11.7 跨部门的概率-interactive

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  跨部门的概率

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  Consider a sample space with event $\begin{align*}A\end{align*}$ and event $\begin{align*}B\end{align*}$ . The shaded section of the below is the outcomes shared by events $\begin{align*}A\end{align*}$ and $\begin{align*}B.\end{align*}$ It is called the intersection of events $\begin{align*}A\end{align*}$ and $\begin{align*}B,\end{align*}$ $\begin{align*}A \cap B.\end{align*}$ Note that $\begin{align*}B \cap A\end{align*}$ is equivalent to $\begin{align*}A \cap B.\end{align*}$
  ::以事件A和事件B为样本空间。 下面的阴影部分是事件A和事件B的共同结果。 它称为事件A和事件B的交叉点,AB.注意BA相当于AB。

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  Sometimes you can calculate $\begin{align*}P(A \cap B)\end{align*}$ directly, especially if you know all of the outcomes in the sample space . Other times, you will only have partial information about the sample space and the events.
  ::有时您可以直接计算 P( AB) , 特别是如果您知道样本空间的所有结果。 有时, 您将只获得关于样本空间和事件的部分信息 。

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  If two events $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ are dependent, then you can find the of $\begin{align*}A\end{align*}$ given $\begin{align*}B\end{align*}$ (or the conditional probability of $\begin{align*}B\end{align*}$ given $\begin{align*}A\end{align*}$ ):
  ::如果两个事件A和B是依附的,那么您可以找到给定 B (或给定 B 的有条件概率) :

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  $\begin{align*}P(A|B)& =\frac{P(A \cap B)}{P(B)} \\ P(B|A) & =\frac{P(A \cap B)}{P(A)}\end{align*}$
  ::P(AB)=P(AB)P(B)P(B)P(BA)=P(AB)P(A)

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  You can rewrite the above equations to solve for $\begin{align*}P(A \cap B)\end{align*}$ :
  ::您可以重写上面的方程式以解析 P( AB) :

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  $\begin{align*}P(A \cap B) & =P(A|B)P(B) \\ P(A \cap B) & =P(B|A)P(A)\end{align*}$
  ::P(AB)=P(AB)P(B)P(AB)=P(BB)P(A)

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  These formulas are known as the Multiplication Rule.
  ::这些公式称为乘法规则。

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  What if the events are independent? The Multiplication Rule will still work, but it can be simplified. Recall that if two events are independent, then the result of one event has no impact on the result of the other event. For $\begin{align*}A\end{align*}$ and $\begin{align*}B,\end{align*}$ $\begin{align*}P(A|B)=P(A)\end{align*}$ and $\begin{align*}P(B|A)=P(B).\end{align*}$ By substitution, the above formulas become:
  ::如果事件是独立的呢? 乘法规则仍然有效, 但它可以简化。 回顾如果两个事件是独立的, 那么一个事件的结果对另一个事件的结果没有影响。 对于 A 和 B, P (AB) = P (A) 和 P (BA) = P (B) = P (B)。 替换后, 上述公式变成 :

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  FOR INDEPENDENT EVENTS:
  ::独立活动:

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  $\begin{align*}P(A \cap B)& =P(A) P(B) \\ P(A \cap B)& =P(B)P(A)\end{align*}$
  ::P(AB)=P(A)P(B)P(B)P(AB)=P(B)P(A)

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  Note that now, these two formulas are identical.
  ::请注意,现在这两个公式是相同的。

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  In the interactive below, click on the different size and color buttons to see how the mutually inclusive probability of choosing certain marbles from a jar changes. Use the Multiplication Rule to calculate the intersection and compare this with the union .
  ::在下文互动部分,单击不同的大小和颜色按钮,以了解从罐体变化中选择某些大理石的相互包容概率。使用乘法规则计算交叉点并与联盟进行比较。

