Help me understand this example of applying Bayes' Theorem

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The discussion centers on a specific example of Bayes' Theorem and its application in conditional probability. The user is confused about the equivalence of two expressions, specifically how the third expression equals the second despite having the same numerator, P(X,Y,Z). Clarification is sought on the underlying principles that establish this equality. The conversation emphasizes understanding the nuances of Bayes' Theorem in relation to conditional probability. Overall, the focus is on resolving the mathematical relationship between the expressions presented.
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I'm reviewing some notes regarding probability, and the section regarding Conditional Probability gives the following example:

HB3ZTal.gif


The middle expression is clearly just the application of Bayes' Theorem, but I can't see how the third expression is equal to the second. Can someone please clarify how the two are equal?
 

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The numerator is P(X,Y,Z) in both cases.
 
Thank you!
 

Thread 'Hypothesis testing: Defining H0, HA hypotheses so that ( H_A)_A' makes sense'

The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...
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