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

小熊猫 · 2023年06月09日

h问从图上看答案为什么不一样?

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((AD)C)Pr({(A∪D)}^C)

h. Pr((ACDC))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((AD)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%.

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?

1 个答案

李坏_品职助教 · 2023年06月09日

嗨,努力学习的PZer你好:


g和h问的结果是一样的,g和h这两问是等价的。


首先注意A+ B+C+D并不等于全集,全集在ABCD之外还有一些空白区域。你用7%+6%+15%只是非a和非b交集的一部分,空白区域被你漏掉了。


所以g问:1-P(A U D) = 1-(30%+36%-9%)= 43%

h问:非A且非D = 1-P(A U D) =43%

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

  • 1

    回答
  • 0

    关注
  • 306

    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%. 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