Section: 计算概率的规则 | 12.11 计算概率的计算规则

Section outline

Rebecca loves green Skittles more than all the other colors: red, yellow, orange, and purple. She wonders if she places a Skittle of each color in a bowl, five Skittles total, and pulls one Skittle, replaces it, then pulls one again, what are her chances of pulling a green Skittle each time.
::Rebecca比其他颜色更喜欢绿色小木板:红色、黄色、橙色和紫色。她想知道她是否把每件颜色的小木板放在碗里,总共五个小木板,拉一个小木板,换一个,再拉一个,她每次拉一个绿色小木板的机会有多大。

In this concept, you will learn how to calculate the probability of two separate events producing favorable outcomes .
::在这个概念中,你将学会如何计算产生有利结果的两个单独活动的概率。

Using the Counting Principle to Calculate Probabilities
::使用计算原则计算概率

You can use the Counting Principle to find the number of total outcomes for two separate events, for example, two spins with two spinners. Considering the spinners below, what is the probability that you spin each spinner and they both land on the same color?
::您可以使用计数原则来找到两个不同事件的总结果数, 比如, 两个旋转和两个旋转器。考虑到下面的旋转器, 您将每个旋转器都旋转成同一个颜色的概率是多少 ?

There are four outcomes on one spinner and three outcomes on the other spinner.
::一个脊柱上有四个结果,另一个脊柱上有三个结果。

$\begin{aligned} Total outcomes & = 4 outcomes \cdot 3 outcomes \\ = 12 outcomes \end{aligned}$

::结果=4成果=4成果=3成果=12成果

Now list those 12 outcomes and mark the outcomes that are the same color for both spins.
::现在列出这12个结果, 并标记两个旋转的颜色相同的结果。

$\begin{aligned} red-red & blue-red & yellow-red & green-red \\ red-blue & blue-blue & yellow-blue & green-blue \\ red-green & blue-green & yellow-green & green-green \end{aligned}$

::红-红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 蓝色- 蓝- 蓝- 蓝- 蓝- 蓝- 蓝- 蓝色- 绿色- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 蓝- 红- 红- 红- 红- 红- 红- 红- 红- 红- 红- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿- 绿-绿- 绿-绿-绿-绿-绿-绿-绿-绿-绿-绿-绿-绿-绿-绿-绿- 绿- 绿-绿-绿-绿- 绿-绿-绿- 绿-

Since there are 3 outcomes that have the same color for both spins:
::因为有三种结果对两个旋转都有相同的颜色:

$\begin{aligned} P (same) = \frac{3}{12} = \frac{1}{4} \end{aligned}$

::P(相同)=312=14

The probability of both spinners landing on the same color is $\begin{aligned} \frac{1}{4} \end{aligned}$ .
::两个旋翼者在同一颜色上着陆的概率为14。

Let's look at another example.
::让我们再举一个例子。

Anna flips a coin 3 times in a row. What is the probability that she will get heads all 3 times?
::Anna连续三次翻硬币。

First, find the number of total outcomes by multiplying the number of outcomes for each flip:
::首先,通过乘以每个翻转的结果数来查找总结果数 :

$\begin{aligned} Total outcomes & = 2 outcomes \cdot 2 outcomes \cdot 2 outcomes \\ = 8 outcomes \end{aligned}$

::结果=2 结果=8结果

Next, list all 8 outcomes and find the number of ways Anna can get heads all 3 times. There is only one arrangement in which all 3 flips result in heads.
::接下来,列出所有八项结果, 并找到安娜可以三次得到头的次数。只有一种安排, 即所有三个翻转都会导致头部。

$\begin{aligned} heads-heads-heads & tails-heads-heads \\ heads-heads-tails & tails-heads-tails \\ heads-tails-heads & tails-tails-heads \\ heads-tails-tails & tails-tails-tails \end{aligned}$

::头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、头尾、尾尾、尾尾尾、尾尾尾

Then, find the ratio of favorable outcomes to total outcomes:
::然后,找到有利结果与总结果的比例:

