Course: 11.6 工会的概率-interactive

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum

工会的概率

Consider a sample space with events

\begin{aligned} A \end{aligned}

and

\begin{aligned} B . \end{aligned}

Recall that the union of events

\begin{aligned} A \end{aligned}

and

\begin{aligned} B \end{aligned}

is an event that includes all the outcomes in either event $\begin{aligned} A \end{aligned}$ , event $\begin{aligned} B \end{aligned}$ , or both . The symbol

\begin{aligned} \cup \end{aligned}

represents union . Below,

\begin{aligned} A \cup B \end{aligned}

is shaded.

How do you find the number of outcomes in a union of events? If you find the sum of the number of outcomes in event $\begin{aligned} A \end{aligned}$ and the number of outcomes in event $\begin{aligned} B, \end{aligned}$ you will have counted some of the outcomes twice. In fact, you will have counted the outcomes that are in both event $\begin{aligned} A \end{aligned}$ and event $\begin{aligned} B \end{aligned}$ twice. Therefore, in order to correctly count the number of outcomes in the union of two events , you must count the number of outcomes in each event separately and subtract the number of outcomes shared by both events (so these are not counted twice). Generalizing to probability :
::您如何在事件联盟中找到结果数量 ? 如果您发现事件A的结果数量和事件B的结果数量的总和, 您将会两次计算其中的一些结果。事实上, 您将会两次计算事件A和事件B的结果。因此, 为了正确计算两个事件联合中的结果数量, 您必须分别计算每个事件的结果数量, 并减去两个事件共享的结果数量( 因此这些结果不会两次计算 )。概括为概率 :

$\begin{aligned} P (A \cup B) = P (A) + P (B) - P (A \cap B) . \end{aligned}$ This is called the Addition Rule for Probability.
::P(AB) = P(A)+P(B)-P(AB) 。这被称为 " 概率补充规则 " 。

Note that $\begin{aligned} (A \cap B) \end{aligned}$ is the intersection of the two events . It contains all the outcomes that are shared by both events and is the intersection of the two circles in the .
::请注意 (AB) 是两个事件的交叉点。它包含两个事件共有的所有结果, 并且是两个圆圈的交叉点。

Suppose that in your class of 30 students , 8 students are in band, 15 students play a sport, and 5 students are both in band and play a sport. Let $\begin{aligned} A \end{aligned}$ be the event that a student is in band and let $\begin{aligned} B \end{aligned}$ be the event that a student plays a sport . Create a Venn diagram that models this situation.
::假设在你的30名学生班里,有8名学生在乐队里,15名学生在玩运动,5名学生在乐队里,还有5名学生在玩运动。让学生在乐队里,让B参加学生在运动的活动。制作一个能模拟这种情况的文恩图。

In order to fill in the Venn diagram, remember that the total of the numbers in circle $\begin{aligned} A \end{aligned}$ must be 8 and the total of the numbers in circle $\begin{aligned} B \end{aligned}$ must be 15. The intersection of the two circles must contain a 5.
::为了填入文恩图,请记住圆A的总数必须是8个,圆B的总数必须是15个。两个圆圈的交叉点必须包含5个。

$\begin{aligned} P (A \cup B) \end{aligned}$ is the probability that a student is in band or plays a sport or both . With the help of the Venn diagram, this is not too difficult to calculate:
::P(AB) 是学生参加乐队或运动或两者兼而有之的概率。在文恩图的帮助下, 这并不难计算 :

$\begin{aligned} P (A \cup B) = \frac{3 + 10 + 5}{30} = \frac{18}{30} = \frac{3}{5} \end{aligned}$
::P(AB)=3+10+530=1830=35

You could also compute this probability using the Addition Rule:
::您也可以使用添加规则来计算此概率 :

Note that by using the Addition Rule, you avoid having to determine that there are 3 people who are in band and don't play a sport and 10 people who play a sport but are not in band. The Addition Rule is easier when you have not created a Venn diagram.
::请注意,通过使用“ 附加规则 ” , 您可以避免确定有 3 个人在乐队里, 不玩运动, 10 个人在玩运动, 但不是乐队里。如果您没有创建 Venn 图表, “ 附加规则” 更容易。

