Course: 11.2 计算结果 | The Way To Learn

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计算结果
Terry is building his own mountain bike at the bike shop. He has a number of choices to make. First he must choose the color: red, green, blue or purple. Next, he must choose either the normal or extra cushion seat. Then, he must choose his handlebar style, either straight or curved. Terry takes out a piece of paper and tries drawing a picture of the bike with all of his options. How many options does he have?
::泰瑞在自行车店建造自己的山地自行车。他有很多选择。首先, 他必须选择颜色 : 红色、绿色、蓝色或紫色。接下来, 他必须选择正常的或额外的坐椅。然后, 他必须选择自己的把手栏风格, 无论是直的还是弯的。泰瑞拿出一张纸, 尝试用所有选项绘制自行车的图片。他有多少选项 ?

In this concept, you will learn to calculate outcomes with and without the Counting Principle .
::在这个概念中,你将学会计算结果与和不计算原则。

Probability
::概率

Probability is a mathematical way of calculating how likely an event is to occur. An event is a result of an experiment or activity that might include such things as flipping a coin, rolling a die or picking a card from a deck. The number of favorable outcomes is the number of choices (such as rolling a two on a die) and the probability is found by comparing this to the total number of outcomes (six in the case of a die).
::概率是计算事件发生可能性的一种数学方法。事件是实验或活动的结果,其中可能包括抛硬币、滚死或从甲板上选一张牌。有利结果的数量是选择的数量(比如在死亡时滚动一个两个),而概率是通过将它与结果总数(在死亡情况下为六个)进行比较而发现的。

Arithmetic can be used to calculate outcomes.
::测量可被用于计算结果。

Let’s look at an example.
::让我们举个例子。

Ichiro’s Car Wash offers three different wash services –basic outer body wash, interior cleaning, and finally, custom hand-detailing. Customers order a wash choice plus wax or no wax. You can use a tree diagram to find that there are 6 different choices, or outcomes, for a car wash.
::一郎的车牌洗涤提供了三种不同的洗涤服务 — — 基本外部洗涤、内部清洁,最后是定制的手工细节。客户要求洗涤选择加上蜡或没有蜡。你可以用树图来发现洗车有6种不同的选择或结果。

Look at the situation in terms of outcomes.
::从结果的角度来审视情况。

For the first choice there are 3 different outcomes.
::第一种选择有三种不同的结果。

For the second choice there are 2 different outcomes.
::第二种选择有两种不同的结果。

The total number of outcomes is $\begin{align*}3 \times 2 = 6\end{align*}$ .
::结果总数为3x2=6。

What if you changed the options? Would this method still work? Let’s add on to the car wash problem and see.
::换个选项会怎么样? 这个方法还能有效吗? 让我们来补充一下洗车问题,看看吧。

Consider Ichiro’s oil change choices. Customers can get standard or synthetic oil, filter or no filter, and pay with a coupon or pay the regular price. For each of the three choices there are 2 different outcomes.
::考虑一郎的石油变化选择。客户可以得到标准或合成石油、过滤器或没有过滤器,也可以用优惠券支付或支付正常价格。对于这三种选择中的每一种,都有两种不同的结果。

The total number of outcomes is $\begin{align*}2 \times 2 \times 2 = 8\end{align*}$ .
::结果总数为 2x2x2=8。

This method of calculating the number of total outcomes can be stated as a general rule called the Counting Principle . The Counting Principle states that the number of choices or outcomes for two independent events, $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ taken together, is the product of the total number of outcomes for each event.
::这种计算总结果数的方法可以称为一项通则,称为 " 计数原则 " 。 " 计数原则 " 规定,两项独立活动,即A和B的总和,的选择或结果数是每项活动总结果数的产物。

$\begin{aligned} Total outcomes for A and B = (number of outcomes for A) \times (number of outcomes for B) \end{aligned}$
$\begin{align*}\text{Total outcomes for } A \text{ and } B = (\text{number of outcomes for } A) \times (\text{number of outcomes for } B)\end{align*}$
::A和B的总结果合计=(A结果数目)x(B结果数目)

