Section: 独立事件的概率 | 12.20独立事件的概率

Section outline

At the end of the school year, Mr. Fike collected 80 surveys to find out which sports had been the most popular during the school year. On the survey he asked students to list their three favorite sports. He found that football was chosen as a favorite sport by 60 out of 80 students. Basketball was chosen as a favorite by 40 students. He was surprised to find out that 20 students chose tennis as a favorite while only 15 chose baseball. Mr. Fike wondered what the likelihood was of a student choosing football, basketball, and tennis as their favorite sports. How can Mr. Fike figure out the probability of a student selecting all three?
::Fike先生收集了80项调查,以了解哪些体育是该学年最受欢迎的运动;在调查中,他要求学生列出他们最喜爱的三种运动;他发现足球被80名学生中的60名选为最受欢迎的运动;篮球被40名学生选为最喜爱的运动;他惊讶地发现有20名学生选择网球为最受欢迎的运动,而只有15名学生选择了棒球;Fike先生想知道学生选择足球、篮球和网球为他们最喜欢的运动的可能性如何;Fike先生如何能发现学生选择所有三种运动的可能性?

In this concept, you will learn how to calculate the probability of .
::在此概念中, 您将学习如何计算 . 的概率。

Calculating the Probability of Independent Events
::计算独立活动的概率

An independent event is an event that does not depend on another event to determine its outcome . When there are two independent events, one outcome does not impact the outcome of the second event. Think about choosing a card from a deck of cards. There are 52 cards in a deck. You can choose a card and there is a certain probability that it will be a red card. Then you can put the card back and choose again. The first outcome of choosing a card has nothing to do with the second outcome because they are independent events.
::独立事件是一个不取决于其它事件来决定其结果的事件。如果有两个独立事件, 一个结果不会影响第二个事件的结果。考虑从牌牌牌牌中选择一张牌。牌牌里有52张牌。您可以选择一张牌, 并且有一定的可能性它会是一个红色卡片。然后您可以将卡片放回去重新选择。选择卡片的第一个结果与第二个结果无关, 因为它们是独立事件。

You can calculate the probability of an independent event occurring.
::您可以计算独立事件发生的概率。

Let’s say there are four tiles face down on a table.
::假设一张桌子上面有四张瓷砖。

On the side you cannot see are different images. Three of them have a sun and one of them has a four-leaf clover.
What is the probability of selecting a sun on the first flip?
First, because this is an independent event, nothing else impacts the outcome except the flip. If there are three suns then you have a 3 out of 4 chance of flipping a sun.
::在你看不到的侧面是不同的图像。其中三张是太阳, 其中一张是四叶三叶三叶。选择太阳第一个翻转的概率是多少? 首先, 因为这是个独立事件, 除了翻转, 没有其他东西会影响结果。如果有三颗太阳, 你就会有三颗太阳翻转的四分之一机会。

$\begin{aligned} P = \frac{3}{4} \end{aligned}$

You can also have two independent events. The events have nothing to do with each other, and yet you can calculate the probability of them occurring.
::您也可以有两个独立的事件。事件与彼此无关, 但是您可以计算出事件的发生概率。

Take a look at the following example.
::以下面的例子为例。

There are 9 marbles in a bag. There are three blue, three green and three orange. What is the probability of selecting one blue marble, then putting that one back and selecting one green marble?
::袋子里有9个大理石,有3个蓝色,3个绿色,3个橙色。选择一个蓝色大理石,然后放回去,选择一个绿色大理石的概率是多少?

There are two independent events in this example. The first is choosing one blue marble. The next is choosing one green marble.
::这个例子有两个独立事件。第一个是选择一个蓝色大理石。下一个是选择一个绿色大理石。

Also notice that the first marble is put back into the bag before the next one is taken out. This is a key for independent events.
::也注意到第一颗大理石在下一颗大理石被取出之前被放回袋子里。这是独立事件的关键。

What is the probability of choosing a blue one first?
::首先选择蓝色的概率是多少?

There are 9 marbles and 3 are blue. That means 3 out of 9 chances.
::9颗大理石和3颗蓝色,这意味着9个机会中有3个是蓝色的。

$\begin{aligned} \frac{3}{9} = \frac{1}{3} \end{aligned}$

What is the probability of choosing a green one next?
::下一个选择绿色的概率是多少?

There are 9 marbles and 3 are green.
::有9颗大理石和3颗绿色

$\begin{aligned} \frac{3}{9} = \frac{1}{3} \end{aligned}$

To find the likelihood of these two independent events occurring, multiply the two probabilities.
::为了发现发生这两种独立事件的可能性,将两种可能性乘以。

$\begin{aligned} P (B and G) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \end{aligned}$

Take another look at the example, this time with three independent events occurring.
::再看看这个例子,这次发生了三个独立事件。

There are 9 marbles in a bag. There are three blue, three green and three orange. What is the probability of selecting one blue marble, then putting that one back and selecting one green marble, then putting that one back and selecting one orange marble?
::袋子里有9个大理石,有3个蓝色,3个绿色,3个橙色。选择一个蓝色大理石,然后将它放回去,然后选择一个绿色大理石,然后将它放回去,然后选择一个橙色大理石的概率是多少?