$\begin{aligned} P (head-head-head) = \frac{1}{8} \end{aligned}$

::P(头头头)=18

The answer is the probability of Anna flipping heads all three times is $\begin{aligned} \frac{1}{8} \end{aligned}$ .
::答案是安娜翻头三次的概率是18

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Rebecca's curiosity about pulling out her favorite color Skittle.
::之前有人给你一个问题关于Rebecca的好奇心关于拉出她最喜欢的彩色滑雪机

She will replace the Skittle after the first pull before pulling the second time. The colors in the bowl are red, orange, yellow, green, and purple. What is the probability that Anna will pull the green Skittle both times?
::在第二次拉之前,她将换掉《滑雪车》。碗中的颜色是红色、橙色、黄色、绿色和紫色。安娜两次拉绿色滑雪车的概率是多少?

First, find the number of total outcomes by multiplying the number of outcomes for each pull:
$\begin{aligned} Total outcomes & = 5 outcomes \cdot 5 outcomes \\ = 25 outcomes \end{aligned}$
Next, list all 25 outcomes and find the number of ways Rebecca can pull the green Skittle both times. There is only one arrangement in which boht pulls result in green Skittles.
::首先,通过乘以每次拉动的结果数量来找到总结果的数量: 总计结果=5 结果=5 结果=5 结果=25 结果=25 下一个, 列出所有25个结果并找到Rebecca两次拉动绿色滑板的方式的数量。只有一种安排, boht拉动会导致绿色滑板。

red-red red-orange red-yellow red-green
::红红色红-红红红-红红红-红红-红红-红-红-红-黄-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-绿-绿-绿-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红-红

red-purple
::红色

orange-orange orange-red orange-yellow orange-green
::橙橙红色橙橙橙橙綠

orange-purple
::橙黄色

yellow-yellow yellow-orange yellow-red yellow-green
::黄黄-黄黄-黄黄-红黄-红黄-黄绿色

yellow-purple
::黄黄色

green-green green-yellow green-orange green-red
::绿绿绿绿绿黄绿绿红

green-purple
::绿色净绿色

purple-purple purple-green purple-yellow purple-orange
::紫-紫-紫-紫-紫-绿色紫-紫-黄色紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫-紫

purple-red
::紫红色

Then, find the ratio of favorable outcomes to total outcomes:
::然后,找到有利结果与总结果的比例:

$\begin{aligned} P (green-green) = \frac{1}{25} \end{aligned}$

::P(绿色绿色)=125

The answer is the probability of pulling a green Skittle both pulls is $\begin{aligned} \frac{1}{25} \end{aligned}$ .
::答案是绿色滑雪机双向拉动的概率是125

Example 2
::例2

If Sam flips a coin five times. What is the probability of having it come up heads all at the same time?
::如果萨姆翻了5次硬币,它同时浮出头来的可能性有多大?

First, figure out the total number of outcomes:
::首先,找出成果总数:

There can be two possible outcomes each time Sam flips the coin.
::每次Sam翻硬币时可能会有两种结果

Total outcomes = 2 outcomes x 2 outcomes x 2 outcomes x 2 outcomes x 2 outcomes = 32 outcomes
::总结果=2项成果x2项成果x2项成果x2项成果x2项成果x2项成果x2项成果x2项成果x2项成果x2项成果=32项成果

There are 32 possible outcomes.
::有32项可能的结果。

Next, list all of the outcomes:
::下面列出所有结果:

