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复合活动联盟
If you think it through, it should make sense that the probability of pulling one Queen at random from a standard deck is $\begin{align*}\frac{4}{52}\end{align*}$ or $\begin{align*}\frac{1}{13}\end{align*}$ , since there are 4 Queens in a standard 52 card deck. How then would you calculate the probability of pulling a Queen OR a King from the same deck?
::如果你仔细想一想,那么从标准甲板上随机拉动一名女王的概率应该是452或113,因为标准52牌板上有4个皇后。那么,你将如何计算从同一甲板拉动一名女王或国王的概率?

Union of Compound Events
::复合活动联盟

When multiple may occur during a particular experiment , there are a couple of different types of outcomes you may need to consider:
::当在特定试验期间可能发生多重情况时,可能需要考虑几种不同类型的结果:

Intersection : the probability of both or all of the events you are calculating happening at the same time (less likely).
::交点 : 您同时计算的所有事件或同时发生事件的概率( 可能性较小 ) 。

Union : the probability of any one of multiple events happening at a given time (more likely).
::联盟:在特定时间(更有可能)发生任何一起多事件的概率。

In this lesson, we will focus on union . Calculating the union is relatively easy, you just add up the individual probabilities of the events:
::在这个教训中,我们将集中关注工会。计算工会相对容易,你只是将事件的个人概率加在一起:

">
$\begin{aligned} P (x o r y) = P (x) + P (y) \end{aligned}$
$\begin{align*}P(x \ or \ y)=P(x)+P(y)\end{align*}$
::P(x 或 y) = P(x)+P

This can also be thought of as:
::这也可以认为是:

$\begin{aligned} P (x o r y) = \frac{(number of outcomes where x is true) + (number of outcomes where y is true)}{total number of possible outcomes} \end{aligned}$
$\begin{align*}P(x \ or \ y)=\frac{(\text{number of outcomes where} \ x \ \text{is true}) +(\text{number of outcomes where} \ y \ \text{is true})}{\text{total number of possible outcomes}}\end{align*}$
::P(x 或 y) = (在x为真的情况下结果数)+(在y为真时结果数) 可能结果总数

It is really just that simple! It is intuitive also, assuming there is no overlap (which we will consider later), it just makes sense to think that if you have a 20% probability of one thing happening, and a 30% probability of another, then you have a 50% probability of one of the two of them happening during a given experiment.
::这其实很简单! 它也是直觉的,假设没有重叠(我们稍后会考虑), 想象一下,如果你有20%发生一件事的概率, 30%发生另一件事情的概率, 那么在特定实验中,你有一个50%发生其中一件的概率。

Calculating Probability
::计算概率

1. You are given a bag containing 15 equally sized marbles. You know there are 5 yellow marbles, 5 blue marbles, and 5 green marbles in the bag. What is the statistical probability that you would pull a yellow or green marble out, if you reach in the bag and grab a marble at random?
::1. 给您一个装有15个相同大小的大理石的袋子,你知道包里有5个黄色大理石、5个蓝色大理石和5个绿色大理石。

Recall the formula for the union of simple probabilities:
::回顾简单概率结合的公式:

$\begin{aligned} P (x o r y) = \frac{(number of outcomes where x is true) + (number of outcomes where y is true)}{total number of possible outcomes} \end{aligned}$
$\begin{align*}P(x \ or \ y)=\frac{(\text{number of outcomes where} \ x \ \text{is true}) +(\text{number of outcomes where} \ y \ \text{is true})}{\text{total number of possible outcomes}}\end{align*}$
::P(x 或 y) = (在x为真的情况下结果数)+(在y为真时结果数) 可能结果总数

In this case, we have:
::在这种情况下,我们有:

$\begin{aligned} P (y e l l o w o r g r e e n) = \frac{5 y e l l o w m a r b l e s + 5 g r e e n m a r b l e s}{15 t o t a l m a r b l e s} \end{aligned}$
$\begin{align*}P(yellow \ or \ green)=\frac{5 \ yellow \ marbles+5 \ green \ marbles}{15 \ total \ marbles}\end{align*}$
::P(黄或绿)=5黄色大理石+5绿色大理石+5绿色大理石15

Which would reduce to:
::这将减少:

$\begin{aligned} P (y e l l o w o r g r e e n) = \frac{2}{3} o r 66.6 \bar{6} % \end{aligned}$
$\begin{align*}P(yellow \ or \ green)=\frac{2}{3} \ or \ 66.6 \bar{6} \%\end{align*}$
::P(黄绿色)=23或66.66

2. What is the probability of rolling an odd or even number on a standard six-sided die?
::2. 在标准的六方死亡上滚动一个奇数或甚至数字的可能性有多大?