Understanding Probability
::理解概率

Two events $\begin{aligned} C \end{aligned}$ and $\begin{aligned} D \end{aligned}$ are disjoint. Explain why $\begin{aligned} P (C \cup D) = P (C) + P (D) . \end{aligned}$
::两个事件C和D是脱节的。解释为什么P( CD) = P( C)+P( D) 的原因。

If two events are disjoint (also known as mutually exclusive), then they share no outcomes. Therefore, the probability of both events occurring simultaneously is 0 $\begin{aligned} (P (C \cap D) = 0) . \end{aligned}$ By the Addition Rule:
::如果两个事件脱节(也称为相互排斥),那么它们不会产生任何结果。因此,两个事件同时发生的可能性是0(P(CD)=0)。根据补充规则:

$\begin{aligned} P (C \cup D) & = P (C) + P (D) - P (C \cap D) \\ P (C \cup D) & = P (C) + P (D) - 0 \\ P (C \cup D) & = P (C) + P (D) \end{aligned}$
::P(CD)=P(C)+P(D)-P(CD)=P(C)+P(D)=P(C)+P(D)-0P(CD)=P(C)+P(D)

The outcomes of rolling a die are disjoint. In the interactive below, click the button and observe what the probability is of rolling an even or and odd number. That is, what is $\begin{aligned} P (Even \cup Odd) . \end{aligned}$
::滚动死亡的结果是脱节的。在下面的交互中,单击按钮并观察滚动偶数或奇数的概率。这就是 P( EvenOdd) 。

Real-World Application: Weather Probability
::真实世界应用程序: 天气概率

Suppose that today there is a 90% chance of snow, a 20% chance of a strong winds, and a 15% chance of both snow and strong winds. What is the probability of snow or strong winds?
::假设今天有90%的降雪机率,20%的强风机率,15%的强风机率。雪或强风机率是多少?

Use the Addition Rule:
::使用添加规则:

$\begin{aligned} P (Snow \cup Strong Winds) & = P (Snow) + P (Strong Winds) - P (Snow \cap Strong Winds) \\ P (Snow \cup Strong Winds) & = 0.90 + 0.20 - 0.15 \\ P (Snow \cup Strong Winds) & = 0.95 \end{aligned}$
::P(Snow强风)=P(Snow)+P(强风)-P(Snow强风)P(Snow强风)=0.90+0.20-0.15P(Snow强风)=0.95。

There is a 95% chance of either snow, strong winds, or both.
::有95%的雪强风或两者兼而有之的机会

Now suppose that today there is a 60% chance of snow, an 85% chance of snow or strong winds, and a 25% chance of snow and strong winds. What is the chance of strong winds?
::现在假设今天有60%的降雪机率,85%的降雪机率或强风机率,25%的降雪机率和强风机率。强风机率有多大?

Once again you can use the Addition Rule, because it relates the probabilities in the problem.
::您可再次使用“附加规则”,因为它与问题的概率有关。

$\begin{aligned} P (Snow \cup Strong Winds) & = P (Snow) + P (Strong Winds) - P (Snow \cap Strong Winds) \\ 0.85 & = 0.60 + P (Strong Winds) - 0.25 \\ 0.50 & = P (Strong Winds) \end{aligned}$
::P(Snow强风)=P(Snow)+P(强风)-P(Snow强风)0.85=0.60+P(强风)-0.250.50=P(强风)

There is a 50% chance of strong winds.
::有50%的强风概率。

On any given night, the probability that Nick has a cookie for dessert is 10%. The probability that Nick has ice cream for dessert is 50% . The probability that Nick has a cookie or ice cream is 55%. What is the probability that Nick has a cookie and ice cream for dessert?
::在任何特定夜晚,尼克有一个饼干用于甜点的概率是10%。尼克有一个冰淇淋用于甜点的概率是50%。尼克有一个饼干或冰淇淋的概率是55%。尼克有一个饼干或冰淇淋用于甜点的概率是55%。尼克有一个饼干和冰淇淋用于甜点的概率是多少?