H-H-H-H-H    H-H-H-H-T H-H-H-T-H H-H-H-T-T H-H-T-H-H
::H -H -H -H -H -H -H

H-H-T-H-T H-H-T-T-H H-H-T-T-T H-T-H-H-H H-T-H-H-T
::H -H -T -H -T -T -H H -H -T

H-T-H-T-H H-T-H-T-T H-T-T-H-H H-T-T-H-T H-T-T-T-H
::H -T -T -T -T -T -T

H-T-T-T-T
::H-T-T-T-T-T

T-H-H-H-H T-H-H-H-T T-H-H-T-H T-H-H-T-T T-H-T-H-H
::-H -H -H -H -H -H

T-H-T-H-T T-H-T-T-H T-H-T-T-T T-T-H-H-H T-T-H-H-T
::T -H -T -T -T -T -T -T -T

T-T-H-T-H T-T-H-T-T    T-T-T-H-H T-T-T-H-T T-T-T-T-H
::T -T -T -T -T -T -T -T -T

T-T-T-T-T
::T -T -T -T

Then, determine how many times all heads appears and create a ratio of favorable outcomes to total outcomes:
::然后,决定所有头部出现多少倍, 并创造出一个有利结果与总结果的比例:

H-H-H-H-H only appears one time out of 32 total outcomes = $\begin{aligned} \frac{1}{32} \end{aligned}$ .
::H-H-H-H-H-H-H在32项总结果中只出现一次=132。

The answer is the probability of Sam flipping all heads over five flips is $\begin{aligned} \frac{1}{32} \end{aligned}$ .
::答案是萨姆用五个翻五转翻五翻五翻五翻五翻五翻一翻的概率是132

Example 3
::例3

Abra flips a coin 2 times. What is the probability that both flips will match?
::Abra翻了2次硬币。两种硬币的几率有多大?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:
::首先,通过乘以每个翻转的结果数来查找总结果数 :

$\begin{aligned} Total outcomes & = 2 outcomes \cdot 2 outcomes \\ = 4 outcomes \end{aligned}$
Next, list all 4 outcomes and find the number of times Abra can get matching flips . There are two outcomes in which the flips will match:
$\begin{aligned} heads-heads & tails-heads \\ heads-tails & tails-tails \end{aligned}$
Then, find the ratio of favorable outcomes to total outcomes: $\begin{aligned} P (heads-heads or tails-tails) = \frac{2}{4} = \frac{1}{2} \end{aligned}$
::结果总数=2 结果=4 结果=4 成果Next,列出所有4个结果并找到Abra可以得到匹配翻转的次数。在两种结果中,翻转将匹配:头尾-头尾-头尾-头尾-尾尾尾-尾尾尾尾-尾尾尾,然后找到有利结果与总结果的比率:P(头尾或尾尾尾)=24=12

The answer is the probability of Abra having matching flips is $\begin{aligned} \frac{1}{2} \end{aligned}$ .
::答案是 Abra 的相匹配翻转概率是 12 。

Example 4
::例4

Abra flips a coin 2 times. What is the probability that both flips will NOT match?
::Abra翻了2次硬币。两种硬币的几率有多大?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:
$\begin{aligned} Total outcomes & = 2 outcomes \cdot 2 outcomes \\ = 4 outcomes \end{aligned}$
Next, list all 4 outcomes and find the number of times the outcomes will not result in matching flips . There are two outcomes in which the flips will not match:
$\begin{aligned} heads-heads & tails-heads \\ heads-tails & tails-tails \end{aligned}$
Then, find the ratio of favorable outcomes to total outcomes: $\begin{aligned} P (heads-tails or tails-heads) = \frac{2}{4} = \frac{1}{2} \end{aligned}$
::首先,通过将每个翻转的结果数乘以倍数来查找总结果的数量: 总计结果=2 结果=2 结果=4 结果Next, 列出所有四个结果并找到结果不会导致匹配翻转的次数。有两种结果, 翻转不会匹配: 头尾- 头尾- 头尾- 头尾- 尾尾尾- 尾尾尾尾尾尾- 尾尾尾尾尾- 尾尾尾尾巴- 尾尾巴- 尾巴- 尾巴- 尾巴- 之后, 找到有利结果与总结果的比例 : P( 头尾尾或尾尾尾头) =24=12

The answer is the probability of Abra not having matching flips is $\begin{aligned} \frac{1}{2} \end{aligned}$ .
::答案是 Abra 没有匹配翻转的概率是 12 。

Example 5
::例5

Abra flips a coin 3 times. What is the probability that all 3 flips will match?
::Abra翻了3次硬币。所有3个硬币的概率有多大?

