Conditional Probability (confusing)

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Discussion Overview

The discussion revolves around the validity of several statements related to conditional probability, including the relationships between probabilities of events A, B, C, and D. The scope includes theoretical aspects of probability and mathematical reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that the statement P(A | B) + P(A c | B) = 1 is true, given that P(B) > 0.
  • Another participant questions the validity of the statement P(A | B) + P(A | B c) = 1, stating it is not true under certain conditions, specifically when A is independent of B.
  • A participant suggests that if P(A) = x and A is independent of B, then P(A | B) = P(A | B c) = x, implying that the second statement does not hold.
  • One participant believes the third statement, P(C ∪ D | B) = P(C | B) + P(D | B) − P(C ∩ D | B), is true.
  • A later reply provides formulas for conditional probability and the union of probabilities, but does not clarify their relevance to the statements under discussion.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the second statement, with some asserting it is false while others do not provide a definitive stance. The first and third statements have mixed support, indicating that the discussion remains unresolved.

Contextual Notes

Participants have not provided detailed steps or proofs for their claims, and there are assumptions regarding independence that remain unexamined. The discussion does not clarify the conditions under which the statements may hold true or false.

sampahmel
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Dear all,

P (A |B) + P (A c|B) = 1 [A c] denotes complement of set A and of course P (B)>0

Is the above statement true?


How about the following two:
P (A |B) + P (A |B c) = 1


P (C ∪ D|B) = P (C |B) + P (D|B) − P (C ∩ D|B)
 
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What steps have you taken in proving/disproving these?
 
first statement is true

P (A |B) + P (A |B c) = 1 is not true

if P(A) = x and A is independent of B, then P(A|B) = P(A|Bc) = x

i think the third statement is true
 
TIP:
Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)} \quad Pr(B)>0
Pr(A \cup B)=Pr(A)+Pr(B)-Pr(A \cap B)
 

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