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  If you don't know whether or not two events are independent or dependent, you can always use the Multiplication Rule for calculating the probability of the intersection of the two events. $\begin{align*}P(A \cap B)= P(A)P(B)\end{align*}$ is just a special case of the Multiplication Rule.
  ::如果您不知道两个事件是独立的还是依赖性的, 您总是可以使用乘法规则来计算两个事件交叉的概率。 P( AB) = P( A)P( B) 只是乘法规则的一个特例。

  播放段落 If the events are... ::如果事情是这样的话...	播放段落 You can use the formula... ::您可以使用公式...
  播放段落 Independent ::独立	$\begin{align*}P(A \cap B)=P(A)P(B)\end{align*}$
  播放段落 Independent or Dependent ::独立或依赖	播放段落 $\begin{align*}P(A \cap B)=P(A|B)P(B)\end{align*}$ ::P(AB)=P(AB)P(B) 播放段落 or ::或 播放段落 $\begin{align*}P(A \cap B)=P(B|A)P(A)\end{align*}$ ::P(AB)=P(BA)P(A)

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  Remember that often you will be able to calculate $\begin{align*}P(A \cap B)\end{align*}$ directly. However, sometimes you will only be given information about the conditional probabilities of the events or the probabilities of the individual events. In those cases, the Multiplication Rule is helpful.
  ::记住您通常能够直接计算 P( AB) 。 但是, 有时您只能得到关于事件有条件概率或个别事件概率的信息。 在这种情况下, 乘法规则很有帮助 。

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  Let's look at a few problems involving intersections
  ::让我们来看看几个交叉问题

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  1. If Mark goes to the store, the probability that he buys ice cream is 30%. The probability that he goes to the store is 10%. What is the probability of him going to the store and buying ice cream?
  ::1. 如果Mark去商店,他买冰淇淋的概率是30%,去商店的概率是10%,去商店买冰淇淋的概率是10%,他去商店买冰淇淋的概率是多少?

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  The first sentence of the problem is a statement of conditional probability. You could restate it as “the probability of Mark buying ice cream given that Mark has gone to the store is 30%”. Let $\begin{align*}S\end{align*}$ be the event that Mark goes to the store. Let $\begin{align*}I\end{align*}$ be the event that Mark buys ice cream. Rewrite the statements and question in the problem in terms of $\begin{align*}S\end{align*}$ and $\begin{align*}I\end{align*}$ :
  ::问题的第一句是有条件概率声明。你可以把它重述为“马克购买冰淇淋的概率,因为马克去店铺的概率是30%”。让马克去店铺的事件为S。让我来一次马克买冰淇淋的事件。用S和I来重写声明和问题:

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  $\begin{align*}P(I|S) &= 30\%=0.30\\ P(S) &= 10\%=0.10\\ P(S \cap I) &= ?\end{align*}$
  ::PIS=300.30P(S)=100.10P(SI)=?

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  In order to figure out the probability of the intersection of the events, use the Multiplication Rule.
  ::为了确定事件交错的概率,请使用乘法规则。

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  $\begin{align*}P(S \cap I)=P(I \cap S) &= P(I|S)P(S)\\ &=( 0.30 )\cdot (0.10)\\ &= .03=3\%\end{align*}$
  ::P(SI) = P(IS) = P(IS) = P(IS) P(S) = (0.30) (0.0.10) =.03= 3%

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  There is a 3% chance that Mark will go to the store and buy ice cream.
  ::马克去商店买冰淇淋的可能性是3%

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  2. Consider the experiment of choosing a card from a deck, keeping it, and then choosing a second card from the deck. Let event $\begin{align*}A\end{align*}$ be getting a diamond in the first draw and event $\begin{align*}B\end{align*}$ be getting a red card in the second draw. Find $\begin{align*}P(B \cap A).\end{align*}$
  ::2. 考虑从甲板上选择一张牌、保留它、然后从甲板上选择第二张牌的实验。让事件A在第一局得到一颗钻石,事件B在第二局得到一张红卡。找到 P(B)A。

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  By the Multiplication Rule, $\begin{align*}P(B \cap A)=P(B|A)P(A).\end{align*}$ Consider each of these probabilities separately.
  ::根据乘法规则,P(B)-A=P(B)-A)-P(A)-P(A)-分别考虑这些概率。