A standard die has three odd numbers (1, 3, 5) and three even numbers (2, 4, 6). Therefore, the probability of rolling an odd or even number is:
::标准死亡有三个奇数(1、3、5)和三个偶数(2、4、6),因此,一个奇数或偶数的滚动概率是:

$\begin{aligned} P (o d d o r e v e n) = \frac{3 o d d + 3 e v e n}{6 t o t a l} = \frac{6}{6} \end{aligned}$
$\begin{align*}P(odd \ or \ even)=\frac{3 \ odd + 3 \ even}{6 \ total}=\frac{6}{6}\end{align*}$
::P(dd或甚至)=3 奇特+3偶6 总计=66

Reducing to:
::减少:

$\begin{aligned} P (o d d o r e v e n) = 1 o r 100 % \end{aligned}$
$\begin{align*}P(odd \ or \ even)=1 \ or \ 100 \%\end{align*}$
::P(dd或甚至)=1或100%

3. If Lawrence is playing with a standard 52-card deck, what is the probability of pulling a 2, a 4, or a 6 out of the deck at random?
::3. 如果劳伦斯在玩标准的52卡牌甲板,那么随机拉出2张、4张或6张的概率是多少?

Let’s solve this one as the total of the individual probabilities. Lawrence’s probability of pulling a 2, 4, or 6 is the same as the union of the probability of each possible outcome :
::让我们用个人概率的总和来解决这个问题。劳伦斯的拉拉2、4或6的概率与每种可能结果概率的结合相同:

$\begin{aligned} P (2, 4, o r 6) = P (2) + P (4) + P (6) = \frac{1}{13} + \frac{1}{13} + \frac{1}{13} = \frac{3}{13} o r 23.1 % \end{aligned}$
$\begin{align*}P(2, 4, \ or \ 6)=P(2)+P(4)+P(6)=\frac{1}{13}+\frac{1}{13}+\frac{1}{13}=\frac{3}{13} \ or \ 23.1 \%\end{align*}$
::P(2、2、4或6)=P(2)+P(4)+P(6)=113+113+113+113=313或23.1%

Earlier Problem Revisited
::重审先前的问题

It should make sense now that the probability of pulling one Queen at random from a standard deck is $\begin{align*}\frac{4}{52}\end{align*}$ or $\begin{align*}\frac{1}{13}\end{align*}$ , since there are 4 Queens in a standard 52 card deck. How then would you calculate the probability of pulling a Queen OR a King from the same deck?
::现在应该可以理解的是,任意从标准甲板上拉走一名女王的概率是452或113,因为标准52牌牌牌上有4个皇后。那么,如何计算从同一甲板拉动一名女王或国王的概率?

Remember that the union of multiple probabilities is simple the total sum of all of the individual probabilities:
::记住多种概率的结合简单简单于所有个体概率的总和:

$\begin{aligned} P (Q u e e n o r K i n g) = \frac{4 Q u e e n s + 4 K i n g s}{52 C a r d s} = \frac{8}{52} o r \frac{4}{26} o r \frac{2}{13} = 15.4 % \end{aligned}$
$\begin{align*}P(Queen \ or \ King)=\frac{4 \ Queens+4 \ Kings}{52 \ Cards}=\frac{8}{52} \ or \ \frac{4}{26} \ or \ \frac{2}{13}=15.4 \%\end{align*}$
::P(女王或国王)=4 皇后+4 Kings52卡=852或426或213=15.4%

Examples
::实例

Example 1
::例1

What is the statistical probability of pulling either the only red or the only blue marble out of a bag with 12 marbles in it?
::从装有12颗大理石的袋子里拉出唯一的红色大理石或唯一的蓝色大理石的统计概率是多少?

$\begin{align*}P(red \ or \ blue)=\frac{1\text{ red marble} + 1\text{ blue marble}}{12\text{ total marbles}}=\frac{2}{12} \ or \ \frac{1}{6} \ or \ 16.6 \%\end{align*}$
::P(红或蓝)=1红色大理石+1蓝色大理石12 总计大理石=212或16或16.6%

Example 2
::例2

What is the probability of a spinner landing on “2”, “3”, or “6” if there are 6 equally spaced points on the spinner?
::如果在“2”、“3”或“6”上方有6个同样空格的点,旋转器降落在“2”、“3”或“6”上的概率是多少?

$\begin{align*}P(2 \ or \ 3 \ or \ 6)=\frac{1 \text{ number }6 + 1\text{ number }2 + 1\text{ number }3}{6\text{ total numbers}}=\frac{3}{6} \ or \ \frac{1}{2} \ or \ 50 \%\end{align*}$
::P(2或3或3或6)=1 6+1编号 2+1编号 36 总计= 36 或 12 或 50%

Example 3
::例3

What is the probability of pulling a red or black card at random from a standard deck?
::从标准甲板随意拉红卡或黑卡的概率是多少?

