Course: 11.13 有条件概率的理解

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理解有条件的概率
On Thursday, Carey was in charge of answering the phones and booking appointments for bike repairs. The bike shop repairs bikes on Monday, Tuesday and Wednesday mornings and on Thursday and Friday afternoons. All appointments are booked randomly. The person making the appointment can choose or the person answering the phone can choose. Carey booked two appointments right away.
::星期四,Carey负责接听电话和为修车预定预约时间;星期一、星期二和星期三上午以及星期四和星期五下午,自行车店修理自行车;所有预约都随机预订;预约的人可以选择,接电话的人可以选择;Carey立即预订了两个预约时间。

What are the chances that both of these appointments were booked for a Monday, Tuesday or Wednesday morning?
::这两个预约在星期一、星期二或星期三上午的预定时间有多少机会?

In this concept, you will learn to understand conditional probability.
::在这个概念中,你会学会理解有条件的概率。

Conditional Probability
::有条件概率

Sometimes the outcome that you get when figuring out a probability is conditional. A conditional outcome means that you will only get an outcome if the conditions are designed to cause a specific result.
::有时,当发现概率时,结果是有条件的。有条件的结果意味着只有条件设计是为了产生特定结果时,你才会得到结果。

Let’s look at an example.
::让我们举个例子。

Consider a jar with 4 black marbles and 6 white marbles. If you randomly pull out 2 marbles from the jar, one at a time, without replacing the first marble, what is the probability that both marbles will be white?
::考虑一个罐子,它有4个黑色大理石和6个白色大理石。如果你随机地从罐子中抽出2个大理石,一个一个一个,不取代第1个大理石,那么两个大理石都是白色的概率是多少?

First, start by approaching the problem the same as you would with dependent events. There are 6 white marbles and there are 10 marbles in the jar. The probability of the first marble being white is:
::首先,首先处理的问题与你所依赖的事件一样。有6颗白色弹珠,罐子里有10颗弹珠。第一颗弹珠是白色的概率是:

$\begin{aligned} \begin{array}{rcl} P (1^{s t} white marble) & = & \frac{6}{10} \\ = & \frac{3}{5} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P( 1^{st} \ \text{white marble}) &=& \frac{6}{10} \\ &=& \frac{3}{5} \end{array}\end{align*}$

Next, having removed the first marble from the jar, now instead of 6 white marbles out of 10 total marbles, there are only 5 white marbles left out of 9 total marbles. The probability of the second marble being white is:
::其次,从罐子上移走第一个大理石之后,现在在总共10个大理石中,没有6个白色大理石,在9个大理石中,只有5个白色大理石遗漏。

$\begin{aligned} P (2^{n d} white marble) = \frac{5}{9} \end{aligned}$
$\begin{align*}P( 2^{nd} \ \text{white marble}) = \frac{5}{9}\end{align*}$
::P(第2次白色大理石白大理石)=59

Then, calculate the probability of both events occurring.
::然后计算两个事件的发生概率。

$\begin{aligned} \begin{array}{rcl} P (white then white) & = & P (1^{s t} white marble) \times P (2^{n d} white marble) \\ = & \frac{3}{5} \times \frac{5}{9} \\ = & \frac{15}{45} \\ = & \frac{1}{3} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{white then white}) &=& P( 1^{st} \ \text{white marble}) \times P( 2^{nd} \ \text{white marble}) \\ &=& \frac{3}{5} \times \frac{5}{9} \\ &=& \frac{15}{45} \\ &=& \frac{1}{3} \end{array}\end{align*}$

The answer is $\begin{align*}\frac{1}{3}\end{align*}$ .

The probability of picking two white marbles, without replacement, from this jar is $\begin{align*}\frac{1}{3}\end{align*}$ .
::从这个罐子里摘取两颗白色大理石的概率是13颗

The same general method works for calculating any two (or more) dependent events.
::计算任何两个(或两个以上)依附事件所用的通用方法相同。

Now let’s look at conditional probability and outcomes .
::现在让我们看看有条件的概率和结果。

Conditional probability involves situations in which you determine the probability of an event based on another event having occurred.
::有条件的概率涉及一种情况,即您根据另一事件的发生情况确定事件的概率。

Let’s look at an example.
::让我们举个例子。

In a class of 25 students, 14 like pizza and 16 like fruit juice. One student likes neither and 6 students like both. If one student is chosen at random, what is the probability that this student likes pizza given that he/she likes fruit juice?
::在25个班级的学生中,14个是比萨饼,16个是果汁。一个学生不喜欢两个学生,6个学生喜欢两个学生。如果一个学生是随机挑选的,那么这个学生喜欢比萨饼的概率是多少,因为他/她喜欢果汁?

