Post History
#2: Post edited
I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid Ceq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below. How do these two probabilities prove that B > C DEPENDS ON C > D?- >32. Consider four nonstandard dice (the Efron dice), whose sides are labeled as follows
- (the 6 sides on each die are equally likely).
- >
- >A: 4; 4; 4; 4; 0; 0
- B: 3; 3; 3; 3; 3; 3
- C: 6; 6; 2; 2; 2; 2
- D: 5; 5; 5; 1; 1; 1
- >
- >These four dice are each rolled once. Let A be the result for die A, B be the result for
- die B, etc.
- >
- >(a) Find P(A > B), P(B > C), P(C > D), and P(D > A).
- >
- >(b) Is the event A > B independent of the event B > C? Is the event B > C independent
- of the event C > D? Explain.
- >
- >## Solution:
- >
- >![Image alt text](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6)
- Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.
- I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C
- eq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below, colored in blue. How do these two probabilities prove that B > C DEPENDS ON C > D?
- >32. Consider four nonstandard dice (the Efron dice), whose sides are labeled as follows
- (the 6 sides on each die are equally likely).
- >
- >A: 4; 4; 4; 4; 0; 0
- B: 3; 3; 3; 3; 3; 3
- C: 6; 6; 2; 2; 2; 2
- D: 5; 5; 5; 1; 1; 1
- >
- >These four dice are each rolled once. Let A be the result for die A, B be the result for
- die B, etc.
- >
- >(a) Find P(A > B), P(B > C), P(C > D), and P(D > A).
- >
- >(b) Is the event A > B independent of the event B > C? Is the event B > C independent
- of the event C > D? Explain.
- >
- >## Solution:
- >
- >![Image alt text](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6)
- Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.
#1: Initial revision
How does $P(C > D \mid C = 2) \neq P(C > D \mid C \neq 2)$ prove that B > C depends on C > D?
I grok that $\color{limegreen}{P(C > D \mid C = 2) = P(D = 1 \mid C = 6) = 1/2}$, and $\color{red}{P(C > D \mid C \neq 2) = P(C > D \mid C = 6) = 1}$. But I don't grok the last sentence in the quotation below. How do these two probabilities prove that B > C DEPENDS ON C > D? >32. Consider four nonstandard dice (the Efron dice), whose sides are labeled as follows (the 6 sides on each die are equally likely). > >A: 4; 4; 4; 4; 0; 0 B: 3; 3; 3; 3; 3; 3 C: 6; 6; 2; 2; 2; 2 D: 5; 5; 5; 1; 1; 1 > >These four dice are each rolled once. Let A be the result for die A, B be the result for die B, etc. > >(a) Find P(A > B), P(B > C), P(C > D), and P(D > A). > >(b) Is the event A > B independent of the event B > C? Is the event B > C independent of the event C > D? Explain. > >## Solution: > >![Image alt text](https://math.codidact.com/uploads/fzxk7HMTvrBLDuo6o9LHfkA6) Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 30, p 88. pp 13-14 in the publicly downloadable PDF of curbed solutions.