Probability of Independent Sets: A,B in Sample Space Z

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In the discussion, participants explore the conditions under which the sets {a, b} and {a, c} are independent within the sample space Z = {a, b, c}. The key point is that for independence, the probability of the intersection of the sets must equal the product of their individual probabilities. There is confusion regarding the interpretation of independence, particularly since the sets share the element 'a', which complicates their independence. The conversation highlights the need for clarity in defining "they" and understanding the nature of independence in this context. Ultimately, the discussion emphasizes the importance of correctly applying probability principles to sets and events.
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In a sample space Z = {a, b, c} is the probability distribution of the numbers x, y, 0.3 (in that order). What should be they that any {a, b} and {a, c} are independent?

maybe i need a suggestion if the type i want to use down is correcty or not because i stucked...



I know that for A1,A2,An we have this type : P(A1andA2and...andAn)=P(A1)P(A2)...P(An) but how i can use this if i have sets?
 
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You need a to be independent of b and c, but b and c need not be independent of each other?

What is the definition of "independent" in this context? (Hint: in terms of probabilities.)
You have:
P(A1andA2and...andAn)=P(A1)P(A2)...P(An) but how i can use this if i have sets?
... you don't have to apply the relation to whole sets, the A1,A2,... form a set {A1, A2, ...}.

if A and B are independent, then P(A)+P(B)=?
 
If I'm interpreting this correctly, you need to find x and y such that P[f∈{a,b}|f∈{a,c}] = P[f∈{a,b}]. Is that it?
 
Well if you are interpreting it correctly - then how would the relation P[f∈{a,b}|f∈{a,c}] = P[f∈{a,b}] possibly hold? After all, the two sets have a member in common.

It boils down to how you read this bit:
What should be they that any {a, b} and {a, c} are independent?
... which is "they" and what is "independent" of what?

Presumably "they" is x and y.
"{a,b} is independent", then I'd read that as outcome a is independent of outcome b... I don't think that {a,b} can be independent of {a,c} since they both contain a.

I read the question as saying that {a,b,c} may or may not not be independent, but {a,b} and {a,c} are.
Given that P(c)=0.3, what is P(a) and P(b)?

However: you are closer to the course than I am so you may have an extra insight.
 
Simon Bridge said:
Well if you are interpreting it correctly - then how would the relation P[f∈{a,b}|f∈{a,c}] = P[f∈{a,b}] possibly hold? After all, the two sets have a member in common.
There is exactly one setting of x and y that makes it true.
"{a,b} is independent", then I'd read that as outcome a is independent of outcome b.
Aren't a, b and c atomic events? If so, they are therefore mutually exclusive, not independent. In probability space terms, a set of possible outcomes, like {a, b} constitutes a (nonatomic) event, and you want the events {a, b}, {a, c} to be independent.
 
Because Z is a sample space? That would make sense OK.
I still think the phrasing is sloppy... which is why I'm uncertain about how I was reading the question.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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