First, to solve this problem, you could draw a to represent the data.
::首先,为了解决这个问题,你可以画一个来代表数据。

Next, of the 16 who like fruit juice, 6 like pizza. Therefore the probability is:
::接下来是16个喜欢果汁的人中 6个喜欢披萨

$\begin{aligned} \begin{array}{rcl} P (pizza | fruit juice) & = & \frac{6}{16} \\ = & \frac{3}{8} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{pizza } \big | \text{ fruit juice}) &=& \frac{6}{16} \\ &=& \frac{3}{8} \end{array} \end{align*}$

The answer is $\begin{align*}\frac{3}{8}\end{align*}$ .
::答案是38岁

If a student chosen likes fruit juice, then the probability that they also like pizza is $\begin{align*}\frac{3}{8}\end{align*}$ .
::如果被选中的学生喜欢果汁那么他们也喜欢披萨的概率是38

Notice that you write the conditional probability as:
::请注意,您将条件概率写为 :

$\begin{align*}P\text{(second choice} \ \big | \ \text{first choice)}\end{align*}$ .
::P(第二选择- 首选)。

You read this as the probability of the second choice given the first choice.
::你读这个是第二选择的概率从第一选择。

Another way to look at the conditional probability formula is:
::另一种观察条件概率公式的方法是:

$\begin{aligned} P (second | first) = \frac{P (first choice and second choice)}{P (first choice)} \end{aligned}$
$\begin{align*}P(\text{second} \ \big| \ \text{first} )= \frac{P(\text{first choice and second choice})}{P \text{(first choice)}}\end{align*}$
::P(第二选择)=P(第一选择和第二选择)P(第一选择)

Let’s look at an example.
::让我们举个例子。

The probability that it is Friday and that a student is absent is 0.05. Since there are 5 school days in a week, the probability that it is Friday is $\begin{align*}\frac{1}{5}\end{align*}$ or $\begin{align*}0.2\end{align*}$ . What is the probability that a student is absent given that today is Friday?
::周五和学生缺课的概率是0.05。由于每周有5个学日,周五的概率是15或0.2。鉴于今天是星期五,学生缺课的概率是多少?

First, set up the conditional probability formula.
::首先,设定有条件的概率公式。

$\begin{aligned} P (absent | Friday) = \frac{P (Friday and absent)}{P (Friday)} \end{aligned}$
$\begin{align*}P(\text{absent} \ \big| \ \text{Friday}) = \frac{P(\text{Friday and absent})}{P \text{(Friday)}}\end{align*}$
::P(缺席星期五) = P( 星期五和缺席) P( 星期五)

Next, fill in the probabilities and solve.
::下一个,填充概率和解决。

$\begin{aligned} \begin{array}{rcl} P (absent | Friday) & = & \frac{P (Friday and absent)}{P (Friday)} \\ = & \frac{0.05}{0.2} \\ = & \frac{1}{4} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{absent} \ \big| \ \text{Friday}) &=& \frac{P(\text{Friday and absent})}{P \text{(Friday)}} \\ &=& \frac{0.05}{0.2} \\ &=& \frac{1}{4} \end{array}\end{align*}$
::P(缺席星期五) = P( 星期五和缺席) P( 星期五) = 0.050.2 = 14

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

The probability that a student is absent given that it is Friday is $\begin{align*}\frac{1}{4}\end{align*}$ .
::学生缺席的概率,因为星期五是14岁。

Let’s look at another example.
::让我们再看看另一个例子。

Jar A contains 2 red and 3 blue marbles. Jar B contains 4 red and 1 blue marble. Thomas selects a jar by tossing a coin and takes a marble from that jar. What is the probability that given the marble is red that it came from jar B?
::Jar A含有2个红色和3个蓝色大理石。Jar B含有4个红色和1个蓝色大理石。Thomas通过扔硬币来选择一个罐子,并从罐子上取一个大理石。如果大理石是红色的,那么它来自罐子B的概率是多少?

