Section: 概率分布 | 7.3 概率分布

Section outline

Assume you have an unfair coin that is weighted to land on heads 65% of the time. If you flip that coin 3 times and let $\begin{align*}T\end{align*}$ represent the number of tails you get, what is the probability distribution for $\begin{align*}T\end{align*}$ ?
::假设你有一个不公平的硬币, 它在65%的时间里被加权在头顶上。如果你翻了3次硬币,让T代表尾巴的数量,那么T的概率分布是多少?

Look to the end of the lesson for the answer.
::寻找教训的结尾以找到答案。

Probability Distribution
::概率分布

A probability distribution is a list of each value a random variable can attain, along with the probability of attaining each value. In other words, the probability distribution of an event is sort of a map of how each possible outcome relates to the chance it will happen.
::概率分布是随机变量可以达到的每个值的列表,以及达到每个值的概率。换句话说,事件概率分布是每个可能的结果与它发生的机会之间关联的地图。

For instance, the probability distribution of flipping a coin twice is:
::例如,翻硬币两次的概率分布是:

heads, heads = 25%, heads, tails = 25%, tails, heads = 25%, and tails, tails = 25%.
::头部=25%,头部=25%,尾部=25%,尾部=25%,尾部=25%,尾部=25%。

If we define the random variable $\begin{align*}X\end{align*}$ to be the number of heads you get when you flip a coin twice, we could create the following probability distribution table for $\begin{align*}X\end{align*}$ :
::如果我们将随机变量 X 定义为您翻硬币两次时获得的头数, 我们可以为 X 创建以下概率分布表 :

$\begin{align}X\end{align}$	$\begin{align}0\end{align}$	$\begin{align}1\end{align}$	$\begin{align}2\end{align}$
$\begin{align}P(X)\end{align}$	$\begin{align}\frac{1}{4}\end{align}$	$\begin{align}\frac{1}{2}\end{align}$	$\begin{align}\frac{1}{4}\end{align}$

There are various ways of visualizing a probability distribution, and we will review that concept in another lesson. For now, we focus on identifying what a probability distribution is, and how to calculate it for a particular event.
::概率分布有多种可视化的方法,我们将在另一个教训中审视这一概念。目前,我们的重点是确定概率分布是什么,以及如何计算特定事件的概率分布。

Creating Probability Distributions
::创建概率分布

1. In Chi’s class, 4 students have one parent, 7 have two parents, and 1 student lives with his uncle. Let $\begin{align*}P\end{align*}$ be the number of parents of a randomly selected student from the class. Create a probability distribution for $\begin{align*}P\end{align*}$ .
::1. 在Chi班中,有4名学生有一个父母,7名学生有两个父母,1名学生与其叔叔住在一起。

Set random variable $\begin{align*}P\end{align*}$ to be the number of parents:
::设定随机变量P为父母数 :

$\begin{align*}P(P)=\% \ probability \ that \ a \ student \ has \ P \ parents\end{align*}$
::P(P) 学生父母为P(P) 的概率

Now find the probability of each $\begin{align*}P\end{align*}$ , noting that there are 12 students total:
::现在找到每个P的概率, 指出共有12名学生:

$\begin{align*}\text{1 student has 0 parents: } & P(0)=\frac{1}{12} \ or \ 8.3 \overline{3} \% \\ \text{4 students have 1 parent: } & P(1)=\frac{4}{12} \ or \ 33.3 \overline{3} \% \\ \text{7 students have 2 parents: } & P(2)=\frac{7}{12} \ or \ 58.3 \overline{3} \% \\\end{align*}$
::1名学生父母为0:P(0)=112或8.3:3% 4名学生父母为1:P(1)=412或33.3:3% 7名学生父母为2:P(2)=712或58.3:3%

2. Roll two fair six-sided dice. Let $\begin{align*}D\end{align*}$ equal the sum of the dice. Create a probability distribution for $\begin{align*}D\end{align*}$ .
::2. 滚动两个公平的六面骰子。让 D 等于骰子的总和。为 D 创建概率分布。

Make a list of the individual probabilities of each of the 36 possible outcomes :
::列出36项可能结果中每项结果的个别概率:

