12.14 独立、有条件和相互排斥的活动

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section 独立、有条件和相互排斥的活动

  Collapse Expand

  独立、有条件和相互排斥的活动

  播放段落

  Freezy's Ice Cream Stand doesn't think it has enough information to decide if it should add Pumpernickel Brickel or Dandy Cotton Candy to its menu. Therefore, it conducts another poll of its customers. It finds that the probability that a customer will like both flavors is 0.33 and the probability that a customer will like the Cotton Candy flavor is 0.8. What is the probability a customer will like the Pumpernickel flavor?
  ::Freezy的冰淇淋冰淇淋站认为它没有足够的信息来决定它是否应该在其菜单中添加 pumpernickel Brickel 或 Dandy Cotton Candy 。 因此, 它会对其客户进行另一次民意测验。 它发现顾客喜欢两种口味的概率是0.33, 顾客喜欢棉花糖果口味的概率是0.8。 顾客喜欢松饼口味的概率是多少?

  播放段落

  Independent Conditional and Mutually Exclusive Events
  ::独立有条件和相互排斥的活动

  播放段落

  We have already discussed the formula for conditional probabilities: $\begin{array}{r}P(A|B)=\frac{P(A\cap B)}{P(B)}\end{array}$ . These events are not independent, they are conditional because the outcome of event $\begin{array}{r}B\end{array}$ affects the outcome of event $\begin{array}{r}A\end{array}$ . When events are independent, $\begin{array}{r}P(A|B)=P(A)\end{array}$ , meaning that it doesn’t matter that event $\begin{array}{r}B\end{array}$ has occurred, the result of event $\begin{array}{r}B\end{array}$ does not affect the result of event $\begin{array}{r}A\end{array}$ . Now we can replace $\begin{array}{r}P(A|B)\end{array}$ with $\begin{array}{r}\frac{P(A\cap B)}{P(B)}\end{array}$ in the previous statement to get $\begin{array}{r}\frac{P(A\cap B)}{P(B)}=P(A)\end{array}$ . Finally, multiply both sides by $\begin{array}{r}P(B)\end{array}$ to get $\begin{array}{r}P(A\cap B)=P(A)\times P(B)\end{array}$ for $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ . We can use this rule to determine if events are independent or to find the intersection of known independent events.
  ::我们已经讨论过有条件概率的公式:P(AB) = P(AB) = P(AB) P(B) 。这些事件不独立,是有条件的,因为事件B的结果影响事件A的结果。当事件独立时,P(AB) = P(A) = P(A) ,这意味着事件B已经发生并不重要,事件B的结果并不影响事件A的结果。现在,我们可以在上一个语句中用P(AB) = P(B) = P(A) 。最后,用P(B) 乘两边来获得事件P(AB) = P(A) x(B) 用于 A和 B。我们可以使用这一规则来确定事件是否独立或找到已知独立事件的交叉点。

  播放段落

  It is also possible for two events to have not intersection, or $\begin{array}{r}P(A\cap B)=0\end{array}$ . When this occurs we say that the events (or sets) are Mutually Exclusive . If one has occurred, then the other cannot occur. Examples of Mutually Exclusive sets are Boys and Girls, Juniors and Seniors-it is not possible to be both. It is important to note here that cannot be independent unless the probability of one of the events is zero since for independent events $\begin{array}{r}P(A\cap B)=P(A)\times P(B)\end{array}$ and the only way a product can equal zero is if one of the factors is equal to zero.
  ::也可以有两个事件没有交叉点, 或者 P( AB) = 0。 当发生这种情况时, 我们就会说事件( 或组) 是相互排斥的。 如果已经发生, 那么另一个就不能发生。 相互排斥的组别是男孩和女孩、 少年和老年人不可能同时出现。 在此必须指出的是, 除非其中一个事件的概率是零, 因为独立事件P( AB) = P( A) xP( B) , 而产品等于零的唯一方式是其中一个因素等于零, 否则无法独立 。

  播放段落

  Let's solve the following problems.
  ::让我们解决以下的问题。

  播放段落
  1. Given two events, $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ , such that $\begin{array}{r}P(A)=0.3\end{array}$ , $\begin{array}{r}P(B)=0.5\end{array}$ and $\begin{array}{r}P(A\cup B)=0.65\end{array}$ , find $\begin{array}{r}P(A\cap B)\end{array}$ .
     ::根据两项活动,A和B,例如P(A)=0.3,P(B)=0.5和P(AB)=0.65,见P(AB)。