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  • $\begin{align*}P(A)\end{align*}$ is the probability that the first card is a diamond. There are 13 diamonds in the deck of 52 cards, so $\begin{align*}P(A)=\frac{13}{52}.\end{align*}$
    ::P(A) 是第一张卡片是钻石的概率。52张卡片的甲板上有13颗钻石, 所以P(A)=1352。
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  • $\begin{align*}P(B|A)\end{align*}$ is the probability that the second card is a red card given that the first card was a diamond. After the first card was chosen, there are 51 cards left in the deck. 25 of them are red since the first card was a diamond. Therefore, $\begin{align*}P(B|A)=\frac{25}{51}.\end{align*}$
    ::P(BA) 是第二张卡片是红色卡片的概率, 因为第一张卡片是钻石。 在第一张卡片被选中后, 甲板上还剩下51张卡片。 其中25张是红色的, 因为第一张卡片是钻石。 因此, P(BA)=2551。 因此, P(BA)=2551。

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  $\begin{align*}P(B \cap A) &= P(B|A)P(A)\\ &= \frac{25}{51} \cdot \frac{13}{52}\\ &= \frac{25}{204} \approx 12\%\end{align*}$
  ::P(BA) = P(BA) = P(BA) P(A) = 2551/1352= 25204= 12%

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  3. The chance of heavy snow tomorrow is 50%, the chance of strong winds is 40%, and the chance of heavy snow or strong winds is 60%. What is the chance of a blizzard, which is heavy snow and strong winds ?
  ::3. 明天大雪的几率是50%,强风的几率是40%,大雪或强风的几率是60%。 暴风雪的几率是多少,大雪和强风的几率是多少?

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  This question asks for $\begin{align*}P(\text{heavy snow} \cap \text{strong winds}) ; \end{align*}$ however, you were not given any conditional probabilities or indication that the events are independent. Remember that you can sometimes use the to solve for intersection probabilities.
  ::这个问题要求 P( 重雪/ 强风) ; 但是, 您没有得到任何有条件的概率或事件独立的迹象 。 请记住, 有时您可以使用它来解决交叉概率 。

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  $\begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{align*}$
  ::P(AB)=P(A)+P(B)-P(AB)

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  You can rewrite this rule to solve for $\begin{align*}P(A \cap B).\end{align*}$
  ::您可以重写此规则以解析 P( AB) 。

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  $\begin{align*}P(A \cap B)=P(A)+P(B)-P(A \cup B)\end{align*}$
  ::P(AB)=P(A)+P(B)-P(AB)

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  For this problem:
  ::对此问题:

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  $$\begin{align*}P(\text{heavy snow} \cap \text{strong winds}) &= P(\text{heavy snow})+P(\text{strong winds})-P(\text{heavy snow} \cup \text{strong winds})\\ P(\text{heavy snow} \cap \text{strong winds})&= 0.50+0.40-0.60\\ &= 0.30\end{align*}$$
  ::P(重雪强风)=P(重雪)+P(强风)-P(重雪强风)P(重雪强风)P(重雪强风)=0.50+0.40-0.60=0.30

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  The chance of a blizzard is 30%.
  ::暴风雪的几率是30%

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  10% of the emails that Michelle receives are spam emails. Her spam filter catches spam 95% of the time. Her spam filter misidentifies non-spam as spam 2% of the time. ( Let $\begin{align*}A\end{align*}$ be the event that an email is spam and let $\begin{align*}B\end{align*}$ be the event that the spam filter identifies an email as spam. ) What is the probability of an email chosen at random being spam and being correctly identified as spam by her spam filter ?
  ::米歇尔收到的电子邮件中有10%是垃圾邮件。 她的垃圾邮件过滤器在95%的时间里捕捉垃圾邮件。 她的垃圾邮件过滤器错误地将非垃圾邮件误认为2%的时间是垃圾邮件。 (如果电子邮件是垃圾邮件,让B误认为邮件是垃圾邮件,那么垃圾邮件过滤器将电子邮件确定为垃圾邮件的可能性有多大? ) 随机选择的电子邮件是垃圾邮件,并由垃圾邮件过滤器正确识别为垃圾邮件的可能性是多少?

