两骰子的概率, 并在每张骰子上获得不同的数字。您如何找到答案而无需列举所有的可能性 ?

This lesson is about a shortcut to the calculation of some probabilities. We’ll return to this question after the lesson.
::这是计算概率的捷径。课后我们会再讨论这个问题。

Finding Probability Through Complements
::通过补充物寻找概率

Sometimes the probability of an event is difficult or impossible to calculate directly. Particularly when calculating the probability of “at least one…” types of problems, or when the sample space of the complement is smaller than that of the event, it may be worth looking first for the probability of the complement of the event you are trying to measure.
::有时事件概率很难或不可能直接计算。特别是当计算“ 至少一个 ” 类型问题的概率时,或者当补充物的样本空间小于事件的概率时,也许值得首先寻找您试图测量的事件补充概率。

Recall that the complement of an event is the sample space containing all the outcomes that are not a part of the event itself. That means that the probability of an event + the probability of the complement = 100% or 1.00, or, to say the same thing as a formula: $\begin{align*}P(A)+P(A^\prime)=1\end{align*}$ . Once you know the probability of the complement, you can just subtract it from 1 to find the probability of the event.
::回顾事件的补充是包含所有非事件本身一部分的结果的样本空间。这意味着事件概率+补充概率=100%或1.00,或者,比如一个公式:P(A)+P(A)=1. 一旦知道补充概率,您就可以从 1 中将其从 1 中减去,以找到事件的概率。

Calculating Probability
::计算概率

1. What is the probability of randomly drawing a king other than the King of Hearts from a bag containing one king of each standard suit: hearts, clubs, diamonds, and spades.
::1. 从装有每件标准西装(红心、梅花、钻石和黑桃)的国王的袋子中随机将国王而不是红心国王从国王身上画出来的可能性有多大?

Let’s say that $\begin{align*}P(KH)\end{align*}$ is the probability that you do draw the King of Hearts. In that case, $\begin{align*}P(KH^\prime)\end{align*}$ is the probability of not drawing the King of Hearts, the complement of $\begin{align*}P(KH)\end{align*}$ . Since there are 4 kings and only 1 is the King of Hearts, we can say:
::假设P(KH)是你画红心之王的概率。在这种情况下,P(KH)就是不画红心之王的概率,而P(KH)是P(KH)的补充。既然有4个国王,只有1个是红心之王,我们可以说:

$\begin{align*}P(KH)=\frac{1}{4}\end{align*}$
::P(KH)=14

therefore:
::因此:

$\begin{align*}P(KH^\prime)=1-\frac{1}{4} \ or \ \frac{3}{4}=75 \%\end{align*}$
::P(KH`)=1-14或34=75%

So the probability of not drawing the King of Hearts is 75%
::所以不画心之王的概率是75%

2. What is the probability of getting a cherry sour from a bag that starts with 9 cherry sours, 5 lemon sours, and 8 lime sours, given that you keep anything you choose and you choose up to 3 times?
::2. 以9个樱桃酸、5个柠檬酸和8个石灰酸开始的包包中樱桃酸的几率是多少,因为你保留自己选择的东西,选择最多3次?

This problem combines conditional probabilities with complements. A key point to this problem is the “keep anything you choose” part. Because we are not putting back the candy after each pull, the probability changes each time, and the chance to get a cherry improves each time we don’t choose one. Let’s deal with each pull separately at first:
::这个问题将有条件的概率和补充性结合起来。这一问题的一个关键点是“保留你所选择的任何东西 ” 。因为我们不会在每次拉动后把糖果放回去,每次的概率都会发生变化,而每次我们不选择樱桃时,获得樱桃的机会就会得到改善。让我们先单独处理每一次拉动:

$\begin{align*}P(C1)=\frac{9 \ cherry}{22 \ candies}=\frac{9}{22} \ or \ 41 \%\end{align*}$ First pull:
::P(C1) = 9 樱桃22 糖果= 922 或 41% 第一拉动 :

$\begin{align*}P(C2)=\frac{9 \ cherry}{21 \ candies}=\frac{9}{21} \ or \ 43 \%\end{align*}$ Second pull:
::P(C2) = 9 樱桃21 罐头= 921 或 43% 第二拉动 :

$\begin{align*}P(C3)=\frac{9 \ cherry}{20 \ candies}=\frac{9}{20} \ or \ 45 \%\end{align*}$ Third Pull:
::P(C3) = 9 樱桃20 罐头= 920 或45% 三号拉力:

It is tempting to think that the probability of getting at least 1 cherry is just the sum of the three probabilities, but obviously there can’t be a $\begin{align*}41 \%+43 \%+45 \%=129 \%\end{align*}$ chance. The twist is that we only choose a 2 ^nd or 3 ^rd time if we don’t get a cherry the time before. That means we need to calculate the chance of choosing a 2 ^nd or 3 ^rd time and multiply the probability of a cherry on that pull by the chance of pulling that time at all.
::想想至少获得一樱桃的可能性只是三种可能性的总和,这是很诱人的。但显然不可能有414345129%的概率。转折是,如果我们以前没有樱桃,我们只选择第二次或第三次机会。这意味着我们需要计算选择第二次或第三次机会,并乘以这一机会的樱桃概率。

Fortunately we can use the complement rule to save time. The chance of needing a 2 ^nd pull is the same as the chance that we don’t pull a cherry on the 1 ^st pull, in other words:
::幸运的是,我们可以利用补充规则来节省时间。需要第二拉一拉一拉一拉一拉一的机率与我们没有第一拉一拉一把樱桃的机率相同,换句话说:

$\begin{align*}P(2nd \ pull)=P(C1^\prime)=1-41 \%=59 \%\end{align*}$
::P( 2nd pull) = P( C1_ ) = 1 - 41 @ 59%

The chance of needing a 3 ^rd pull is the same as not getting a cherry the 2 ^nd time:
::需要第三次拉力的机会与第二次得不到樱桃相同

$\begin{align*}P(3rd \ pull)=P(C1^\prime) \times P(C2^\prime)=59 \% \times 57 \%=34 \% \end{align*}$
::P( 第3次拉动) = P( C1}) × P( C2} ) = 59 57 34%

Now we can find the overall probability of getting a cherry in three pulls or less:
::现在,我们可以发现三下或更少获得樱桃的总体概率:

$\begin{align*} P&(cherry) =\text{(chance of cherry of pull 1)} \\ + \text{(chance of }& \text{needing pull 2} \times \text{chance of cherry on pull 2)} \\ + \text{(chance of }& \text{needing pull 3} \times \text{chance of cherry on pull 3)} \\ P(cherry)&=41 \%+(59 \% \times 43 \%)+(34 \% \times 45 \%)=82 \%\end{align*}$
::P(cherry) = (樱樱桃的顺序拉拉 1+(樱桃的顺序需要拉 2×樱桃的顺序拉 2+(樱桃的顺序拉 2)+(樱桃的顺序需要拉 3) P(樱) =41(5943%) +(3445%) =82%

3. What is the probability of rolling two dice and at least one die showing a factor of 6?
::3. 滚动两骰子和至少一骰子死亡的概率是6系数是多少?

This is an “at least one…” problem. To satisfy the requirements, either one of the two dice or both need to land on 1, 2, 3, or 6. That is quite a few possibilities to solve for! However, there are many fewer possible outcomes where neither die shows 1, 2, 3, or 6 – in other words, where both dice show 4 or 5! There are only 4 ways that could happen:
::这是一个“至少一个... ” 的问题。为了满足要求,两只骰子中的一个或两个都需要降落在1、2、3或6号上,这是相当少的几种可能解决的办法。然而,如果双方没有死亡,结果就会少得多 — — 换句话说,两只骰子都显示4或5号!只有4种方法可以发生:

1 ^st die rolls 5 and 2 ^nd rolls 4
::第一死5号第二死4号死5号死4号死4号死5号死2号死4号死4号死

1 ^st die rolls 4 and 2 ^nd rolls 5
::第一死4号第二死5号死4号死5号死2号死5号死5号死4号死4号死2号死5号死5号死1号死4号死4号死4号死2号死2号死5号死1号死4号死4号死4号死2号死4号死4号死4号死4号死4号死2号死2号死5号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死2号死2号死2号死5号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死2号死2号死2号死5号死5号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死4号死2号死2号死2号死2号死2号死2号死2号死5号死5号死5号死

Both dice roll 4
::两个骰子卷4

Both dice roll 5
::两个骰子滚5

There are $\begin{align*}6 \times 6=36 \end{align*}$ total possible outcomes. If we say that the event “at least one die shows a factor of 6” is $\begin{align*}A\end{align*}$ , then $\begin{align*}A^\prime\end{align*}$ would be the complement, so we can say:
::有6x6=36的总结果。如果我们说“至少一个死亡的因数为6”是A,那么A`就是补充,所以我们可以说:

$\begin{align*}P(A^\prime)=\frac{4 \ favorable \ outcomes}{36 \ total \ outcomes}=\frac{1}{9}\end{align*}$
::P(A`)=4 有利成果36 总结果=19

If the complement of the event we want to calculate is $\begin{align*}\frac{1}{9}\end{align*}$ , then the event itself can be calculated as:
::如果我们想要计算的事件的补充是19, 那么该事件本身可以被计算为:

$\begin{align*}P(A)=1-P(A^\prime)=\frac{8}{9} \ or \ 88.9 \%\end{align*}$
::P(A)=1-P(A’)=89或88.9%

Therefore, we can say that the probability of rolling two dice and getting at least one factor of six is $\begin{align*}\frac{8}{9}\end{align*}$ .
::因此,我们可以说,滚动两骰子和获得至少六分之一的概率是89。

Earlier Problem Revisited
::重审先前的问题

Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?
::假设你被要求计算滚动两骰子的概率, 并在每张骰子上获得不同的数字。您如何找到答案而无需列举所有的可能性 ?