Since you already know the probability of the first two independent events, find the probability of choosing an orange one next.
::既然您已经知道前两次独立事件的概率, 找到下一个选择橙色事件的概率。

There are 9 marbles and 3 are orange.
::有9颗大理石和3颗橙色

$\begin{aligned} \frac{3}{9} = \frac{1}{3} \end{aligned}$

Next, figure the likelihood of three independent events occurring by multiplying each probability.
::下一位,请用乘以每个概率来计算发生三个独立事件的可能性。

$\begin{aligned} P (B, then G, then O) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{27} \end{aligned}$
There is a very small likelihood that this event will occur.

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Mr. Fike and his sports survey.
::早些时候,你对菲克先生和他的体育调查感到疑惑。

Mr. Fike surveyed a representative sample of students from grades 6, 7, and 8 to find out what their three favorite sports were at school. He collected 80 surveys and found that football was chosen as a favorite sport by 60 students. Basketball was chosen as a favorite by 40 students. And tennis was the next top favorite with 20 votes.
::Fike先生对六、七、八年级的学生进行了代表性抽样调查,以了解他们在学校最喜欢的三种运动是什么,他收集了80项调查,发现足球被60名学生选为最受欢迎的运动,篮球被40名学生选为最受欢迎的运动,网球是第二最受欢迎的运动,有20票。

Mr. Fike wants to know the likelihood of a student choosing football, basketball, and tennis as their favorite sports.
::Fike先生想知道学生选择足球、篮球和网球作为他们最喜欢的运动的可能性。

First, figure out the probability of the first event, 60 out of 80 students chose football.
::首先,找出第一场比赛的概率,80名学生中有60人选择足球。

$\begin{aligned} \frac{60}{80} = \frac{3}{4} \end{aligned}$

Next, figure out the probability of the second event, 40 out of 80 students chose basketball.
::接下来,找出第二次活动的概率,80名学生中有40人选择篮球。

$\begin{aligned} \frac{40}{80} = \frac{1}{2} \end{aligned}$

Then, figure out the probability of the third event, 20 out of 80 students chose tennis.
::80名学生中有20人选择网球。

$\begin{aligned} \frac{20}{80} = \frac{1}{4} \end{aligned}$

Finally, multiply the probabilities of the three independent events.
::最后,将三次独立活动的概率乘以倍增。

$\begin{aligned} \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{3}{32} \end{aligned}$

There is a 3 out of 32 chance that a student would choose all three.
::32个学生中有3个可能选择所有3个。

You can look at this as a percent too to get an idea of how small the likelihood is.
::你可以把这个也看成是百分之一来了解可能性有多小

$\begin{aligned} \frac{3}{32} = 9.38 % \end{aligned}$
The answer is there is less than a 1 in 10 chance of a student choosing all three sports as a favorite.

In the following examples, calculate the likelihood of the independent event using a percent.
::在以下例子中,使用百分比计算独立事件的可能性。

Example 2
::例2

Find the probability of the independents events occurring in the following scenario.
::在以下假设情景中找到独立事件的概率。

Out of 100 students at Riverview Middle School, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf?
::在河景中学的100名学生中,有50人享受游泳,在100名学生中,有25人享受高尔夫。学生享受游泳和高尔夫的可能性有多大?

$\begin{aligned} \frac{50}{100} = \frac{1}{2} \end{aligned}$

Next, figure out the probability of the second event, 25 out of 100 students enjoy golf.
::接下来,找出第二次活动的概率,100名学生中有25人享受高尔夫球。

$\begin{aligned} \frac{25}{100} = \frac{1}{4} \end{aligned}$

Then, multiply the probabilities of the two independent events.
::然后,乘以这两个独立事件的概率。

$\begin{aligned} \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \end{aligned}$

The answer is there is a one out of eight chance that this will happen.
::答案是,八分之一的概率将发生这种情况。

Now, let’s add a third sport, basketball.
::现在,让我们加上第三项运动,即篮球。

Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf?
::在河景中游的100名学生中,有50人享受游泳;在同样的100名学生中,有25人享受高尔夫。学生享受游泳和高尔夫的可能性有多大?

Let’s add to this problem that 40 out of 100 enjoy basketball.
::更何况每100人中就有40人享受篮球。

What is the likelihood that a student would enjoy all three?
::学生享受这三样东西的可能性有多大?