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  The probability that an email chosen at random, is spam and was identified as spam is .
  ::随机选择的电子邮件是垃圾邮件并被识别为垃圾邮件的概率 。

  ---

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  Examples
  ::实例实例实例实例

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  Example 1
  ::例1

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  0.1% of the population is said to have a new disease. A test is developed to test for the disease. 98% of people without the disease will receive a negative test result. 99.5% of people with the disease will receive a positive test result. A random person who was tested for the disease is chosen. What is the probability that the chosen person does not have the disease and got a negative test result?
  ::据说,0.1%的人口患有新疾病。为检测该疾病,进行了测试。98%的无该疾病者将获得负面测试结果。99.5%的患者将获得正测试结果。随机选择了接受该疾病检测的人。被选中的人没有该疾病并获得负测试结果的概率是多少?

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  Let $\begin{align*}A\end{align*}$ be the event that a random person has the disease. Let $\begin{align*}B\end{align*}$ be the event that a random person gets a positive test result. Rewrite the statements and question in the problem in terms of $\begin{align*}A\end{align*}$ and $\begin{align*}B.\end{align*}$
  ::以随机者患上该疾病为例。 以随机者患上该疾病为例。 以随机者获得正测试结果为例, 以 A 和 B 重写声明和问题 。

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  • $\begin{align*}P(A)=0.1\%=0.001\end{align*}$
    ::P(A)=0.10.001
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  • $\begin{align*}P(B^\prime|A^\prime)=98\%=0.98\end{align*}$
    ::P(BA)=980.98
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  • $\begin{align*}P(B|A)=99.5\%=0.995\end{align*}$
    ::P(BA)=99.50.995
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  • $\begin{align*}P(A^\prime \cap B^\prime)=?\end{align*}$
    :PAB)=? (PAB)=?

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  By the Multiplication Rule, $\begin{align*}P(A^\prime \cap B^\prime)=P(B^\prime|A^\prime)P(A^\prime).\end{align*}$ You know $\begin{align*}P(B^\prime|A^\prime)=0.98\end{align*}$ but you were not given $\begin{align*}P(A^\prime)\end{align*}$ directly. $\begin{align*}P(A^\prime)+P(A)=100\%\end{align*}$ because these two events are complements. Therefore, $\begin{align*}P(A^\prime)=99.9\%=0.999.\end{align*}$
  ::根据乘法规则,P(A)B=P(B)A(A)P(A)。你知道P(B)A(A)=0.98,但你没有直接获得P(A)。P(A)+P(A)=100%,因为这两个事件是相辅相成的。因此,P(A)=99.90.999。

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  $\begin{align*}P(A^\prime \cap B^\prime) &= P(B^\prime|A^\prime) P(A^\prime)\\ &= (0.98) \cdot (0.999)\\ &= .97902\\ &= 97.902\%\end{align*}$
  ::P(AB})=P(BAA})P(A})=(0.98)(0.9999)=9.7902=97.902%

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  This means that the probability that a random person who had this test done doesn't have the disease and got a negative test result is 97.902%. Most of the people who took the test for the disease will not have it and will get a negative test result.
  ::这意味着随机进行这种测试的人没有病的概率是97.902 % 。 接受这种测试的大多数人没有病的概率是97.902 % , 测试结果为负。

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  Example 2
  ::例2

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  a) $\begin{align*}P(C|D)=0.8\end{align*}$ and $\begin{align*}P(D)=0.5.\end{align*}$ What is $\begin{align*}P(C \cap D)?\end{align*}$
  :a) P(CD)=0.8和P(D)=0.5。

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  $\begin{align*}P(C \cap D) &= P(C|D)P(D)\\ &= (0.8) \cdot (0.5)\\ &= 0.4\end{align*}$
  ::P(CD)=P(CD)P(D)=(0.8)(0.5)=0.4

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  b) If $\begin{align*}P(D|C)=0.6\end{align*}$ what is $\begin{align*}P(C)?\end{align*}$
  :b) 如果P(DC)=0.6,什么是P(C)?