$\begin{align*}& \text{1 possibility with a sum of 2: } && P(D= 2) = \frac{1}{36} = 0.0278\\ & \text{2 possibilities with a sum of 3: } && P(D= 3) = \frac{2}{36} = 0.0556\\ & \text{3 possibilities with a sum of 4: } && P(D= 4) = \frac{3}{36} = 0.0833\\ & \text{4 possibilities with a sum of 5: } && P(D= 5) = \frac{4}{36} = 0.1111\\ & \text{5 possibilities with a sum of 6: } && P(D= 6) = \frac{5}{36} = 0.1389\\ & \text{6 possibilities with a sum of 7: } && P(D= 7) = \frac{6}{36} = 0.1667\\ & \text{5 possibilities with a sum of 8: } && P(D= 8) = \frac{5}{36} = 0.1389\\ & \text{4 possibilities with a sum of 9: } && P(D= 9) = \frac{4}{36} = 0.1111\\ & \text{3 possibilities with a sum of 10: } && P(D= 10) = \frac{3}{36} = 0.0833\\ & \text{2 possibilities with a sum of 11: } && P(D= 11) = \frac{2}{36} = 0.0556\\ & \text{1 possibility with a sum of 12: } && P(D= 12) = \frac{1}{36} = 0.0278 \end{align*}$
::1种可能性,总和2P(D=2)=136=0.2782可能性,总和3:P(D=3)=236=0.05563可能性,总和3:P(D=4)=336=0.08334可能性,总和5:P(D=4)=336=0.08334可能性,总和5:P(D=5)=436=436=0.8334可能性,总和6:P(D=6)=536=5136=0.13896可能性,总和7:P(D=7)=636=0.16675可能性,总和8:P(D=8)=536=0.13894可能性,总和9:P(D=9)=436=0.11113可能性,总和:P(D=10)=336=0.08332可能性,总和11:P(D=11)=236=236=0.05561可能性,总和:P(D=12)=136=0.0278

Evaluating Probabilities
::评价概率

Janie wants to evaluate the probabilities of pulling various cards from a deck. She sets the discrete random variable $\begin{align*}C\end{align*}$ to be the number of diamonds she gets over the course of three trials, if each trial consists of pulling, recording, and replacing one random card from a standard deck. What is the probability distribution of $\begin{align*}C\end{align*}$ ?
::Janie想评估从甲板上拉出各种牌的概率。她设置了离散随机变量C, 即三次试金刚石的数量, 如果每次试金刚石包括从标准牌上拉、记录和替换一张随机卡。 C的概率分布是什么?

To evaluate the probability distribution of $\begin{align*}C\end{align*}$ , Janie needs to identify the probability of each of the possible values of $\begin{align*}C\end{align*}$ . Note that the chance she will pull a diamond is $\begin{align*}\frac{13}{52}\end{align*}$ or $\begin{align*}0.25\end{align*}$ , meaning that the chance she will not pull a diamond is $\begin{align*}1-0.25=0.75\end{align*}$ :

::为了评估C的概率分布,Janie需要确定C每种可能价值的概率。注意,她拉钻石的机会是1352或0.25,这意味着她不拉钻石的机会是1-0.25=0.75:

For $\begin{align*}C=(0),\end{align*}$ the total probability is: $\begin{align*}0.42\end{align*}$ or $\begin{align*}42\%\end{align*}$
- Other, other, other: $\begin{align*}0.75\times0.75\times0.75=0.42\end{align*}$
  ::其他,其他,其他: 0.75x0.75x0.75=0.42
::对于C=(0),总概率是:0.42或42% 其他,其他,其他:0.75×0.75×0.75=0.42。

For $\begin{align*}C=(1)\end{align*}$ , the total probability is: $\begin{align*}0.14+0.14+0.14=0.42 \ or \ 42 \%\end{align*}$ (see the three possible outcomes resulting in $\begin{align*}C=1\end{align*}$ below)
- Diamond, other, other : $\begin{align*}0.25 \times 0.75 \times 0.75=0.14\end{align*}$
  ::钻石,其他,其他:0.25×0.75×0.75=0.14
- Other, Diamond, other : $\begin{align*}0.75 \times 0.25 \times 0.75=.14\end{align*}$
  ::其他,钻石,其他:0.75x0.25x0.75=14。
- Other, other, Diamond : $\begin{align*}0.75 \times 0.75 \times 0.25=0.14\end{align*}$
  ::其他,其他,钻石:0.75×0.75×0.0.25=0.14
::C=(1)的总概率是:0.14+0.14+0.14+0.14=0.42或42%(见下文C=1的三种可能结果)。钻石,其他,其他:0.25×0.75×0.75=0.14。其他,钻石,其他:0.75×0.25×0.75=0.14。其他,其他,钻石:0.75×0.75×0.25=0.14。

For $\begin{align*}C=(2)\end{align*}$ , the total probability is: $\begin{align*}0.047+0.047+0.047=0.141 \ or \ 14.1 \%\end{align*}$
- Diamond, Diamond, other : $\begin{align*}0.25 \times 0.25 \times 0.75=0.047\end{align*}$
  ::钻石、钻石、其他:0.25×0.25×0.75=0.047
- Diamond, other, Diamond : $\begin{align*}0.25 \times 0.75 \times 0.25=0.047\end{align*}$
  ::钻石,其他,钻石:0.25×0.75×0.25=0.047
- Other, Diamond, Diamond : $\begin{align*}0.75 \times 0.25 \times 0.25=0.047\end{align*}$
  ::其他,钻石,钻石:0.75×0.25×0.25=0.047
::C=(2)的总概率是:0.047+0.047+0.047+0.047=0.141或14.1%钻石、钻石、其他:0.25×0.25×0.75=0.047钻石、其他:0.25×0.75×0.75=0.047钻石、其他:0.25×0.75×0.25=0.047其他、钻石、钻石:0.75×0.25×0.25=0.047

