Using conditional probability, it is found that P(X⏐Y) = 0.
Conditional probability is the probability of one event happening, considering a previous event. The formula is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which:
In this problem, we have that X and Y are mutually exclusive events, hence [tex]P(X \cap Y) = 0[/tex], hence:
[tex]P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.32} = 0[/tex]
More can be learned about conditional probability at https://brainly.com/question/14398287