Are these 3 events independent?

[Guest]

Let us consider 3 events A,B,C such that:

$$P((A \cap B )\cup C)=P(A)*P(B)*P(C)$$

Notice that the second term is a union and not an intersection.

Are they independent?

And what if the assumption was: $$P(A \cap( B \cup C))=P(A)*P(B)*P(C)$$?

I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well.

But I do not know how to prove that they are/they are not independent

Thank you.

[Guest]

Let us consider 3 events A,B,C such that:

$$P((A \cap B )\cup C)=P(A)*P(B)*P(C)$$

Notice that the second term is a union and not an intersection.

Are they independent?

And what if the assumption was: $$P(A \cap( B \cup C))=P(A)*P(B)*P(C)$$?

I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well.

But I do not know how to prove that they are/they are not independent

Thank you.

[tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P((A \cup B) \cup (B \cup C)) - P(A \cup B) - P(B \cup C)[/tex].

Hence, [tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P(A \cup B) * P(B \cup C)[/tex] if sets [tex]A \cup B[/tex] and [tex]B \cup C[/tex] are independent.

Therefore, [tex]P(A \cap B \cap C)=P(A)*P(B)*P(C)[/tex] if only if the sets A, B, C are mutually independent.

Relevant Reference Links:

'Algebra of Sets',

https://en.wikipedia.org/wiki/Algebra_of_sets;

'Further Concepts in Probability',

https://www.wyzant.com/resources/lessons/math/statistics_and_probability/probability/further_concepts_in_probability.

[Guest]

Let us consider 3 events A,B,C such that:

$$P((A \cap B )\cup C)=P(A)*P(B)*P(C)$$

Notice that the second term is a union and not an intersection.

Are they independent?

And what if the assumption was: $$P(A \cap( B \cup C))=P(A)*P(B)*P(C)$$?

I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well.

But I do not know how to prove that they are/they are not independent

Thank you.

[tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P((A \cup B) \cup (B \cup C)) - P(A \cup B) - P(B \cup C)[/tex].

Hence, [tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P(A \cup B) * P(B \cup C)[/tex] if sets [tex]A \cup B[/tex] and [tex]B \cup C[/tex] are independent.

Therefore, [tex]P(A \cap B \cap C)=P(A)*P(B)*P(C)[/tex] if only if the sets A, B, C are mutually independent.

Relevant Reference Links:

'Algebra of Sets',

https://en.wikipedia.org/wiki/Algebra_of_sets;

'Further Concepts in Probability',

https://www.wyzant.com/resources/lessons/math/statistics_and_probability/probability/further_concepts_in_probability.

A Minor Remark: [tex]P((A \cup B) \cup (B \cup C)) = P(A \cup B \cup C)[/tex].

[Guest]

Hmm.

[tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P(A \cup B) * P(B \cup C)[/tex] if sets, [tex]A \cup B[/tex] and [tex]B \cup C[/tex], are independent.

However, the sets, [tex]A \cup B[/tex] and [tex]B \cup C[/tex], are not independent. The set B is common to both of those sets even if A and B are mutually independent sets.

Therefore, [tex]P((A \cap B )\cup C)= P((A \cup B) \cap (B \cup C)) = P((A \cup B) \cup (B \cup C)) - P(A \cup B) - P(B \cup C)[/tex]
[tex]= P(A \cup B \cup C) - P(A \cup B) - P(B \cup C)[/tex].