First, find the number of total outcomes by multiplying the number of outcomes for each flip:
::首先,通过乘以每个翻转的结果数来查找总结果数 :

$\begin{aligned} Total outcomes & = 2 outcomes \cdot 2 outcomes \cdot 2 outcomes \\ = 8 outcomes \end{aligned}$
Next, list all 8 outcomes and find the number of ways Abra can get matching flips all 3 times. There are two arrangements in which all 3 flips result in a match.
::总结果=2 结果=2 结果=2 结果=8 结果=8 下一个,列出所有8个结果并找到Abra可以三次匹配翻转的次数。有两种安排,所有3个翻转都导致匹配。

$\begin{aligned} heads-heads-heads & tails-heads-heads \\ heads-heads-tails & tails-heads-tails \\ heads-tails-heads & tails-tails-heads \\ heads-tails-tails & tails-tails-tails \end{aligned}$
Then, find the ratio of favorable outcomes to total outcomes:
::然后,找到有利结果与总结果的比率:

$\begin{aligned} P (heads-heads-heads or tails-tails-tails) = \frac{2}{8} = \frac{1}{4} \end{aligned}$

::P(头头头或尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾尾)=28=14

The answer is the probability of Anna flipping heads all three times is $\begin{aligned} \frac{1}{4} \end{aligned}$ .
::答案是安娜翻头三次的概率是14

Review
::回顾

Find probabilities using the Counting Principle.
::使用计算原则查找概率。

Abra flips a coin 3 times. What is the probability that all 3 flips will NOT match?
::Abra翻了3次硬币。所有3个硬币的概率有多大?

Abra flips a coin 3 times. What is the probability that heads will come up exactly 2 times?
::Abra翻了3次硬币,头的概率是2倍?

Billy spins the spinner twice. What is the probability that blue will come up both times?
::Billy翻了两次旋转器蓝色的几率有多大?

Billy spins the spinner twice. What is the probability that blue will come up at least one time?
::Billy把旋转器翻了两次蓝色至少一次的概率是多少?

Billy spins the spinner twice. What is the probability that blue will come up exactly one time?
::Billy翻了两次旋转器蓝色的概率是多少?

Billy spins the spinner twice. What is the probability that both spins will match?
::Billy两次旋转旋转旋转器。两种旋转的概率有多大?

Cindy tosses a number cube two times. What is the probability that both tosses will match?
::Cindy两次抛出一个数字立方体。两个抛出的概率有多大?

Cindy tosses a number cube two times. What is the probability that both tosses will NOT match?
::Cindy 两次抛出一个数字立方体。两个抛出的概率有多大 ?

Cindy tosses a number cube two times. What is the probability that the sum of the two tosses will be 7?
::Cindy两次抛出一个数字立方体。两次抛出的总和是7的概率是多少?

Tyler rolls a number cube three times. What is the probability that he will roll a two?
::泰勒将一个数字立方体滚动了三次。他将滚动一个2的概率是多少?

Tyler rolls a number cube three times. What is the probability that he will roll an odd number?
::泰勒将一个数字立方体滚动了三次。他会滚动一个奇数的概率是多少?

Tyler rolls a number cube three times. What is the probability that he will roll an even number?
::泰勒将一个数字立方体滚动了三次。他会卷动一个偶数的概率是多少?

If Tyler rolls a number cube once, what is the probability of rolling a two or a five?
::如果泰勒一次滚动一个数字立方体, 滚动两个或五个的概率是多少?

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。

Resources
::资源

Section outline

Using the Counting Principle to Calculate Probabilities ::使用计算原则计算概率

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

Review (Answers) ::回顾(答复)

Resources ::资源

Using the Counting Principle to Calculate Probabilities
::使用计算原则计算概率

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

Review (Answers)
::回顾(答复)

Resources
::资源