First, let’s draw a tree diagram . Remember there are 5 marbles in each jar. Since Thomas flips a coin he has a 1 in 2 chance of choosing each jar.

::首先,让我们绘制一幅树图。记住每个罐子里有5颗大理石。自托马斯翻硬币以来,他每罐中就有1分之一的机会选择。

Next, calculate the probability that the marble chosen is red. Notice that there are two ways you can choose a red marble, from jar A or from Jar B.
::接下来,计算选择的大理石为红色的概率。请注意,从罐子A或罐子B,有两种方法可以选择红大理石。

$\begin{aligned} \begin{array}{rcl} P (red) & = & P_{Jar A} + P_{Jar B} \\ = & \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{4}{5} \\ = & \frac{2}{10} + \frac{4}{10} \\ = & \frac{6}{10} \\ = & \frac{3}{5} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{red}) &=& P_{\text{Jar A}} + P_{\text{Jar B}} \\ &=& \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{4}{5} \\ &=& \frac{2}{10} + \frac{4}{10} \\ &=& \frac{6}{10} \\ &=& \frac{3}{5} \end{array}\end{align*}$

Then, calculate the probability of choosing a marble from Jar B given that it is red. Use algebra to simplify.
::然后计算从 Jar B 中选择大理石的概率, 因为它是红色。使用代数来简化。

$\begin{aligned} \begin{array}{rcl} P (Jar B | red) & = & \frac{P (red and Jar B)}{P (red)} \\ = & \frac{\frac{1}{2} \times \frac{4}{5}}{\frac{3}{5}} \\ = & \frac{\frac{4}{10}}{\frac{3}{5}} \\ = & \frac{4}{10} \times \frac{5}{3} \\ = & \frac{20}{30} \\ = & \frac{2}{3} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{Jar B} \ \big| \ \text{red}) &=& \frac{P \text{(red and Jar B)}}{P \text{(red)}} \\ \\ &=& \frac{\frac{1}{2} \times \frac{4}{5}} {\frac{3}{5}} \\ \\ &=& \frac{\frac{4}{10}}{\frac{3}{5}} \\ \\ &=& \frac{4}{10} \times \frac{5}{3} \\ \\ &=& \frac{20}{30} \\ \\ &=& \frac{2}{3} \end{array}\end{align*}$
::P(Jar B 红) = P(red 和 Jar B) P(red) = 12x4535= 41035= 410x53= 2030=23

The answer is $\begin{align*}\frac{2}{3}\end{align*}$ .
::答案是23岁

The probability of choosing a red marble from Jar B is $\begin{align*}\frac{2}{3} \end{align*}$ .
::从Jar B中选择红大理石的概率是23。

Examples
::实例

Example 1
::例1

Earlier, you were given a problem about the appointments at the bike shop.
::早些时候,你遇到了一个问题有关在自行车店的预约。

Carey is randomly booking appoints at the bike shop for Monday, Tuesday, and Wednesday mornings as well as Thursday and Friday afternoons. You need to find the probability that the first two appointments booked will be in the morning.
::Carey正在自行车店随机预订周一、周二、周三上午以及周四和周五下午的预约时间。您需要找到头两次预约时间是早上的可能性。

First, to work on this probability, you must determine the probability of the first appointment booked being on a Monday, Tuesday or Wednesday. There are five possible days for appointments, but three favorable outcomes.
::首先,为了研究这种可能性,你必须确定第一次预约是在星期一、星期二或星期三的概率。有五天的预约时间,但有三个有利的结果。

Probability of the first appointment being on Mon, Tues, or Wed $\begin{align*}= \frac{3}{5} \end{align*}$ .
::第一次任命是在蒙、图斯或韦德=35。

Probability of the second appointment being on Mon, Tues, or Wed $\begin{align*}= \frac{2}{4} \end{align*}$ or $\begin{align*}\frac{1}{2}\end{align*}$ .
::第二次任命是在Mon、Tues或Wed=24或12的任命的可能性。

