If X,Y and Z are three exhaustive and mutually exclusive events, so the total probability is
$P(X)+P(Y)+P(Z)=1 --\rightarrow(1)eq$
Given data is $P(X)=0.5P(Y)$ and $P(Z)=0.3P(Y)$
Substitute the values in the above equation (1)
$0.5P(Y)+P(Y)+0.3P(Y) =1$
$1.8P(Y)=1$
$P(Y)=1/1.8=0.5555$