For $\begin{align*}C=(3)\end{align*}$ , the probability is : $\begin{align*}0.25 \times 0.25 \times 0.25=0.016 \ or \ 1.6 \%\end{align*}$
- Diamond, Diamond, Diamond: $\begin{align*}0.25 \times 0.25 \times 0.25=0.016\end{align*}$
  ::钻石、钻石、钻石:0.25×0.25×0.25=0.016
::C=(3)的概率为:0.25×0.25×0.25=0.016或1.6% 钻石、钻石、钻石:0.25×0.25×0.25=0.016

Earlier Problem Revisited
::重审先前的问题

If each throw has a 65% chance of heads, then it has a 35% chance of tails:
::如果每次抛球都有65%头的机率, 那么它有35%尾巴的机率:

For $\begin{align*}T=1\end{align*}$ , we could have THH, HTH, or HHT. Each of those has a $\begin{align*}.35 \times .65 \times .65=.15\end{align*}$ chance of occurring, so $\begin{align*}P(T=1)=.15 \times 3=.45 \ or \ 45 \%\end{align*}$
::T=1 1时,我们可以有THH、HTH或HHT。每一个有35x. 65x. 65=15的发生概率, 所以P( T=1)=15x3=45或45%

For $\begin{align*}T=2\end{align*}$ , we could have TTH, THT, or HTH. Each has a $\begin{align*}.35 \times .35 \times .65=.08\end{align*}$ chance, so $\begin{align*}P(T=2)=.08 \times 3=.24 \ or \ 24 \%\end{align*}$
::T=2 T=2, 我们可以拥有TTTT、TTTT或HTTH。每个人都有35x35x65=0. 08的机会, 所以P(T=2)=. 08x3=24 或24%

For $\begin{align*}T=3\end{align*}$ , we could have only TTT, with a chance of $\begin{align*}.35 \times .35 \times .35=.043 \ or \ 4.3 \%\end{align*}$
::T=3, T=3, 我们只能有TTT, 概率为35×35×35=0. 043或4.3%

Examples
::实例

Example 1
::例1

Create a probability distribution for number of heads when you flip a coin 3 times.
::当翻硬币三次时, 为头数创建概率分布。

Write out all the possibilities:
::写出所有的可能性:

$\begin{align*}& \text{TTT has 0 heads.} && \text{TTH has 1 heads.} && \text{THT has 1 heads.}\\ & \text{THH has 2 heads.} && \text{HTT has 1 heads.} && \text{HTH has 2 heads.}\\ & \text{HHT has 2 heads.} && \text{HHH has 3 heads.} \end{align*}$
::TTT有0个头。TT有1个头。TT有1个头。TH有2个头。HTT有1个头。HT有2个头。HHT有2个头。HHH有3个头。

$\begin{align*}&\text{So we have 1 possibility with 0 heads:} && P(0) = \frac{1}{8} = 0.125\\ &\text{3 possibilities with 1 heads:} && P(1) = \frac{3}{8} = 0.375\\ &\text{3 possibilities with 2 heads:} && P(2) = \frac{3}{8} = 0.375\\ &\text{1 possibility with 3 heads:} && P(3) = \frac{1}{8} = 0.125\end{align*}$
::因此,我们有一个可能:头0:P(0)=18=0.1253;头1:P(1)=38=0.3753;头2:P(2)=38=0.3751;头3:P(3)=18=0.125。

Example 2
::例2

Let $\begin{align*} C\end{align*}$ be the number of cholate chip cookies you get if you randomly pull and replace two cookies from a jar containing 6 chocolate chip, 4 peanut butter, 8 snickerdoodle, and 12 sugar cookies. Create a probability distribution for $\begin{align*} C\end{align*}$ .
::让C成为巧克力曲奇饼的数量, 如果你随机抽取和替换两个曲奇饼, 从一个罐子里含有 6 个巧克力曲奇饼, 4 个花生酱, 8 个饼干, 8 个饼干, 12个糖饼干。创建 C 的概率分布。

There are a total of 30 cookies, the probability of pulling a chocolate chip cookie is $\begin{align*}\frac{6}{30}=.20\end{align*}$ , so the probability of not pulling a chocolate chip is $\begin{align*}\frac{24}{30}=.80\end{align*}$
::总共有30个饼干,拉巧克力曲奇的概率是630=20, 所以不拉巧克力曲奇的概率是2430=80。