Next, multiply the two probabilities together.
::接下来,将两种概率相乘。

$\begin{aligned} \begin{array}{rcl} P (morning then morning) & = & P (1^{s t} appointment) \times P (2^{n d} appointment) \\ = & \frac{3}{5} \times \frac{1}{2} \\ = & \frac{3}{10} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{morning then morning}) &=& P(1^{st} \ \text{appointment}) \times P(2^{nd} \ \text{appointment}) \\ &=& \frac{3}{5} \times \frac{1}{2} \\ &=& \frac{3}{10} \end{array}\end{align*}$

The answer is $\begin{align*}\frac{3}{10}\end{align*}$ .
::答案是310

There is a $\begin{align*}\frac{3}{10}\end{align*}$ or a 30% chance that the first two appointments booked would be on a Monday, Tuesday or Wednesday morning.
::预定的前两次预约时间是星期一、星期二或星期三上午,有310次或30%的可能性。

Example 2
::例2

Jack’s Catering Service is accepting weekday appointments for Monday through Thursday, and weekend appointments for Friday through Sunday. If appointment dates are made randomly, what is the probability that 2 weekdays will be the first 2 days to be booked?
::杰克餐饮服务处接受周一至周四的日间预约和周五至周日的周末预约。如果任命日期是随机的,那么2个工作日是头两天预订的概率是多少?

First, calculate the probability that the first day will be a weekday. Jack only books appointments 4 of the seven days in the week.
::首先,计算第一天是每周一天的概率。 Jack每周只有7天中的4天预约。

$\begin{aligned} P (1^{s t} weekday) = \frac{4}{7} \end{aligned}$
$\begin{align*}P(1^{st}\text{weekday}) = \frac{4}{7}\end{align*}$
::P(第一个星期日)=47

Next, calculate the probability that the second booked day will also be a weekday. Remember now, there are only six days left to choose from and three of these will be workdays.
::接下来,计算第二个预订日的概率。记住, 现在只有六天可以选择, 其中三天是工作日。

$\begin{aligned} \begin{array}{rcl} P (2^{n d} weekday) & = & \frac{3}{6} \\ P (2^{n d} weekday) & = & \frac{1}{2} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(2^{nd}\ \text{weekday}) &=& \frac{3}{6} \\ P(2^{nd}\ \text{weekday}) &=& \frac{1}{2} \end{array} \end{align*}$
::P( 第2周) = 36P( 第2周) = 12

Then, calculate the probability that the two days chosen will be weekdays.
::然后计算选择的两天为工作日的概率。

$\begin{aligned} \begin{array}{rcl} P (weekday and weekday) & = & P (1^{s t} weekday) \times P (2^{n d} weekday) \\ P (weekday and weekday) & = & \frac{4}{7} \times \frac{1}{2} \\ P (weekday and weekday) & = & \frac{4}{14} \\ P (weekday and weekday) & = & \frac{2}{7} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{weekday and weekday}) &=& P( 1^{st} \ \text{weekday}) \times P( 2^{nd} \ \text{weekday}) \\ P(\text{weekday and weekday}) &=& \frac{4}{7} \times \frac{1}{2} \\ P(\text{weekday and weekday}) &=& \frac{4}{14} \\ P(\text{weekday and weekday}) &=& \frac{2}{7} \end{array} \end{align*}$
::P(周日和周日)=P(第1周日和周日)xP(第2周日)P(周日和周日)=47x12P(周日和周日)=414P(周日和周日)=27

A jar contains four blue marbles and eight red marbles.
::一罐装有四颗蓝色大理石和八颗红色大理石

Example 3
::例3

What is the probability of taking out a blue, then a red marble without replacing the first marble?
::拿出蓝色,然后红色大理石而不取代第一颗大理石的概率是多少?

First, start by approaching the problem the same as you would with dependent events.
::首先,首先处理问题的方式和处理事件时一样。

There are 4 blue marbles and 8 red marbles in the jar. The probability of the first marble being blue is:
::罐子里有4颗蓝色大理石和8颗红色大理石。

$\begin{aligned} \begin{array}{rcl} P (1^{s t} blue marble) & = & \frac{4}{12} \\ = & \frac{1}{3} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(1^{st}\ \text{blue marble}) &=& \frac{4}{12} \\ &=& \frac{1}{3} \end{array}\end{align*}$