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  Remember that $\begin{align*}P(C \cap D)=P(C|D)P(D)\end{align*}$ and $\begin{align*}P(C \cap D)=P(D|C)P(C).\end{align*}$ Here, use the second formula.
  ::记住 P( CD) = P( CD) P( D) 和 P( CD) = P( DC) P( C) 。 在此使用第二个公式 。

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  $\begin{align*}P(C \cap D) &= P(D|C)P(C)\\ (0.4 )&= (0.6) \cdot P(C)\\ P(C) & \approx 0.67\end{align*}$
  ::P(CD) = P(DC) P(C) (0.4) = (0.6) P(C) P(C) 0.67

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  CK-12 PLIX Interactive: Additive and Multiplicative Rules for Probability
  ::CK-12 PLIX 互动:关于概率的增倍和倍增规则

  Summary

  播放段落 The probability of a particular dependent event given the outcome of the event on which it occurs. To calculate the conditional probability of A given B: $$\begin{align*}&P(A\cap B)=P(A|B)P(B) && \text{Independent or Dependent}\\ &P(A\cap B)=P(A)P(B) && \text{Independent}\\\end{align*}$$   ::特定依赖事件概率, 取决于事件发生结果的概率。 要计算给定 B 的有条件概率 : P( AB) = P( AB) = P( AB) P( B) 独立或依赖 P( AB) = P( A) P( A) P( B) 独立

  ---

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  Review
  ::审查审查审查审查

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  1. What is the Multiplication Rule and when is it used?
  ::1. 乘法规则是什么,何时使用?

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  2. If events $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ are disjoint, what is $\begin{align*}P(A \cap B)?\end{align*}$
  ::2. 如果事件A和B脱节,什么是P(AB)?

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  3. If events $\begin{align*}A\end{align*}$ and $\begin{align*}B\end{align*}$ are independent, what is $\begin{align*}P(A \cap B)?\end{align*}$
  ::3. 如果事件A和事件B是独立的,那么P(AB)是什么?

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  4. Why does the Multiplication Rule work for events whether or not they are dependent?
  ::4. 为什么《乘法规则》适用于事件,而不论事件是否依附关系?

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  5. $\begin{align*}P(C)=0.4\end{align*}$ and $\begin{align*}P(B|C)=0.2\end{align*}$ . What is $\begin{align*}P(C \cap B)?\end{align*}$
  ::5. P(C)=0.4和P(BC)=0.2. 什么是P(CB)?

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  6. $\begin{align*}P(A)=0.5, \ P(B)=0.7, \ P(B|A)=0.6.\end{align*}$ What is $\begin{align*}P(A|B)?\end{align*}$ Hint: first find $\begin{align*}P(A \cap B).\end{align*}$
  
  ::6. P(A)=0.5,P(B)=0.7,P(B)=0.6。P(A)=0.6是什么?Hint:首先找到P(A)B。

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  7. $\begin{align*}P(D)=0.3, \ P(E)=0.9, \ P(E|D)=0.7.\end{align*}$ What is $\begin{align*}P(D|E)?\end{align*}$
  ::7. P(D)=0.3,P(E)=0.9,P(E)=0.7。

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  0.05% of the population is said to have a new disease. A test is developed to test for the disease. 97% of people without the disease will receive a negative test result. 99% of people with the disease will receive a positive test result. A random person who was tested for the disease is chosen.
  ::据说,0.05%的人口患有新疾病,为检测该疾病而进行了测试。 97%的未患该疾病的人将获得负面检测结果。 99%的患者将获得阳性检测结果。随机选择了接受该疾病检测的人。

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  8. What is the probability that the chosen person does not have the disease?
  ::8. 被选中的人没有这种疾病的概率是多少?