Once again, the Counting Principle requires that you take the number of choices or outcomes for two and multiply them together. The product of these outcomes will give you the total number of outcomes for each event.
::计算原则再次要求您选择或结果的数量或结果的数量, 以两种方式计算, 并同时乘以这些结果。这些结果的产物将为您提供每个事件的结果的总数。

You can use the Counting Principle to find probabilities of events. The probability of any event is equal to the ratio of favorable outcomes to the total number of equally likely possible outcomes.
::您可以使用计算原则来寻找事件的概率。任何事件的概率都等于有利结果与同样可能的结果总数的比例。

$\begin{aligned} P (event) = \frac{# of favorable outcomes}{total outcomes} \end{aligned}$
$\begin{align*}P(\text{event}) = \frac{\# \text{ of favorable outcomes}}{\text{total outcomes}}\end{align*}$
::P(活动)++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Favorable outcomes are the outcomes you are looking for.
::喜好的结果是你们正在寻找的结果。

Let’s look at an example.
::让我们举个例子。

Suppose you wanted to know the probability of rolling two number cubes and coming up with a sum of 5.
::假设你想知道滚动两个数字立方体的概率并得出5总和的概率

First, find the total number of outcomes.
::首先,找到结果的总数。

To find the number of total outcomes for the two tosses, you can use the Counting Principle. Since each toss of a number cube has 6 different outcomes (1, 2, 3, 4, 5, and 6):
::要找到两个组合的总结果数, 您可以使用“ 计数原则 ” 。因为数字立方体的每个投出有 6 种不同的结果(1、 2、 3、 4、 5、 6) :

$\begin{aligned} \begin{array}{rcl} Total outcomes & = & 6 \times 6 \\ = & 36 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \text{Total outcomes} &=& 6 \times 6\\ &=& 36 \end{array}\end{align*}$
::总结果=6x6=36

Next, find the number of favorable outcomes. In this case the favorable outcomes are outcomes that have a sum of 5.
::下一步, 找到有利结果的数量。在此情况下, 有利结果是总和为 5 的结果。

To figure out this particular problem you must add another step. Now list those 36 outcomes and mark the outcomes that result in a sum of 5.
::要解决这个特殊的问题, 您必须添加另一步。现在列出这36个结果, 并标出导致5个结果的总和。

From the table above, you can see that there are 4 possible ways that you can roll two die and get a sum of 5.
::从上表可以看出,有4种可能的方法, 你可以把2人卷死,然后得到5人的总和。

Then, use the $\begin{align*}P (\text{event})\end{align*}$ formula to find the probability of rolling a sum of 5.
::然后,使用 P( 活动) 公式来找到滚动总和为 5 的概率。

$\begin{aligned} \begin{array}{rcl} P (5) & = & \frac{4}{36} \\ = & \frac{1}{9} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(5) &=& \frac{4}{36}\\ \\ &=& \frac{1}{9} \end{array}\end{align*}$
::P(5)=436=19

The answer is $\begin{align*}\frac{1}{9}\end{align*}$ .
::答案是19个

The probability of rolling two number cubes and getting a sum of 5 is $\begin{align*}\frac{1}{9}\end{align*}$ .
::滚动两个数字立方体和获得5个总和的概率是19个。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about Terry building a bike. Terry is building a bike and can choose from 4 colors (red, green, blue, and purple), 2 seat choices (normal and extra cushion), and 2 types of handle bars (straight and curved).
::早些时候,你曾遇到过Terry制造自行车的问题。 Terry正在建造自行车,可以选择4种颜色(红色、绿色、蓝色和紫色)、2个座位选择(正常和额外的垫子)和2种手柄(直线和弯曲)。

First, notice how many of each option Terry has to choose from.
::首先,注意每个选项中多少特瑞必须选择。

Colors: 4
::颜色: 4

Seats: 2
::座位:2个

Handle Bars: 2
::处理栏: 2

Next, use the Counting Principle to find the number of possible outcomes he has.
::其次,利用计算原则找出他可能取得的结果的数目。