The probability that Nick has a cookie and ice cream for dessert is .
::Nick有饼干和冰淇淋做甜点的概率是

Examples
::实例实例实例实例

Example 1
::例1

Consider the experiment of rolling a pair of dice. There are 36 outcomes in the sample space . You are interested in the sum of the numbers. Let $\begin{aligned} A \end{aligned}$ be the event that the sum is even and let $\begin{aligned} B \end{aligned}$ be the event that the sum is less than 5.
::考虑一下滚动骰子的实验。样本空间里有36个结果。您对数字的总和感兴趣。如果数字是偶数, 则让B来计算总和小于5。

a) Create a Venn diagram that models this situation. The Venn diagram should contain 36 numbers.
:a) 创建一个能模拟这种情况的文恩图,文恩图应包含36个数字。

Find all 36 outcomes, and then find the sum of each pair of numbers. Sort the numbers into the Venn diagram so that even numbers are in circle $\begin{aligned} A \end{aligned}$ and numbers less than 5 are in circle $\begin{aligned} B . \end{aligned}$ Any other numbers should appear outside of the circles.
::查找所有36个结果, 然后找到每对数字的总和。在 Venn 图表中排序数字, 使偶数在圆 A 中, 数字少于 5 的数字在圆 B 中。任何其他数字都应该出现在圆外。

b) Find $\begin{aligned} P (A \cup B) \end{aligned}$ using the Venn diagram.
:b) 使用Venn图查找P(AB)。

There are 20 numbers within the circles and 36 numbers total. Therefore,
::圈内有20个数字,总共36个数字。

$\begin{aligned} P (A \cup B) = \frac{14 + 4 + 2}{36} = \frac{20}{36} = \frac{5}{9} . \end{aligned}$
::P(AB)=14+4+236=2036=59。

Example 2
::例2

Find $\begin{aligned} P (A \cup B) \end{aligned}$ using the Addition Rule and explain why the answer makes sense.
::使用添加规则查找 P( AB) 并解释为什么答案是有道理的。

To use the Addition Rule, you need to know $\begin{aligned} P (A), P (B) \end{aligned}$ and $\begin{aligned} P (A \cap B) . \end{aligned}$
::要使用添加规则,你需要知道P(A)、P(B)和P(AB)。

$\begin{aligned} P (A) = \frac{18}{36} = \frac{1}{2} \\ P (B) = \frac{6}{36} = \frac{1}{6} \\ P (A \cap B) = \frac{4}{36} = \frac{1}{9} \end{aligned}$
::P(A)=1836=12P(B)=636=16P(AB)=436=19

Now, use the Addition Rule: $\begin{aligned} P (A \cup B) \end{aligned}$ $\begin{aligned} = P (A) + P (B) - P (A \cap B) = \frac{18}{36} + \frac{6}{36} - \frac{4}{36} = \frac{20}{36} = \frac{5}{9} . \end{aligned}$ This answer is the same as the answer to Example #2, as it should be. This calculation makes sense because both $\begin{aligned} P (A) \end{aligned}$ and $\begin{aligned} P (B) \end{aligned}$ include the 4 numbers in the intersection of the circles. You need to subtract $\begin{aligned} P (A \cap B), \end{aligned}$ the probability of those 4 numbers, so that you do not count the probability of those numbers twice.
::现在,使用“附加规则” : P(A) = P(A)+P(B) = P(A) +P(B)- P(A) = 1836+636- 436= 2036= 59。这个答案与例2 的答案相同。这个计算是有道理的, 因为 P(A) 和 P(B) 都在圆圈的交叉点中包含 4 个数字。您需要减去 P(A) +P(B) , 这4个数字的概率, 这样您就不会两次计算这些数字的概率。

CK-12 PLIX Interactive: Probability of Unions
::CK-12 PLIX 互动:工会的概率

Summary

The addition rule for probability is:

$\begin{aligned} P (A \cup B) = P (A) + P (B) - P (A \cap B) \end{aligned}$

::增加概率的规则是:P(AB)=P(A)+P(B)-P(AB)

$\begin{aligned} P (A \cap B) \end{aligned}$ is the intersection of of the two events
::PAB是两个事件的交叉点

If two events are disjoint (also known as mutually exclusive), then they share no outcomes so,

$\begin{aligned} P (A \cup B) = P (A) + P (B) \end{aligned}$

::如果两个事件脱联(也称为相互排斥),那么它们不会产生任何结果,P(AB)=P(A)+P(B)

Review
::审查审查审查审查

1. State the Addition Rule for probability and explain when it is used.
::1. 说明概率的附加规则,并解释何时使用该规则。

2. What happens to the Addition Rule when the two events considered are disjoint?
::2. 如果审议的两个事件脱节,《附加规则》会如何?

3. Use a Venn diagram to help explain why there is subtraction in the Addition Rule.
::3. 使用文恩图帮助解释为何在《附加规则》中减去。

4. Sarah tells her mom that there is a 40% chance she will clean her room, a 70% she will do her homework, and a 24% chance she will clean her room and do her homework. What is the probability of Sarah cleaning her room or doing her homework?
::4. Sarah告诉她妈妈说,她有40%的机会打扫房间,70%可以做功课,24%可以打扫房间和做功课。

5. You dad only ever makes one meal for dinner. The probability that he makes pizza tonight is 30%. The probability that he makes pasta tonight is 60%. What is the probability that he makes pizza or pasta?
::5. 你爸爸晚餐只做一顿饭,今晚做披萨的概率是30%,今晚做意大利面的概率是60%。

6. After your little sister has gone trick-or-treating for Halloween, your mom says she is allowed to have 2 pieces of candy. The probability of her having a Snickers is 50%. The probability of her having a peanut butter cup is 60%. The probability of her having a Snickers or a peanut butter cup is 100%. What is the probability of her having a Snickers and a peanut butter cup?
::6. 在你的小妹妹为万圣节玩弄或捣乱之后,你妈妈说,她可以拥有两块糖果,她拥有一个小饼干的概率是50%,她拥有一个花生酱杯的概率是60%,拥有一个小饼干或花生酱杯的概率是100%。她拥有一个小饼干和花生酱杯的概率是多少?

7. Deanna sometimes likes honey or lemon in her tea. There is a 50% chance that she will have honey and lemon, a 95% chance that she will have honey or lemon, and a 80% chance that she will have honey. What is the probability that she will have lemon?
::7. 迪安娜有时喜欢茶里有蜂蜜或柠檬,有50%的可能性有蜂蜜和柠檬,95%的可能性有蜂蜜或柠檬,80%的可能性有蜂蜜。

Consider the experiment of drawing a card from a standard deck. Let $\begin{aligned} A \end{aligned}$ be the event that the card is a diamond. Let $\begin{aligned} B \end{aligned}$ be the event that the card is a Jack. Let $\begin{aligned} D \end{aligned}$ be the event that the card is a four.
::考虑从标准甲板上绘制一张牌的实验。请将卡片是钻石的事件列入 A 。请将卡片是杰克的事件列入 B 。请将卡片是杰克的事件列入 D 。请将卡片是四号的事件列入 D 。

8. Find $\begin{aligned} P (A), P (D), P (A \cap D) \end{aligned}$ .
::8. 查找P(A)、P(D)、P(AD)。

9. Find $\begin{aligned} P (A \cup D) \end{aligned}$ . What does this probability represent compared to $\begin{aligned} P (A \cap D) \end{aligned}$ ?
::9. 查找P(AD). 与P(AD)相比,这种概率代表什么?

10. To find $\begin{aligned} P (B \cup D) \end{aligned}$ , all you need to do is add $\begin{aligned} P (B) \end{aligned}$ and $\begin{aligned} P (D) \end{aligned}$ . Why is this and why do you not have to subtract anything?
::10. 要找到P(BD),你只需要加上P(B)和P(D)。为什么这样,为什么你不必减去什么?

