开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

返回

8527 · 2021年12月22日

b 题型

  1. FRM I
  2. Quantitative

NO.PZ2020010301000005

问题如下：

Based on the probabilities in the plot below, what are the values of the following?

a. Pr(AC)Pr(A^C)

b. Pr(D|A∪B∪C)

c. Pr(A|A)

d. Pr(B|A)

e. Pr(C|A)

f. Pr(D|A)

g. Pr((A∪D)C)Pr({(A∪D)}^C)

h. Pr((AC⋂DC))Pr((A^C\bigcap D^C))

i. Are any of the four events pairwise independent?

选项：

解释：

a. 1 - Pr(A) = 100% - 30% = 70%

b. This value is Pr(D∩(A∪B∪C))/Pr(A∪B∪C). The total probability in the three areas A, B, and C is 73%. The overlap of D with these three is 9% + 8% + 7% = 24%, and so the conditional probability is 24%/73%= 33%.

c. This is trivially 100%.

d. Pr(B∩A) = 9%. The conditional probability is 9%/30% = 30%.

e. There is no overlap and so Pr(C∩A) = 0.

f. Pr(D∩A) = 9%. The conditional probability is 30%.

g. This is the total probability not in A or D. It is 1 – Pr(A∪D) = 1 - (Pr(A) + Pr(D) - Pr(A∩D)) = 100% - (30% + 36% - 9%) = 43%.

h. This area is the intersection of the space not in A with the space not in D. This area is the same as the area that is not in A or D, Pr((A∪D)C)Pr({(A\cup D)}^C) and so 43%.

i. The four regions have probabilities A = 30%, B = 30%, C = 28% and D = 36%. The only region that satisfied the requirement that the joint probability is the product of the individual probabilities is A and B because Pr(A∩B) = 9% = Pr(A)Pr(B) = 30% * 30%.

请问像b小题这种类型在视频里哪里有讲到

添加评论
8527 · 2021年12月22日

根据这道题，麻烦老师再回答一下两个问题： 1. 请问 P(A n B) 的意思是 P（A|B), 还是A和B两合集中intersection的部分 ->P(AB). 如果答案是后者的话，为什么b小题的答案中‘’Pr(D|A∪B∪C)‘’ 等于 Pr(D∩(A∪B∪C))/Pr(A∪B∪C)。（n等于conditional ‘|’的概念） 麻烦老师解释一下什么是pairwise independent。 谢谢~！

1 个答案
已采纳答案

DD仔_品职助教 · 2021年12月22日

嗨，从没放弃的小努力你好：


同学你好，

b问是贝叶斯公式的变形，在基础班section1最后一个视频有讲。


1，P(AnB)A交B，代表他俩的交集。P(AuB)A并B，代表并集，所有面积，概率如下图绿色部分所示：

P(A|B)表示在已知A发生的概率下B发生的概率，具体计算公式有下面两种，这道题考的是第二个公式，第二个公式讲义里没给，补充记忆一下即可：

Pr(D|A∪B∪C)= Pr(D∩(A∪B∪C))/Pr(A∪B∪C)=D与ABC并集的交集/ABC的并集=24%/73%= 33%.


2，pairwise independent的英文解释是each event is independent of of every other possible combination of paired events.

