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Looking at @Ingix's answer at this link, is it possible to have the following event manipulation?

X = find in A day 2,

Y = not find in A day 1,

Z = in A

$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y)}$

Basically, I want to account for the event in A that he left out. However, I am not sure it would be actually like this, when Y and Z not not known to depend on each other.

$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y\mid Z)}$

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As per similar question found here:

pro

By switching the equality above, we have:

enter image description here

The answer should be:

$\mathbf P(X∣Y\cap Z) = \frac{\mathbf P(X\cap Y∣Z)}{P(Y\mid Z)}$

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