First, find the probability of the third event, 40 out of 100 enjoy basketball.
::首先,找到第三次比赛的概率, 每100人中有40人享受篮球比赛。

$\begin{aligned} \frac{40}{100} = \frac{2}{5} \end{aligned}$

Remember the probability of the first and second events.
::记住第一次和第二次事件的概率。

$\begin{aligned} \begin{array}{rcl} \frac{50}{100} = \frac{1}{2} \\ \frac{25}{100} = \frac{1}{4} \end{array} \end{aligned}$

Next, multiply the third event’s probability by the first and second independent events.
::接下来,将第三个事件的概率乘以第一个和第二个独立事件的概率。

$\begin{aligned} \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{5} = \frac{2}{40} = \frac{1}{20} \end{aligned}$

The answer is there is a one out of 20 chance that a student would like all three.
::答案是,每20个机会中就有1个一个学生会喜欢这3个。

You can look at this as a percent too to get an idea of how small the likelihood is.
::你可以把这个也看成是百分之一来了解可能性有多小

$\begin{aligned} \frac{1}{20} = 5 % \end{aligned}$

Example 3
::例3

If 6 out of 24 students enjoy opera in a class, what is the probability that a student selected at random from the class will enjoy the opera?
::如果每24名学生中有6人享受一班的歌剧,从班上随机挑选的学生享受歌剧的可能性有多大?

First, express the probability as a fraction, 6 out of 24, and then simplify.
::首先,将概率表示为分数,24分之6,然后简化。

$\begin{aligned} \frac{6}{24} = \frac{1}{4} \end{aligned}$

Next, divide 1 by 4 to get a percent.
::接下来,一比四分一获得百分之一。

$\begin{aligned} 25 % \end{aligned}$

The answer is $\begin{aligned} 25 % \end{aligned}$ .
::答案是25%。

Example 4
::例4

If 4 out of 12 students like Math class the best, what is the probability that a student selected at random from the class will enjoy Math the best?
::如果在12名数学类学生中,有4名是最好的,那么从该班随机挑选的学生享受数学最佳的可能性有多大?

First, express the probability as a fraction, 4 out of 12, and then simplify.
::首先,将概率作为一个分数表示,12分之4,然后简化。

$\begin{aligned} \frac{4}{12} = \frac{1}{3} \end{aligned}$

Next, divide 1 by 3 to get a percent.
::接下来,一比三分成一比三获得百分之一

$\begin{aligned} 33.33 % \end{aligned}$

The answer is $\begin{aligned} 33.33 % \end{aligned}$ .
::答案是33.33%

Example 5
::例5

If 5 out of 10 like Drama best, what is this probability?
::如果每10个中就有5个像戏剧最棒,那么这种概率是多少?

First, express the probability as a fraction, 5 out of 10, and then simplify.
::首先,将概率表示为分数,10分之5,然后简化。

$\begin{aligned} \frac{5}{10} = \frac{1}{2} \end{aligned}$

Next, divide 1 by 2 to get a percent.
::接下来,一比二分成一比二获得百分之一

$\begin{aligned} 50 % \end{aligned}$

The answer is $\begin{aligned} 50 % \end{aligned}$ .
::答案是50%。

Review
::回顾

Read the following example and then answer the questions about independent events.
::阅读下面的例子,然后回答有关独立事件的问题。

A bag has sixteen marbles in it. There are four of each color. There are four red, four blue, four green and four yellow.
::袋子里有16颗大理石,每颗颜色有4颗,4颗红色,4颗蓝色,4颗绿色,4颗黄色。

What is the probability of pulling a green marble out of the bag?
::把绿色大理石从袋子里拉出来的可能性有多大?

What is the probability of pulling a yellow or a green marble out of the bag?
::从袋子里拉出黄色或绿色大理石的概率是多少?

What is the probability of pulling a red marble out of the bag?
::把红大理石从袋子里拉出来的可能性有多大?

What is the probability of pulling a red, yellow or green marble out of the bag?
::将红色、黄色或绿色大理石从袋子里拉出来的可能性有多大?

What is the probability of not pulling a red marble out of the bag?
::不把红大理石从袋子里拉出来的可能性有多大?

What is the probability of pulling a green marble followed by a red marble if you put the green marble back before you pull the second marble?
::如果你把绿色大理石放回后再拉第二颗大理石,那么拉绿色大理石然后拉红色大理石的概率是多少?

What is the probability of pulling a yellow or blue marble out of the bag?
::从袋子里拉出黄色或蓝色大理石的概率是多少?

What is the probability of pulling a red marble followed by a blue marble if you put the red marble back before you pull the second marble?
::如果你把红大理石放回去,再拉第二大理石,那么拉红色大理石然后拉蓝色大理石的概率是多少?

If a red and a blue marble are removed from the bag, what is the probability of pulling a green marble out of the bag?
::如果从袋子中取出红色和蓝色大理石,从袋子中拉出绿色大理石的可能性有多大?

If the red and blue marbles from the previous question are removed from the bag, what is the probability that you will pull a red marble out of the bag?
::如果把上一个问题的红色和蓝色弹珠从袋子里移走,你从袋子里拉出红色弹珠的概率是多少?

For questions 10-20, look at the probability fractions that you wrote for questions 1-10. Convert each one into a percent. Round your answers to the nearest hundredths place if needed.
::对于问题 10-20, 请查看您为问题 1- 10 撰写的概率分数 1- 10 。将每个人转换为百分之一。如果需要, 请将您的答复转至最接近的百分百位。

Section outline

Calculating the Probability of Independent Events ::计算独立活动的概率

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

Calculating the Probability of Independent Events
::计算独立活动的概率

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