For $\begin{align*}C=0\end{align*}$ we have to pull a non-chocolate chip both times: $\begin{align*}.8 \times .8=.64 \ or \ 64 \%\end{align*}$
::对于 C=0, 我们必须两次拉出一个非巧克力芯片: 0. 8×.8=. 64 或 64%

For $\begin{align*}C=1\end{align*}$ we could either pull the chocolate chip cookie first or second, so we get $\begin{align*}(.2 \times .8)+(.8 \times .2)=.32 \ or \ 32 \% \end{align*}$
::对于 C=1, 我们可以首先或第二次拉巧克力曲奇饼干, 所以我们得到( 2x.8) +( 8x.2) = 32 或 32%

For $\begin{align*}C=2\end{align*}$ we have to pull chocolate chip both times, so we have $\begin{align*}.2 \times .2=.04 \ or \ 4 \%\end{align*}$
::C=2的C=2 我们两次都要拉巧克力芯片所以有2×2=0.04或4%

Example 3
::例3

Let $\begin{align*}S\end{align*}$ be the score of a single student chosen at random from Mr. Spence's class. Create a probability distribution for $\begin{align*}S\end{align*}$ , given the following:
::S 是 Spence 先生的班级随机选择的单个学生的分数。创建 S 的概率分布, 如下 :

Number of Students ::学生人数	Test ::测试测试测试
11	87
7	89
13	92
9	94
6	96

There are a total of 46 students in Mr. Spence’s class, so there are 46 scores. The probability of a random student having score $\begin{align*}S\end{align*}$ is the same as that score’s portion of the total number of scores:
::Spence先生的班级共有46名学生,因此有46人得分。随机学生得S分的概率与得分总数中得分的比例相同:

$\begin{align*}P(S=87)=\frac{11}{46}\end{align*}$
::P(S=87)=1146

$\begin{align*}P(S=89)=\frac{7}{46}\end{align*}$
::P(S=89)=746

$\begin{align*}P(S=94)=\frac{9}{46}\end{align*}$
::P(S=94)=946

$\begin{align*}P(S=96)=\frac{6}{46}\end{align*}$
::P(S=96)=646

Review
::回顾

1. What is a probability distribution?
::1. 什么是概率分布?

2. What is a random variable?
::2. 什么是随机变数?

3. What is the difference between a discrete and a continuous random variable?
::3. 离散变量和连续随机变量之间有什么区别?

For problems 4-7, refer to the following table:
::关于4-7问题,请见下表:

*$\begin{align}S\end{align}$*	2	3	4	5	6	7	8	9	10
$\begin{align}P(S)\end{align}$	.04		.12	.16		.16	.12		.04

4. Assuming the table is a probability distribution for discrete random variable $\begin{align*}S\end{align*}$ , which is the sum of two dice rolled once, how many sides does each die have?
::4. 假设表格是离散随机变量S的概率分布,即两个骰子滚动一次之和,则每个死亡方有多少?

5. What is $\begin{align*}P(3)\end{align*}$ ?
::5. 什么是P(3)?

6. What is $\begin{align*}P(6)\end{align*}$ ?
::6. P(6)是什么?

7. What is $\begin{align*}P(9)\end{align*}$ ?
::7. 什么是P(9)?

8. Roll two seven-sided dice once. Let $\begin{align*}S\end{align*}$ be the sum of the two dice. Create a probability distribution for $\begin{align*}S\end{align*}$ .
::8. 将两张七面骰子放一次。S是两张骰子的总和。为 S 设定一个概率分布。

9. Flip a fair coin 3 times, let $\begin{align*}H\end{align*}$ be the number of heads. Create a probability distribution for $\begin{align*}H\end{align*}$ .
::9. 将一个公平的硬币翻转3次,由H作为头数。

10. Let $\begin{align*}S\end{align*}$ be the sum of two standard fair dice. Create a probability distribution for $\begin{align*}S\end{align*}$ , if the experiment consists of a single roll of both dice.
::10. 如果实验由两张骰子的单卷组成,那么S就是两个标准公平骰子的总和。为S创造概率分布。

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。

$\begin{align}X\end{align}$	$\begin{align}0\end{align}$	$\begin{align}1\end{align}$	$\begin{align}2\end{align}$
$\begin{align}P(X)\end{align}$	$\begin{align}\frac{1}{4}\end{align}$	$\begin{align}\frac{1}{2}\end{align}$	$\begin{align}\frac{1}{4}\end{align}$

Section outline

Probability Distribution ::概率分布

Creating Probability Distributions ::创建概率分布

Evaluating Probabilities ::评价概率

Earlier Problem Revisited ::重审先前的问题

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)