  播放段落

  Since we are not told that the events are independent, we do not know that $\begin{array}{r}P(A\cap B)=P(A)\times P(B)\end{array}$ . However, for all events, independent or otherwise, it is true that $\begin{array}{r}P(A)+P(B)-P(A\cap B)=P(A\cup B)\end{array}$ so
  ::由于我们没有被告知事件是独立的,我们不知道P(AB)=P(A)x(B)x(B).然而,对于所有事件,无论独立与否,P(A)+P(B)-P(AB)=P(A)-B确实如此。

  播放段落

  $$\begin{array}{rl}0.3+0.5-P(A\cap B)& =0.65\\ P(A\cap B)& =0.15\end{array}$$

  
  ::0.3+0.5-P(AB)=0.65P(AB)=0.15

  播放段落

  State with a reason whether the events are independent.
  ::国家有理由说明事件是否独立。

  播放段落

  To determine if the events are independent we will test the rule $\begin{array}{r}P(A\cap B)=P(A)\times P(B)\end{array}$ .
  ::为了确定事件是否独立,我们将测试规则P(AB)=P(A)×P(B)。

  播放段落

  $$\begin{array}{r}P(A)\times P(B)=0.3\times 0.5=0.15=P(A\cap B).\end{array}$$

  
  ::P(A)×P(B)=0.3×0.5=0.15=P(AB)。

  播放段落

  Thus, the events are independent since the product of their probabilities is equal to the probability of their intersection.
  ::因此,这些事件是独立的,因为其概率的产物与其交叉的概率相等。

  播放段落

  State with a reason whether the events are mutually exclusive.
  ::国家有理由说明事件是否相互排斥。

  播放段落

  The events are not mutually exclusive because $\begin{array}{r}P(A\cap B)=0.15\ne 0\end{array}$ .
  ::这些事件并不相互排斥,因为P(AB)=0.150。

  播放段落
  2. Given that $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent events such that $\begin{array}{r}P(A)=0.4\end{array}$ and $\begin{array}{r}P(A\cup B)=0.76\end{array}$ , find  $\begin{array}{r}P(B)\end{array}$ .
     ::鉴于A和B是独立事件,例如P(A)=0.4和P(AB)=0.76,找到P(B)。

  播放段落

  Since we know the two events are independent, we know that $\begin{array}{r}P(A\cap B)=0.4P(B)\end{array}$ . Now we can use the formula for the probability of the union of the two sets and substitute this product for the probability of the intersection:
  ::由于我们知道这两个事件是独立的,我们知道P(AB)=0.4P(B)。现在我们可以使用两种情况结合概率的公式,用这一产品替代交叉概率:

  播放段落

  $$\begin{array}{rl}0.4+P(B)-0.4P(B)& =0.76\\ 0.6P(B)& =0.36\\ P(B)& =0.6\end{array}$$

  
  ::0.4+P(B)-0.4P(B)=0.760.6P(B)=0.36P(B)=0.6

  播放段落

  Probability of $\begin{array}{r}A\end{array}$ or $\begin{array}{r}B\end{array}$ but not both occurring.
  ::A或B的概率,但不同时发生。

  播放段落

  To find the probability of either $\begin{array}{r}A\end{array}$ or $\begin{array}{r}B\end{array}$ occurring but not both, we need to find $\begin{array}{r}P(A\cap B)\end{array}$ and subtract this from $\begin{array}{r}P(A\cup B)\end{array}$ .
  ::要发现发生A或B的概率,但不能同时发生,我们需要找到P(AB),并从P(AB)中减去。

  $$\begin{array}{r}0.76-(0.4)(0.6)=0.76-0.24=0.52\end{array}$$

  播放段落
  3. Events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent such that $\begin{array}{r}P(B\cap A^{\mathrm{\prime }})=0.2\end{array}$ and $\begin{array}{r}P(A\cap B)=0.3\end{array}$ . Find $\begin{array}{r}P(A\cup B)\end{array}$ .
     ::A和B事件是独立的,例如P(B) =0.2和P(A) =0.3.。

  播放段落

  For this problem, a might be useful to illustrate the given information.
  ::对于这个问题,也许有必要说明所提供的资料。

  

  播放段落

  From the diagram we can see $\begin{array}{r}P(B)=0.5\end{array}$ and since we know that the events are independent, we know:
  ::从图表中我们可以看到P(B)=0.5,而且由于我们知道事件是独立的,我们知道:

  播放段落

  $$\begin{array}{rl}P(A)\times P(B)& =P(A\cap B)\\ P(A)\times 0.5& =0.3\\ P(A)& =\frac{0.3}{0.5}=0.6\end{array}$$

