I am going through some exercises for probability and stumbled upon one that I can't wrap my head around.
Question- Three brands of coffee, X, Y , and Z, are to be ranked according to taste by a judge. Define the following events:
A: Brand X is preferred to Y .
B: Brand X is ranked best.
C: Brand X is ranked second best.
D: Brand X is ranked third best.
If the judge actually has no taste preference and randomly assigns ranks to the brands, is event A independent of events B, C, and D?
My Attempt to the answer- Sample space: E1->XYZ, E2->XZY, E3->YXZ, E4->YZX, E5->ZXY, E6->ZYX where XYZ denotes that X is ranked best, Y is second best, and Z is last. Then A->{E1,E2,E5}, B->{E1,E2}, C->{E3,E5}, D->{E4,E6}. Now P(A)=3/6=1/2, P(A)=P(A and B) but how can I calculate P(A|D) or P(A|C) to check for independence? Any pointers to the correct approach here would be greatly appreciated.
You are on the right track with the creation of the sample space. Here the strings $XYZ$, $XZY$, etc. are actually the outcomes. Further we have events like $E_1=\{XYZ\}, E_2=\{XZY\}$, etc., and these $6$ events are equiprobable so have probability $\frac16$.
Also $A,B,C,D$ are events and each of them can be described as a union of the $E_i$.
Also $A\cap D$ and $A\cap C$ are events and can be described as (eventually empty) unions of the $E_i$.
Actually you do not have to calculate $P(A\mid D)$ and $P(A\mid C)$ to check independence.
It is enough to check whether $P(A\cap D)=P(A)P(D)$ and $P(A\cap C)=P(A)P(C)$ are true equalities (then independence) or not (then no independence).