The probability of rolling two not matching numbers is the complement of rolling a matching pair. Since there are only 6 numbers to make matches from, there are only 6 matching pairs. The total number of possible outcomes of two dice is $\begin{align*}6 \times 6=36\end{align*}$ .
::滚动两个不匹配数字的概率是滚动对匹配数字的补充。由于只有6个匹配数字, 只有6个匹配对。两个骰子可能的结果总数是 6x6=36 。

$\begin{align*}\therefore P(matching)=\frac{6\text{ matches}}{36\text{ possibilities}}=\frac{1}{6} \ or \ .167\end{align*}$
::P(配对)=6 匹配36 可能性=16 或 167

That means that the complement, rolling not matching numbers is:
::这意味着补充、滚动和不匹配的数字是:

$\begin{align*}1-.167=.833 \ or \ 83.3 \%\end{align*}$
::1-167=833或83.3%

That was quite a bit faster than trying to enumerate all of the possible non-matching rolls!
::这比试图列举所有可能非匹配的卷子要快一点!

Examples
::实例

Example 1
::例1

What is the probability of rolling a number other than 5 on a number cube?
::在数字立方体上滚动除5之外的数字的概率是多少?

There are six sides on a number cube, so the probability of rolling any single number is $\begin{align*}\frac{1}{6}\end{align*}$ . Therefore, we can say: $\begin{align*}P(5)=\frac{1}{6}\end{align*}$ so $\begin{align*}P(5^\prime)=\frac{5}{6}\end{align*}$ Therefore the probability of not rolling a 5 is $\begin{align*}\frac{5}{6}\end{align*}$ .
::数字立方体有六边, 所以滚动任何一个数字的概率是16。因此,我们可以说:P(5)=16 所以P(5)=56。因此,不滚动5的概率是56。

Example 2
::例2

What is the probability of choosing a card that isn't a club from a standard deck?
::从标准甲板上选择不是俱乐部的卡的概率是多少?

There are four suits, so the probability of choosing a club is $\begin{align*}\frac{1}{4}\end{align*}$ . The complement is the probability of not choosing a club, and it is $\begin{align*}1-\frac{1}{4}=\frac{3}{4} \ or \ 75 \%\end{align*}$
::有四套西装,所以选择俱乐部的概率是14。补充是不选择俱乐部的概率,是1-14=34或75%。

Example 3
::例3

What is the probability of not rolling a factor of 10 on a single roll of a 20 sided die?
::在20边死亡的一卷中不滚动因数为10的概率是多少?

There are 4 factors of 10: 1, 2, 5, and 10. Therefore the probability of rolling a factor of 10 on a 20-sided die is $\begin{align*}\frac{4}{20}\end{align*}$ . The complement is the probability of not rolling one of the four factors: $\begin{align*}1-\frac{4}{20}=\frac{16}{20}\end{align*}$
::4个系数为10:1、2、5和10。因此,20方死亡时滚动系数为10的概率为420。补充是,4个系数之一(1-420=1620)不滚动的概率。

Example 4
::例4

You have a bag of crazy-flavor jellybeans. You know that the flavors are distributed as: 12 earwax, 14 belly-button lint, 26 dog food, 38 gym sock, and 10 of your favorite fruit in a the bag. What is the probability tha tyou will be unhappy with the taste of our choice if you choose one jellybean at random?
::你有一袋疯狂的软糖豆。你知道味道的分布方式是:12个耳盘、14个腹盘、26个狗食品、38个运动袜和10个你最喜欢的果子。如果你随意选择一个果冻,那么你会不会不满意我们选择的味道呢?

Let $\begin{align*}P(F)\end{align*}$ be the probability of pulling a fruit-flavor, then $\begin{align*}P(F^\prime)\end{align*}$ is the probability of not pulling a fruit flavor (and getting something disgusting instead!). There are 10 fruit-flavored beans in the bag of 100 beans, so:
::Let P(F) 可以是拉水果蔬菜的概率,而P(F) 则是不拉水果口味的概率(而得到令人厌恶的东西 ) 。 100 豆袋里有 10 个水果蔬菜豆, 所以:

$\begin{align*}P(F)&=\frac{10}{100} \\ P(F^\prime)&=1-P(F)=\frac{90}{100}\end{align*}$
::P(F)=10100P(F’)=1-P(F)=90100