  
  ::P(A)×P(B)=P(AB)P(A)×0.5=0.3P(A)=0.30.5=0.6

  播放段落

  Now, $\begin{array}{r}P(A\cup B)=0.6+0.5-0.3=0.8\end{array}$ .
  ::现在,P(AB)=0.6+0.5-0.3=0.8。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked to find  the probability that a customer will like the Pumpernickel flavor. 
  ::早些时候,你被要求 找到一个顾客 喜欢南瓜尼可口味的概率。

  播放段落

  Since we know liking the Pumpernickel flavor (B) and liking the Cotton Candy flavor (A) are independent events, we also know that $\begin{array}{r}P(A\cap B)=0.33P(B)\end{array}$ . Now we can use this formula for the probability of the intersection of the two sets to find the information we're seeking.
  ::由于我们知道喜欢南瓜尼可口味(B)和棉花糖果口味(A)是独立事件,我们也知道P(AB)=0.33P(B)。现在我们可以用这个公式来计算两组的交叉概率,以找到我们所寻求的信息。

  播放段落

  $$\begin{array}{r}0.33=0.8P(B)\\ P(B)& =0.4125\end{array}$$

  
  ::0.33=0.8P(B)P(B)=0.4125

  播放段落

  Therefore, the probability that a customer will like the Cotton Candy flavor is 41.25%.
  ::因此,顾客喜欢棉花糖果的口味的概率是41.25%。

  播放段落

  Example 2
  ::例2

  播放段落

  Given two events, $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ , such that $\begin{array}{r}P(A)=0.4\end{array}$ , $\begin{array}{r}P(B)=0.5\end{array}$ and $\begin{array}{r}P(A\cup B)=0.75\end{array}$
  ::根据两个活动,A和B,如P(A)=0.4,P(B)=0.5和P(AB)=0.75

  播放段落

   Find $\begin{array}{r}P(A\cap B)\end{array}$ .
  ::查找 P( AB) 。

  播放段落

  
  

  $$\begin{array}{rl}0.4+0.5-P(A\cap B)& =0.75\\ P(A\cap B)& =0.15\end{array}$$

  
  ::0.4+0.5-P(AB)=0.75P(AB)=0.15

  播放段落

  State with a reason whether the events are independent.
  ::国家有理由说明事件是否独立。

  播放段落

  If the events are independent, $\begin{array}{r}P(A\cap B)=0.4\times 0.5=0.2\end{array}$ .
  ::如果事件是独立的,P(AB)=0.4×0.5=0.2。

  播放段落

  Since $\begin{array}{r}0.2\ne 0.15\end{array}$ , the events are not independent.
  ::自0.20.0.15以来,事件并不独立。

  播放段落

  Find $\begin{array}{r}P(A|B)\end{array}$ .
  ::查找 P( AB) 。

  播放段落

    $\begin{array}{r}P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.15}{0.5}=0.3\end{array}$ .
  ::P(AB) = P(AB) P(B) = 0.1.50.50 = 0.3。

  播放段落

  Example 3
  ::例3

  播放段落

  Given that $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent events such that $\begin{array}{r}P(A)=0.8\end{array}$ and $\begin{array}{r}P(A\cup B)=0.9\end{array}$ , find
  ::鉴于A和B是独立事件,P(A)=0.8和P(AB)=0.9,

  播放段落

  $\begin{array}{r}P(B)\end{array}$
  ::P(B) P(B)

  播放段落

  
  

  $$\begin{array}{rl}0.8+P(B)-0.8P(B)& =0.9\\ 0.2P(B)& =0.1\\ P(B)& =0.5\end{array}$$

  
  ::0.8+P(B)-0.8P(B)=0.90.2P(B)=0.1P(B)=0.5

  播放段落

    $\begin{array}{r}P(B\cap A^{\mathrm{\prime }})\end{array}$
  :B_A_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  $\begin{array}{r}P(B\cap A^{\mathrm{\prime }})=P(B)-P(B\cap A)=0.5-0.8\times 0.5=0.1\end{array}$
  ::P(BA})=P(B)-P(BA)=0.5-0.8×0.5=0.1

  播放段落

  Example 4
  ::例4

  播放段落

  Events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent such that $\begin{array}{r}P(A\cap B^{\mathrm{\prime }})=0.25\end{array}$ and $\begin{array}{r}P(A\cap B)=0.25\end{array}$ . Find $\begin{array}{r}P(A\cup B)\end{array}$ .
  ::A和B事件是独立的,例如P(AB)=0.25和P(AB)=0.25。