翻译一下这个题中的ABCD这四个事件，我们任选两个事件，也就是一对，这对事件相对于其他时间，是不是独立的。

问的还是独立不独立，只不过针对的是两个事件。


----------------------------------------------
努力的时光都是限量版，加油！

  • 1

    回答

  • 0

    关注

  • 489

    浏览
相关问题

NO.PZ2020010301000005 问题如下 Baseon the probabilities in the plot below, whare the values of the following?Pr(AC)Pr(A^C)Pr(AC)Pr(A∪B∪C)Pr(A|A) Pr(B|A)e. Pr(C|A)f. Pr(A)g. Pr((A∪C)Pr({(A∪}^C)Pr((A∪C)h. Pr((AC⋂))Pr((A^C\bigcC))Pr((AC⋂))i. Are any of the four events pairwise inpennt? 1 - Pr(= 100% - 30% = 70% This value is Pr((A∪B∪C))/Pr(A∪B∪C). The totprobability in the three areanC is 73%. The overlof with these three is 9% + 8% + 7% = 24%, anso the contionprobability is 24%/73%= 33%.This is trivially 100%. Pr(B∩= 9%. The contionprobability is 9%/30% = 30%.e. There is no overlanso Pr(C∩= 0. f. Pr(= 9%. The contionprobability is 30%.g. This is the totprobability not in A or It is 1 – Pr(A∪ = 1 - (Pr(+ Pr( - Pr(A∩) = 100% - (30% + 36% - 9%) = 43%.h. This area is the intersection of the spanot in A with the spanot in This area is the same the area this not in A or Pr((A∪C)Pr({(A\cup }^C)Pr((A∪anso 43%.i. The four regions have probabilities A = 30%, B = 30%, C = 28% an= 36%. The only region thsatisfiethe requirement ththe joint probability is the proof the inviprobabilities is A anB because Pr(A∩= 9% = Pr(A)Pr(= 30% * 30%. 如题，第i小题怎么理解，谢谢

2024-02-25 21:46 1 · 回答

NO.PZ2020010301000005 问题如下 Baseon the probabilities in the plot below, whare the values of the following?Pr(AC)Pr(A^C)Pr(AC)Pr(A∪B∪C)Pr(A|A) Pr(B|A)e. Pr(C|A)f. Pr(A)g. Pr((A∪C)Pr({(A∪}^C)Pr((A∪C)h. Pr((AC⋂))Pr((A^C\bigcC))Pr((AC⋂))i. Are any of the four events pairwise inpennt? 1 - Pr(= 100% - 30% = 70% This value is Pr((A∪B∪C))/Pr(A∪B∪C). The totprobability in the three areanC is 73%. The overlof with these three is 9% + 8% + 7% = 24%, anso the contionprobability is 24%/73%= 33%.This is trivially 100%. Pr(B∩= 9%. The contionprobability is 9%/30% = 30%.e. There is no overlanso Pr(C∩= 0. f. Pr(= 9%. The contionprobability is 30%.g. This is the totprobability not in A or It is 1 – Pr(A∪ = 1 - (Pr(+ Pr( - Pr(A∩) = 100% - (30% + 36% - 9%) = 43%.h. This area is the intersection of the spanot in A with the spanot in This area is the same the area this not in A or Pr((A∪C)Pr({(A\cup }^C)Pr((A∪anso 43%.i. The four regions have probabilities A = 30%, B = 30%, C = 28% an= 36%. The only region thsatisfiethe requirement ththe joint probability is the proof the inviprobabilities is A anB because Pr(A∩= 9% = Pr(A)Pr(= 30% * 30%. H这个小问如果不用逻辑用计算的话要怎么算呢？

2023-10-09 23:27 1 · 回答

NO.PZ2020010301000005 问题如下 Baseon the probabilities in the plot below, whare the values of the following?Pr(AC)Pr(A^C)Pr(AC)Pr(A∪B∪C)Pr(A|A) Pr(B|A)e. Pr(C|A)f. Pr(A)g. Pr((A∪C)Pr({(A∪}^C)Pr((A∪C)h. Pr((AC⋂))Pr((A^C\bigcC))Pr((AC⋂))i. Are any of the four events pairwise inpennt? 1 - Pr(= 100% - 30% = 70% This value is Pr((A∪B∪C))/Pr(A∪B∪C). The totprobability in the three areanC is 73%. The overlof with these three is 9% + 8% + 7% = 24%, anso the contionprobability is 24%/73%= 33%.This is trivially 100%. Pr(B∩= 9%. The contionprobability is 9%/30% = 30%.e. There is no overlanso Pr(C∩= 0. f. Pr(= 9%. The contionprobability is 30%.g. This is the totprobability not in A or It is 1 – Pr(A∪ = 1 - (Pr(+ Pr( - Pr(A∩) = 100% - (30% + 36% - 9%) = 43%.h. This area is the intersection of the spanot in A with the spanot in This area is the same the area this not in A or Pr((A∪C)Pr({(A\cup }^C)Pr((A∪anso 43%.i. The four regions have probabilities A = 30%, B = 30%, C = 28% an= 36%. The only region thsatisfiethe requirement ththe joint probability is the proof the inviprobabilities is A anB because Pr(A∩= 9% = Pr(A)Pr(= 30% * 30%. 但从图上看 非a和非b的交集就应该是b和c两个圆里7%+6%+15%的部分啊为什么和g算出来的值不一样？