$\begin{aligned} \begin{array}{rcl} # outcomes & = & 4 \times 2 \times 2 \\ = & 16 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \# \text{ outcomes} &=& 4 \times 2 \times 2\\ &=& 16 \end{array}\end{align*}$
::# 结果=4x2x2=16

The answer is 16.
::答案是16岁

Terry has 16 different options for the bike he is building
::泰瑞有16种不同的选择他正在建造的自行车

Example 2
::例2

A spinner has colors red, blue, green, and yellow. What is the probability of spinning the spinner twice and having it land on the same color both times?
::旋转器有红色、蓝色、绿色和黄色的颜色。旋转器两次旋转和两次以同一颜色降落的概率是多少?

First, rather than draw a tree diagram, use the Counting Principle to find the number of total outcomes. Since each spin has 4 outcomes:
::首先,不是绘制树图,而是使用计数原则寻找总结果的数目。

$\begin{aligned} \begin{array}{rcl} Total outcomes & = & 4 \times 4 \\ = & 16 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \text{Total outcomes} &=& 4 \times 4\\ &=& 16 \end{array}\end{align*}$
::总结果=4x4=16

Next, find the number of ways both spinners can land on the same color both times:
::接下来,找到两个旋翼者可以在同一颜色降落的方法数量, 两次 :

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。

$\begin{aligned} \begin{array}{rcl} P (same color) & = & \frac{4}{16} \\ = & \frac{1}{4} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{same color}) &=& \frac{4}{16}\\ \\ &=& \frac{1}{4} \end{array}\end{align*}$
::P( 相同颜色) = 416= 14

The answer is $\begin{align*}\frac{1}{4}\end{align*}$ .
::答案是14岁

The probability of spinning a spinner with four colors twice and getting the same color both times is $\begin{align*}\frac{1}{4}\end{align*}$ .
::旋转一个旋转器的概率是四色两次同一颜色两次的概率是14

Example 3
::例3

Jake has two shirts, four pairs of pants and three sweaters. How many different outfits can he make?
::Jake有两件衬衫,四条裤子和三件毛衣他能做几件不同的衣服?

First, notice how many of each type of clothing Jake has.
::首先,注意杰克有几件每件衣服

Shirts: 2
::衬衫:2件

Pants: 4
::裤子:4

Sweaters: 3
::喷雾者:3人

Next, use the Counting Principle to find the number of possible outfits.
::接下来,用计数原则找出可能穿几件衣服

$\begin{aligned} \begin{array}{rcl} # outfits & = & 2 \times 4 \times 3 \\ = & 24 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \# \text{ outfits} &=& 2 \times 4 \times 3\\ &=& 24 \end{array}\end{align*}$
::# 服装=2x4x3=24

The answer is 24.
::答案是24个。

Jake has 24 options of outfits to wear.
::Jake有24种衣服可以穿

Example 4
::例4

What if Jake has two shirts, five pairs of pants and four sweaters? How many outfits can he make?
::如果杰克有两件衬衫,五条裤子和四件毛衣呢?

First, notice how many of each type of clothing Jake has.
::首先,注意杰克有几件每件衣服

Shirts: 2
::衬衫:2件

Pants: 5
::裤子:5

Sweaters: 4
::喷雾者:4人

Next, use the Counting Principle to find the number of possible outfits.
::接下来,用计数原则找出可能穿几件衣服

$\begin{aligned} \begin{array}{rcl} # outfits & = & 2 \times 5 \times 4 \\ = & 40 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \# \text{ outfits} &=& 2 \times 5 \times 4\\ &=& 40 \end{array}\end{align*}$
::# 服装=2x5x4=40

The answer is 40.
::答案是40岁

Jake has 40 options of outfits to wear.
::Jake有40种衣服可以穿

Example 5
::例5

What if Jake only chooses one sweater? How many outfits can he create with four pairs of pants and two shirts?
::如果杰克只挑一件毛衣呢?