Next, having removed the first marble from the jar, you now have 11 marbles in the jar. The probability of the second marble being red is:
::接下来,从罐子上移走第一颗大理石之后,你现在在罐子里有11颗大理石。

$\begin{aligned} P (2^{n d} red marble) = \frac{8}{11} \end{aligned}$
$\begin{align*}P(2^{nd}\ \text{red marble}) = \frac{8}{11}\end{align*}$
::P( 2nd 红大理石) = 811

Then, calculate the probability of both events occurring.
::然后计算两个事件的发生概率。

$\begin{aligned} \begin{array}{rcl} P (blue then red) & = & P (1^{s t} blue marble) \times P (2^{n d} red marble) \\ = & \frac{1}{3} \times \frac{8}{11} \\ = & \frac{8}{33} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{blue then red}) &=& P(1^{st}\ \text{blue marble}) \times P(2^{nd}\ \text{red marble}) \\ &=& \frac{1}{3} \times \frac{8}{11} \\ &=& \frac{8}{33} \end{array} \end{align*}$

The answer is $\begin{align*}\frac{8}{33}\end{align*}$ .
::答案是833

The probability of picking a blue then a red marble, without replacement, from this jar is $\begin{align*}\frac{8}{33}\end{align*}$ .
::从这个罐子里摘出蓝色和红色大理石的概率是833。

Example 4
::例4

What is the probability of taking out two red marbles without replacement?
::不换掉两个红弹珠的概率是多少?

First, start by approaching the problem the same as you would with dependent events. There are 4 blue marbles and 8 red marbles in the jar. The probability of the first marble being red is:
::首先,首先处理的问题与处理附带事件的问题一样。罐子里有4个蓝色弹珠和8个红色弹珠。第一道弹珠可能是红色的:

$\begin{aligned} \begin{array}{rcl} P (1^{s t} red marble) & = & \frac{8}{12} \\ = & \frac{2}{3} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(1^{st}\ \text{red marble}) &=& \frac{8}{12} \\ &=& \frac{2}{3} \end{array} \end{align*}$
::P( 1 红色红大理石) = 812= 23

Next, having removed the first marble from the jar, you now have 11 marbles in the jar and only 7 of them are red. The probability of the second marble being red is:
::接下来,从罐子上移走第一颗大理石之后,你现在在罐子里有11颗大理石,其中只有7颗是红色的。第二颗大理石可能是红色的:

$\begin{aligned} P (2^{n d} red marble) = \frac{7}{11} \end{aligned}$
$\begin{align*}P(2^{nd}\ \text{red marble}) = \frac{7}{11}\end{align*}$
::P( 2nd 红大理石) = 711

Then, calculate the probability of both events occurring.
::然后计算两个事件的发生概率。

$\begin{aligned} \begin{array}{rcl} P (red then red) & = & P (1^{s t} red marble) \times P (2^{n d} red marble) \\ = & \frac{2}{3} \times \frac{7}{11} \\ = & \frac{14}{3} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{red then red}) &=& P(1^{st} \ \text{red marble}) \times P(2^{nd}\ \text{red marble}) \\ &=& \frac{2}{3} \times \frac{7}{11} \\ &=& \frac{14}{3} \end{array}\end{align*}$

The answer is $\begin{align*}\frac{14}{33}\end{align*}$ .

The probability of picking two red marbles, without replacement, from this jar is $\begin{align*}\frac{14}{33}\end{align*}$ .
::从这个罐子里摘取两颗红色大理石的概率是1433

Example 5
::例5

At the local high school, the probability that a student takes math and music is $\begin{align*}\frac{1}{4}\end{align*}$ . The probability that a student is taking math is $\begin{align*} \frac{17}{20} \end{align*}$ . What is the probability that a student is taking music given that they are taking math?
::在本地高中,学生学习数学和音乐的概率是14,学生学习数学的概率是1720。学生学习数学的概率是1720。学生接受音乐的概率是多少?