Consider the experiment of flipping three coins and recording the sequence of heads and tails. Let $\begin{aligned} B \end{aligned}$ be the event that there is at least one heads. Let $\begin{aligned} C \end{aligned}$ be the event that the third coin is a tails. Let $\begin{aligned} D \end{aligned}$ be the event that the first coin is a heads.
::想象一下翻三个硬币的实验, 并记录头和尾的顺序。让 B 成为至少有一个头的事件。让 C 成为第三个硬币是尾巴的事件。让 D 成为第一个硬币是头的事件。

11. Find $\begin{aligned} P (C \cup D) \end{aligned}$ . What does this probability represent?
::11. 查找 P(CD). 这种概率代表什么?

12. Create a Venn diagram to show events $\begin{aligned} B \end{aligned}$ and $\begin{aligned} C \end{aligned}$ for this experiment.
::12. 为本实验创建 Venn 图表以显示事件B和C。

13. Find $\begin{aligned} P (B \cup C) \end{aligned}$ and $\begin{aligned} P (B \cap C) \end{aligned}$ . Compare and contrast these two probabilities.
::13. 查找P(BC)和P(BC)。比较和比较这两种可能性。

The Addition Rule can be extended for three events. Consider three events that all share outcomes, as shown in the Venn diagram below.
::如下文文恩图所示,《附加规则》可适用于三场活动。审议三场活动,所有活动都分享成果。

14. Label the shaded part of the diagram in terms of $\begin{aligned} A, B, C \end{aligned}$ .
::14. 用A、B、C表示图的阴影部分贴上标签。

15. Find a rule for $\begin{aligned} P (A \cup B \cup C) \end{aligned}$ in terms of $\begin{aligned} P (A), P (B), P (C), P (A \cap B), P (B \cap C), P (A \cap C), P (A \cap B \cap C) \end{aligned}$ .
::15. 就P(A)、P(B)、P(C)、P(A)B(B)、P(B)、P(B)C、P(B)C、P(A)C、P(A)B)C、P(A)B(C)为P(A)B(C))确定一条规则。

16. In a Venn diagram, P(A) = 0.5, P(A∪B) = 0.8, and P(A∩B) = 0.3. Use the addition rule to find P(B).
::16. 在文恩图中,P(A) = 0.5,P(A) = 0.8,P(A) = 0.8,P(A) = 0.3,使用添加规则查找P(B)。

17. In a survey done in your class, you found that 45% like rap music, 30% like country music, and 15% like both. Draw a Venn diagram to show these results, and then find the probability that a student will like rap music and not country. What is the probability that they will like neither rap nor country music?
::17. 在班级的一项调查中,你发现有45%的人喜欢饶舌音乐,30%的人喜欢乡村音乐,15%的人喜欢乡村音乐。绘制文恩图表来显示这些结果,然后发现学生喜欢饶舌音乐而不是乡村的可能性。他们不喜欢饶舌音乐和乡村音乐的概率是多少?

18. In a survey of 65 people, 28 consider themselves republicans, 27 consider themselves democrats. The rest are considered independent. What is the probability that a person chosen at random will be a democrat or independent?
::18. 在对65人的调查中,28人认为自己是共和党人,27人认为自己是民主党人,其余人被视为独立者,随机挑选的人成为民主党人或独立者的可能性有多大?

19. In a hospital on one evening, there were 8 nurses and 5 doctors. 7 of the nurses and 3 of the doctors were female. What is the probability that a person chosen at random will be male or a nurse?
::19. 一天晚上,医院里有8名护士和5名医生,其中7名护士和3名医生是女性,随意挑选的人是男性或护士的可能性有多大?

Review (Answers)
::审查(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。

Section outline

General

工会的概率

Understanding Probability ::理解概率

Real-World Application: Weather Probability ::真实世界应用程序: 天气概率

Examples ::实例实例实例实例

Example 1 ::例1

Example 2 ::例2

CK-12 PLIX Interactive: Probability of Unions ::CK-12 PLIX 互动:工会的概率

Review ::审查审查审查审查

Review (Answers) ::审查(答复)