问题如下图：
]g.h题目我都看不懂，上课老师好像没有讲这个\是什么意思
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