Therefore, you have a 90% probability of being unhappy with your choice.
::因此,你有90%的概率对你的选择不满意。

The factors of ten or twelve are: 1, 2, 3, 4, 5, 6, 10, and 12. Since all six numbers from a standard die are in that set, the probability of rolling one of them is 100% or 1.0. Therefore, the probability of not rolling one of them is $\begin{align*}1-1=0\end{align*}$ .
::10或12的系数是:1、2、3、4、5、6、10和12。由于标准死亡的所有六个数字都在这个组数中,滚动的概率是100%或1.0。因此,不滚动的概率是1-1=0。

Review
::回顾

If you randomly pull a single card from a standard deck, what is the probability that the card is anything other than a king?
::如果你随机地从标准甲板上拉一张单张卡片, 那么该卡片除了国王之外还有什么其他的概率呢?

What is the probability of not pulling a face card from a standard deck?
::不从标准甲板上拉一张脸牌的概率是多少?

What is the probability that a single roll of a 10-sided die will not land on a 7?
::仅仅一卷十面之死会不会在七点降落呢?

What is the probability that a single roll of a standard die will be 1, 2, 3, 4, or 5?
::一卷标准死亡的几率是 1,2,3,3,4,或5?

A candy jar contains 6 red, 7 blue, 8, green, and 9 yellow candies. What is the probability that choosing a single candy at random will result in a piece that is either red, blue, or green?
::一个糖果罐里有6个红色,7个蓝色,8个绿色,9个黄色糖果。随意选择一个糖罐,会产生一个红、蓝色或绿色的糖罐的可能性有多大?

What is the probability that a single choice from the jar in Q 5 will result in a piece that is either red or yellow?
::从Q5罐子中单挑一个选择会导致一块红的或黄色的碎片的可能性有多大?

What is the probability that a single choice from the same jar will not be blue?
::从同一个罐子中作出单一选择不会是蓝色的概率是多少?

You have $39 in cash, composed of the largest bills possible. What is the probability that a randomly chosen bill from the $39 will not be a $1 bill?
::你们有39美元现金,由可能的最大帐单组成。从39美元中随机选择的帐单不会是1美元帐单的可能性有多大?

There are 450 songs on the .mp3 player you share with your father and sister. If your dad has 125 80’s songs, and your sister has twice that many country music songs, what is the percent probability that a randomly chosen song will not be one of yours?
::. mp3 播放器上有450首歌曲,您与父亲和姐姐分享。如果你爸有12580年代的歌曲,而你姐姐有两倍的乡村音乐歌曲,那么随机选择的歌曲不会成为您的歌曲的概率是多少?

What is the percent probability that a random roll of a fair die will not result in an even or prime number?
::公平死亡的随机滚动不会导致偶数或质数的概率是多少?

The train station has 47 active trains. 5 are late by less than 10 minutes, 4 are between 11 and 15 minutes late, and 10 are more than 15 minutes late. What is the probability that a randomly chosen train will be on time?
::火车站有47辆运行中的列车。 5辆火车晚于10分钟,4辆晚于11至15分钟,10辆晚于15分钟。随机选择的列车是否准时? 5辆晚于10分钟,4辆晚于11至15分钟,10辆晚于15分钟。

The probability that a student has called in sick and that it is Monday is 12%. The probability that it is Monday and not another day of the school week is 20% (there are only five days in the school week). What is the probability that a student has not called in sick, given that it is Monday?
::学生请病假和请病假的概率是12 % 。周一是周一而不是学校周的另一天的概率是20% ( 每周只有5天 ) 。学生请病假的概率是多少? 周一是周一,学生请病假的概率是多少? 学生请病假的概率是多少?

A neighborhood wanted to improve its parks so it surveyed kids to find out whether or not they rode bikes or skateboards. Out of 2300 children in the neighborhood that ride something, 1800 rode bikes, and 500 rode skateboards, while 200 of those ride both a bike and skateboard. What is the probability that a student does not ride a skateboard, given that he or she rides a bike?
::一个街区想改善公园,因此它调查了孩子是否骑过自行车或滑板。在附近骑过车的2300名儿童中,有1800人骑过自行车,500人骑过滑板,其中200人骑过自行车和滑板。鉴于学生骑过自行车,学生不骑过滑板的可能性有多大?

A movie theatre is curious about how many of its patrons buy food, how many buy a drink, and how many buy both. They track 300 people through the concessions stand one evening, out of the 300, 78 buy food only, 113 buy a drink only and the remainder buy both. What is the probability that a patron does not buy a drink if they have already bought food?
::电影院对有多少人购买食物、多少人购买饮料和多少人购买两者都很好奇。他们有一晚通过特许摊位追踪300人,在300人中,78人只购买食物,113人只购买饮料,其余人两者都购买。如果已经购买了食物,那么顾客不购买饮料的可能性有多大?

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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