First, notice how many of each type of clothing Jake has.
::首先,注意杰克有几件每件衣服

Shirts: 2
::衬衫:2件

Pants: 4
::裤子:4

Sweaters: 1
::喷雾者:1

Next, use the Counting Principle to find the number of possible outfits.
::接下来,用计数原则找出可能穿几件衣服

$\begin{aligned} \begin{array}{rcl} # outfits & = & 2 \times 4 \times 1 \\ = & 8 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} \# \text{ outfits} &=& 2 \times 4 \times 1\\ &=& 8 \end{array}\end{align*}$
::# 服装= 2x4x1=8

The answer is 8.
::答案是8岁

Jake has 8 options of outfits to wear.
::Jake有8种衣服可以穿

Review
::回顾

Use the counting principle to determine the number of outcomes.
::使用计数原则来确定结果的数量。

1. Nigel dropped 3 open-faced peanut butter sandwiches that were equally likely to land face-up or face-down. How many different outcomes are there?
::1. 奈杰尔掉了3个露天花生酱三明治,这些三明治同样有可能面朝上或面朝下落地。

2. Movie Star Toothpaste comes in 3 different flavors: sparkle, blast, and stripe, 3 different sizes, and 2 different tube styles. How many toothpaste choices are there?
::2. 电影《星牙膏》有三种不同的口味:火花、爆炸和条纹、3个不同的尺寸和2个不同的管式。

3. Dave made 3 predictions for this Sunday’s football games. How many different outcomes of being right or wrong are there?
::3. Dave对这个星期天的足球赛作了三次预测:对错的结果有多少不同?

4. Bridget made 4 predictions for this Sunday’s football games. How many different outcomes of being right or wrong are there?
::4. Bridget对这个星期天的足球赛作了4次预测:对错的结果有多少不同?

5. How many more outcomes will there be if Bridget adds an extra prediction for problem 4 above?
::5. 如果Bridget增加上文问题4的额外预测,结果还会有多少?

6. Sandy’s Sandals come in 4 different models: sport, super-sport, casual, and chic, 5 different colors, and 9 different sizes. How many choices are there?
::6. 桑迪的桑达尔山地有四种不同的模式:运动、超级运动、随意和奇奇、5种不同的颜色和9种不同的大小。有多少选择?

7. Mike’s Bikes features three different bike styles: mountain, racer, and stunt bike. You can choose from 6 different gear systems, 4 frame alloys, and 5 colors. How many bike choices are there?
::7. 迈克的自行车有三种不同的自行车风格:山车、赛车和特技自行车。你可以选择6种不同的齿轮系统、4个框架合金和5种颜色。有多少自行车选择?

8. Tilly spins a spinner that has red, blue, and yellow sections 3 times. How many different outcomes are possible?
::8. Tilly旋转一个红色、蓝色和黄色部分3倍的旋转器,可以产生多少不同的结果?

9. Tilly spins a spinner that has red, blue, and yellow sections and tosses a number cube. How many different outcomes are possible?
::9. Tilly旋转一个有红色、蓝色和黄色部分的旋转器,并抛出一个数字立方体。

10. Tilly tosses a number cube 2 times. How many different outcomes are possible?
::10. 将一个数字立方体扔出两次,可以产生多少不同的结果?

11. Tilly tosses a number cube 3 times. How many different outcomes are possible?
::11. 将一个数字立方体扔出三次,可以产生多少不同的结果?

Use the counting principle to determine probabilities for a spinner with four colors: red, yellow, blue and green.
::使用计数原则来确定四色(红色、黄色、蓝色和绿色)的脊柱的概率。

12. Kalyani spins the spinner 2 times. What is the probability that the arrow will land on green both times?
::12. Kalyani将旋转器旋转两次,箭头两次降落绿色的可能性有多大?

13. Kalyani spins the spinner 2 times. What is the probability that the arrow will land on the same color both times?
::13. Kalyani两次旋转旋转旋转器,箭头两次降落在同一颜色上的可能性有多大?

14. Kalyani spins the spinner 2 times. What is the probability that the arrow will land on a different color each time?
::14. Kalyani两次旋转旋转旋转器,箭头每次落下不同颜色的概率有多大?

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。

Resources
::资源

Section outline

General

计算结果

Probability ::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

Probability
::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