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  9. What is the probability that the chosen person does not have the disease and received a negative test result?
  ::9. 被选中的人没有这种疾病并获得负面测试结果的概率有多大?

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  10. What is the probability that the chosen person does have the disease and received a negative test result?
  ::10. 被选中的人确患该疾病并获得负测试结果的概率是多少?

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  11. If 1,000,000 people were given the test, how many of them would you expect to have the disease but receive a negative test result?
  ::11. 如果对1 000 000人进行检测,其中有多少人可能患上这种疾病,但得到负面检测结果?

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  12. If Kaitlyn goes to the store, the probability that she buys blueberries is 90%. The probability of her going to the store is 30%. What is the probability of her going to the store and buying blueberries?
  ::12. 如果Kaitlyn去商店,她购买蓝莓的概率为90%,去商店的概率为30%,去商店买蓝莓的概率为30%。

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  13. For three events $\begin{align*}A,B\end{align*}$ and $\begin{align*}C\end{align*}$ , show that $\begin{align*}P(A \cap B \cap C)=P(C|A \cap B)P(A \cap B).\end{align*}$
  ::13. A、B和C三场活动表明P(ABC)=P(CAB)P(AB)。

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  14. Using your answer to #14, show that $\begin{align*}P(A \cap B \cap C)=P(C|A \cap B)P(B|A)P(A).\end{align*}$
  ::14. 使用您对 #14 的回答, 显示 P (ABC) = P (CAB) P (BA) P (A)。

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  15. On rainy weekend days, the probability that Karen bakes bread is 90%. On the weekend, the probability of rain is 50%. There is a 29% chance that today is a weekend day. What is the probability that today is a rainy weekend day in which Karen is baking bread?
  ::15. 在雨天周末,Karen烤面包的概率是90%;在周末,降雨的概率是50%;在今天是周末,29%的概率是29%;今天是Karen烤面包的雨季,其概率是多少?

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  16. A piece of fruit is chosen at random from a fruit basket containing 6 apples and 7 bananas. If event $\begin{align*}A\end{align*}$ represents choosing and apple and event $\begin{align*}B\end{align*}$ represents choosing a banana, then determine:
  ::16. 从含有6个苹果和7个香蕉的水果篮中随机选取一块水果,如果事件A代表选择,而事件B代表选择香蕉,则确定:

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  1. $\begin{align*}P(A)\end{align*}$
     ::P(A)项
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  2. $\begin{align*}P(B)\end{align*}$
     ::P(B) P(B)
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  3. $\begin{align*}P ( A \cup B )\end{align*}$
     ::PA__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  4. $\begin{align*}P ( A \cap B )\end{align*}$
     ::PA__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  5. Are the two event disjoint?
     ::这两件事是不是脱节了?

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  17. A box contains 3 jelly donuts, 5 chocolate donuts, and 7 sour cream glazed donuts. If a person selects a donut randomly from the box, what is the probability that they will choose a jelly donut or a sour cream glazed donut? Are these events disjoint?
  ::17. 盒装3个果冻甜甜圈、5个巧克力甜甜圈和7个酸奶油甜甜圈。 如果一个人随机从盒子中选择一个甜甜圈,那么他们选择果冻甜甜圈或酸奶油甜甜圈的概率是多少?这些事件是否脱节?

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  18. 50 students went on a class trip. Of the two choices of activities one day, 40 decided to go white water rafting and 25 went parasailing. Each student signed up for at least one of these two activities.
  ::18. 50名学生参加了班级旅行,在一天的两次活动选择中,40人决定参加白流水活动,25人参加寄生,每个学生至少报名参加其中一次活动。

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  1. Draw a Venn diagram to find out how many students did both of these activities.
     ::绘制一份文恩图,以了解有多少学生参加了这两项活动。
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  2. How many students went white water rafting but not parasailing?
     ::有多少学生在白水中游荡,但没有在水中游荡?
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  3. How many students went parasailing given that they went white water rafting?
     ::有多少学生去寄生 因为他们去白水漂流?

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  Review (Answers)
  ::审查(答复)

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  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 。