\(n\) - the number of green blocks which is equal to the number of favourable events
\(1+3+n=4n\) - the number of all blocks, which is equal to the number of all possible outcomes
\(P(A)=\dfrac{n}{4+n}=\dfrac{1}{2}\\
1\cdot( 4+n)=2\cdot n\\
4+n=2n\\
2n-n=4\\
n=4\)