  播放段落

  $\begin{array}{r}P(A)=P(A\cap B^{\mathrm{\prime }})+P(A\cap B)=0.25+0.25=0.5\end{array}$
  ::P(A) = P(A) = P(AB) + P(AB) = 0.25+0.25= 0.5

  播放段落

  $\begin{array}{r}P(A\cap B)=0.5P(B)=0.25\end{array}$ and thus $\begin{array}{r}P(B)=0.5\end{array}$
  ::P(AB)=0.5P(B)=0.25,因此P(B)=0.5

  播放段落

  $\begin{array}{r}P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.5-0.25=0.75\end{array}$
  ::P(AB)=P(A)+P(B)-P(AB)=0.5+0.5-0.25=0.75

  播放段落

  Review
  ::回顾

  播放段落
  1. Events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are mutually exclusive. Describe $\begin{array}{r}P(A|B)\end{array}$ .
     ::A和B事件是相互排斥的。
  播放段落
  2. Events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent. Show that $\begin{array}{r}P(B)=P(B|A)\end{array}$ .
     ::事件 A 和 B 是独立的。 显示 P( B) = P( BA) 。

  播放段落

  For problems 3-6, use the given information about events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ to determine whether or not the events are independent.
  ::对于问题3-6,使用关于事件A和事件B的既定信息来确定事件是否独立。

  播放段落
  3. $\begin{array}{r}P(A)=0.6\end{array}$ , $\begin{array}{r}P(B)=0.4\end{array}$ and $\begin{array}{r}P(A\cup B)=0.76\end{array}$
     ::P(A)=0.6,P(B)=0.4和P(AB)=0.76
  播放段落
  4. $\begin{array}{r}P(A)=0.5\end{array}$ , $\begin{array}{r}P(A\cap B)=0.2\end{array}$ and $\begin{array}{r}P(A\cup B)=0.8\end{array}$
     ::P(A)=0.5,P(AB)=0.2和P(AB)=0.8
  播放段落
  5. $\begin{array}{r}P(A)=0.3\end{array}$ , $\begin{array}{r}P(B)=0.4\end{array}$ and $\begin{array}{r}P(A\cup B)=0.55\end{array}$
     ::P(A)=0.3,P(B)=0.4和P(AB)=0.55
  播放段落
  6. $\begin{array}{r}P(A)=0.6\end{array}$ , $\begin{array}{r}P(B\cap A^{\mathrm{\prime }})=0.28\end{array}$ and $\begin{array}{r}P(A\cap B)=0.42\end{array}$
     ::P(A)=0.6;P(B)=0.28;P(A)=0.42

  播放段落

  For problems 7-10, events $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ are independent.
  ::对于7-10问题,事件A和B是独立的。

  播放段落
  7. Given $\begin{array}{r}P(A)=0.8\end{array}$ and $\begin{array}{r}P(A\cup B)=0.88\end{array}$ , find $\begin{array}{r}P(B)\end{array}$ and $\begin{array}{r}P(AorB\text{but not both})\end{array}$ .
     ::给定P(A)=0.8和P(AB)=0.88,找到P(B)和P(A或B),但不包括两者。
  播放段落
  8. Given $\begin{array}{r}P(A\cap B^{\mathrm{\prime }})=0.54\end{array}$ and $\begin{array}{r}P(A\cap B)=0.36\end{array}$ , find $\begin{array}{r}P(B)\end{array}$ and $\begin{array}{r}P(A\cup B)\end{array}$ .
     ::参考P(AB)=0.54和P(AB)=0.36,参见P(B)和P(AB)。
  播放段落
  9. Given $\begin{array}{r}P(B)=0.8\end{array}$ and $\begin{array}{r}P(A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }})=0.04\end{array}$ , find $\begin{array}{r}P(A)\end{array}$ and $\begin{array}{r}P(A^{\mathrm{\prime }}\cup B^{\mathrm{\prime }})\end{array}$ .
     ::鉴于P(B)=0.8和P(AB)=0.04,请见P(A)和P(AB)。
  播放段落
  10. $\begin{array}{r}P(A\cap B)=0.28\end{array}$ and $\begin{array}{r}P(A\cup B)=0.82\end{array}$ , find $\begin{array}{r}P(A)\end{array}$ and $\begin{array}{r}P(B)\end{array}$ .
      ::P(AB)=0.28和P(AB)=0.82,见P(A)和P(B)。

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。