2023-06-09 03:01 1 · 回答

NO.PZ2020010301000005问题如下Baseon the probabilities in the plot below, whare the values of the following?Pr(AC)Pr(A^C)Pr(AC)Pr(A∪B∪C)Pr(A|A) Pr(B|A)e. Pr(C|A)f. Pr(A)g. Pr((A∪C)Pr({(A∪}^C)Pr((A∪C)h. Pr((AC⋂))Pr((A^C\bigcC))Pr((AC⋂))i. Are any of the four events pairwise inpennt? 1 - Pr(= 100% - 30% = 70% This value is Pr((A∪B∪C))/Pr(A∪B∪C). The totprobability in the three areanC is 73%. The overlof with these three is 9% + 8% + 7% = 24%, anso the contionprobability is 24%/73%= 33%.This is trivially 100%. Pr(B∩= 9%. The contionprobability is 9%/30% = 30%.e. There is no overlanso Pr(C∩= 0. f. Pr(= 9%. The contionprobability is 30%.g. This is the totprobability not in A or It is 1 – Pr(A∪ = 1 - (Pr(+ Pr( - Pr(A∩) = 100% - (30% + 36% - 9%) = 43%.h. This area is the intersection of the spanot in A with the spanot in This area is the same the area this not in A or Pr((A∪C)Pr({(A\cup }^C)Pr((A∪anso 43%.i. The four regions have probabilities A = 30%, B = 30%, C = 28% an= 36%. The only region thsatisfiethe requirement ththe joint probability is the proof the inviprobabilities is A anB because Pr(A∩= 9% = Pr(A)Pr(= 30% * 30%. i的思考过程是什么，只能通过计算式逐一验证吗

2023-05-27 21:31 1 · 回答

NO.PZ2020010301000005 问题如下 Baseon the probabilities in the plot below, whare the values of the following?Pr(AC)Pr(A^C)Pr(AC)Pr(A∪B∪C)Pr(A|A) Pr(B|A)e. Pr(C|A)f. Pr(A)g. Pr((A∪C)Pr({(A∪}^C)Pr((A∪C)h. Pr((AC⋂))Pr((A^C\bigcC))Pr((AC⋂))i. Are any of the four events pairwise inpennt? 1 - Pr(= 100% - 30% = 70% This value is Pr((A∪B∪C))/Pr(A∪B∪C). The totprobability in the three areanC is 73%. The overlof with these three is 9% + 8% + 7% = 24%, anso the contionprobability is 24%/73%= 33%.This is trivially 100%. Pr(B∩= 9%. The contionprobability is 9%/30% = 30%.e. There is no overlanso Pr(C∩= 0. f. Pr(= 9%. The contionprobability is 30%.g. This is the totprobability not in A or It is 1 – Pr(A∪ = 1 - (Pr(+ Pr( - Pr(A∩) = 100% - (30% + 36% - 9%) = 43%.h. This area is the intersection of the spanot in A with the spanot in This area is the same the area this not in A or Pr((A∪C)Pr({(A\cup }^C)Pr((A∪anso 43%.i. The four regions have probabilities A = 30%, B = 30%, C = 28% an= 36%. The only region thsatisfiethe requirement ththe joint probability is the proof the inviprobabilities is A anB because Pr(A∩= 9% = Pr(A)Pr(= 30% * 30%. 如第i题

2023-04-26 17:14 1 · 回答