First, set up the conditional probability formula.
::首先,设定有条件的概率公式。

$\begin{aligned} P (music | math) = \frac{P (math and music)}{P (math)} \end{aligned}$
$\begin{align*}P(\text{music}\ \big| \ \text{math}) = \frac{P \text{(math and music)}}{P \text{(math)}}\end{align*}$
::P(音乐) = P(数学和音乐) P(数学)

Next, fill in the probabilities and solve.
::下一个,填充概率和解决。

$\begin{aligned} \begin{array}{rcl} P (music | math) & = & \frac{P (math and music)}{P (math)} \\ = & \frac{\frac{1}{4}}{\frac{17}{20}} \\ = & \frac{1}{4} \times \frac{20}{17} \\ = & \frac{20}{68} \\ = & \frac{5}{17} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{rcl} P(\text{music} \ \big| \ \text{math}) &=& \frac{P \text{(math and music)}}{P \text{(math)}} \\ \\ &=& \frac{\frac{1}{4}} {\frac{17}{20}} \\ \\ &=& \frac{1}{4} \times {\frac{20}{17}} \\ \\ &=& \frac{20}{68} \\ \\ &=& \frac{5}{17} \end{array}\end{align*}$

The answer is $\begin{align*}\frac{5}{17}\end{align*}$ .

The probability that a student is taking music given that they are taking math is $\begin{align*}\frac{5}{17}\end{align*}$ .
::学生接受音乐的概率是517分因为他们接受数学的概率是517分

Review
::回顾

Solve the problems.
::解决问题。

A stack of 12 cards has 4 Aces, 4 Kings, and 4 Queens. What is the probability of picking 2 Aces from the stack at random?
::12张牌堆有4张A,4张K,4张Q。从堆叠中随机选取2张A的可能性有多大?

What is the probability of picking an Ace then a King from the stack above?
::从上面的堆叠中选取王冠和王冠的概率是多少?

What is the probability of picking 3 Queens from the stack above?
::从上面的堆叠中选取3个皇后的概率是多少?

Stoyko’s shirt drawer has 4 colored t-shirts and 4 white t-shirts. If Stoyko picks out 2 shirts at random, what is the probability that they will both be colored?
::斯托伊科的衬衫抽屉有4件彩色T恤和4件白色T恤。如果斯托伊科随意挑出两件衬衫,那么他们两人染色的可能性有多大?

If Stoyko picks out 2 shirts at random from the drawer above, what is the probability that the first one will be colored and the second one will be white?
::如果斯托伊科随机地从上面的抽屉里挑出两件衬衫,那么第一件染色而第二件是白色的概率有多大?

On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions randomly, what is the probability that the first two contestants will get easy questions?
::在一场比赛中,有16个问题:8个简单,5个中度硬,3个硬。如果对参赛者随机提出问题,前两个参赛者获得简单问题的可能性有多大?

On the game show above, what is the probability that the first contestant will get an easy question and the second contestant will get a hard question?
::在以上比赛中,第一位参赛者会得到一个简单的问题, 而第二位参赛者会得到一个难题的概率是多少?

On the game show above, what is the probability that both of the first two contestants will get hard questions?
::上面的比赛节目上, 前两位参赛者都可能会遇到什么难题呢?

For a single toss of a number cube, what is the probability that the cube will land on a number that is both odd and greater than 2?
::对于一个数字立方体的单振标,立方体在奇数和大于2的数值上着陆的概率是多少?

For a single toss of a number cube, what is the probability that the cube will land on a number that is greater than 2 and less than 6?
::对于一个数字立方体的单次转移,立方体在大于2和小于6的数值上着陆的概率是多少?

For a single toss of a number cube, what is the probability that the cube will land on a number that is greater than 1 and less than 6?
::对于一个数字立方体的单次切数,立方体在一个大于1和小于6的数值上着陆的概率是多少?

What is the probability that a sum of a pair of number cubes will be 11 if the first cube lands on 5?
::如果第一个立方体在5点降落,一对数立方体的总和是11的概率是多少?

What is the probability that a sum of a pair of number cubes will be odd if the first cube lands on 2?
::如果第一个立方体在2号上着陆,一对数立方体的总和会是奇异的概率是多少?

What is the probability that a sum of a pair of number cubes will be even and greater than 6 if the first cube lands on 4?
::如果第一个立方体在4上着陆,一对数立方体的总和会大于或等于6的概率是多少?

If you toss a number cube, predict how likely it is to roll a number less than 6?
::如果按下一个数字立方体, 预测其滚动数字小于 6 的可能性 ?

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

理解有条件的概率

Conditional Probability ::有条件概率

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

Conditional Probability
::有条件概率

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