$\begin{align*}P(red \ or \ black)=\frac{26\text{ red cards} + 26\text{ black cards}}{52\text{ total cards}}=\frac{52}{52}=\frac{1}{1} \ or \ 100 \%\end{align*}$
::P(红或黑)=26红卡+26黑卡52共卡=5252=11=11或100%

Example 4
::例4

What probability of picking a red or green marble from a bag with 5 red, 7 green, 6 blue, and 14 yellow marbles in it?
::从包里摘红绿大理石 5个红色 7个绿色 6个蓝色 14个黄色大理石的 5个红色 7个绿色 6个蓝色

$\begin{align*}P(red \ or \ green)=\frac{5\text{ red marbles}+7\text{ green marbles}}{32\text{ total marbles}}=\frac{12}{32} \ or \ \frac{6}{16} \ or \ \frac{3}{8} \ or \ 37.5 \%\end{align*}$
::P(红或绿)=5红色大理石+7绿色大理石+7绿色大理石32 总计大理石=1232或616或38或37.5%

Example 5
::例5

What is the of shaking the hand of a student wearing red if you randomly shake the hand of one person in a room containing the following mix of students?
::如何摇动一个身着红色学生的手如果你随机地摇动一个人的手在一个包含以下学生的房间里

13 female students wearing blue
::13名身着蓝色的女学生

7 male students wearing blue
::7名身穿蓝色的男学生

6 female students wearing red
::6名身穿红色服装的女学生

9 males students wearing red
::9名穿红色衣服的男生

18 female students wearing green
::18名身着绿色绿色服装的女学生

21 male students wearing green
::21名穿绿色绿色制服的男学生

$\begin{align*}P(red)=\frac{6\text{ females wearing red}+9\text{ males wearing red}}{74\text{ total students}}=\frac{15}{74} \ or \ 20.3 \%\end{align*}$
::P(red)=6名女性,身穿红色+9的男性,身着红色+9的男性,身着红色74的男性,学生总数=1574人,占20.3%

Review
::回顾

What is the probability of rolling a standard die and getting between a 1 and 6 (inclusive)?
::滚动标准死亡和在1至6(包括)之间死亡的概率是多少?

What is the probability of pulling one card from a standard deck and it being an 8, a 3, or a queen?
::从标准甲板上拉一张牌的概率是8张卡还是3张卡还是皇后?

What is the probability of rolling a 5 or a 2 on an 8-sided die?
::在8方死亡时滚动5或2的概率是多少?

What is the probability of pulling one card from a standard deck and it being a spade, a diamond, or a club?
::从标准甲板上拉一张牌的概率是多少?

What is the probability of rolling a 1, 3, or 5 on a 7-sided die?
::7方死亡时滚动1、3或5的概率是多少?

What is the probability of pulling one card from a standard deck and it being a king, a 4, or a 8?
::从标准甲板拉一张牌的概率是多少? 它是国王,4张还是8张?

What is the probability of pulling a yellow or blue candy from a bag containing 35 candies equally distributed among yellow, blue, green, red, and brown candies?
::从一个装有35只糖果的袋子中拉出黄色或蓝色糖果的可能性有多大? 糖在黄色、蓝色、绿色、红色和棕色糖果中平均分配。

What is the probability of spinning 2, 4, or 7 on a 10-space spinner (equally spaced)?
::在10个空间旋转器上旋转2、4或7的概率(平均间距)是多少?

What is the probability of rolling a 1, 3, 5, or 6 on a 20-sided die?
::一、三、五、六轮20方死亡的概率是多少?

A car factory creates cars in the following ratio: 3 green, 2 blue, 7 white, 2 black and 1 brown. What is the probability that a randomly selected car will be either blue or brown?
::汽车厂制造汽车的比例如下:3辆绿色汽车、2辆蓝色汽车、7辆白色汽车、2辆黑色汽车和1辆棕色汽车。

There are 4 flavors of donuts on the shelf: glazed, sprinkles, plain, and powdered sugar. If there are equal numbers of each of the non-plain donuts, and half as many plain as any one of the others, what is the probability of randomly choosing a plain donut out of all donuts on the shelf?
::架子上有4种甜甜圈的口味:玻璃、喷洒、平原和粉末糖。如果每个非白板甜甜圈的数量相等,而平面甜甜圈的数量则比其他任何一种都多一半,那么从架子上的所有甜甜圈中随机选择一个平面甜甜圈的概率是多少?

What is the probability of randomly choosing a red Ace or a black King from a standard deck?
::从标准甲板随机选择红A或黑王的概率是多少?

What is the probability of rolling a prime number or an even number on a standard die?
::在标准死亡上滚动一个质数或偶数的概率是多少?

Mr. Spence’s class has 13 students. 4 students are wearing coats, 3 are wearing vests, 3 are wearing hoodies, and the rest are in t-shirts. What is the probability that Mr. Spence will randomly call the name of a student wearing a coat or a vest?
::思朋斯先生的班级有13名学生。 4名学生穿着大衣,3名学生穿着防弹背心,3名学生穿着连帽衫,其余的穿T恤衫。思朋斯先生会随意说出身穿大衣或防弹背心的学生姓名的可能性有多大?

In the same class, what is the probability that Mr. Spence will randomly call the name of a student in a hoodie or t-shirt?
::在同一班级,Spencer先生会随意调用穿连帽衫或T恤的学生的名字的可能性有多大?

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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