QuoteOriginally posted by: PaolosThe table below (basically the same table of Vit2007) partitions the sample space---------------------------------------------------------------------------------------....Sat - Sun.............perfect dependence............ mutual exclusivity--------------------------------------------------------------------------------------- A) Rain - Rain..................... 30%..................................... 0%B) Dry - Rain.........................0% .................................... 30%C) Rain - Dry.......................30% ..................... ...............60%D) Dry - Dry.........................40% ...................... ............. 10% ---------------------------------------------------------------------------------------According to the problem there are 4 possible outcomes. The unknown is prob (D) and the only data known are:prob (A)+prob(C) =60%prob (A)+prob(B) =30%In case of perfect dependence (rain on Sunday implies rain on Saturday) prob (B)=0 then prob (D)=40%In case of mutual exclusivity ((if it rains on Saturday it cannot rain on Sunday) prob (A)=0 then prob (D)=10%I think you meant 'dependence' when you said 'perfect dependence'. Although your table is correct, the reason behind it is not very clear to me. Before filling any element in the table, the only info that we have (or can conclude) are:P1(A)+P1(C)=0.6; P1(A)+P1(B)=0.3 ;P2(A)+P2(C)=0.6; P2(A)+P2(B)=0.3 ;P2(A)=0 ; where, P1(.) = probability for the 'dependent' column & P2(.) = prob for 'mutually exclusive' column.We can easily say that P2(B)=0.3 and P2(C)=0.6, but not so directly for the 'dependent' caseFrom these equations, we can obtain (1 - P1(A) - P1(B) - P1(C)) = (0.4 - P1(B)) (or in terms of P1(A) or P1(C))So, the max value of P(dry) for dependent case = max{0.4 - P1(B)} = 0.4I get the same values by taking conditional probabilities.

QuoteOriginally posted by: panpsi get min=0.1, max=0.7...Argued as follows:A=event rains on sat. P(A)=0.6B=event rains on sun. P(B)=0.3P(dry)=1- (P(A)+P(B)-P(AB))=0.1 + P(AB)Three cases:AB=empty set -> P(dry)=0.1AB=A -> P(dry)=0.7AB=B -> P(dry)=0.4P(AB) = P(B/A).P(A) -(1)P(AB)= P(A/B).P(B) -(2)In the case of P(AB)=P(A), P(A/B)=2 from eq 2 ---> invalid caseSo, P(dry)=0.7 can never occur. Infact, P(dry) cannot exceed 0.4.. else, P(A/B) exceeds 1