问题如下图：
h、i不懂
orange品职答疑助手 · 2020年02月04日
h：A的补集并上D的补集，相当于既不在A中也不在D中的交集，所以是15%+7%+6%+15%（其中15%是要自己算的，它等于不在abcd内的概率，要自己根据1-图中所有概率 来自己算出）
i：问这四个事件中，任选两个事件，这一对事件会不会是独立事件。只有A事件与B事件 组成的这一对事件满足。因为P(A)*P(B)=P(AB)
比如世界 · 2020年03月30日
其中15%是要自己算的，它等于不在abcd内的概率，要自己根据1-图中所有概率 来自己算出，这个是什么意思啊？不在abcd内的概率是啥，不存在吗？老师麻烦再讲一下，谢谢
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
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？
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？
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
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