Hello,

As you said, recall that exp(1) ~ Gamma(1)

Since X+Y ~ Gamma(a+b) if X ~ Gamma(a) and Y ~ Gamma(b), it follows that Z2+Z3 ~ Gamma(2)

Now there's this thing that you should know : if X and Y are independent, and X ~ Gamma(a) and Y ~ Gamma(b), then X/(X+Y) ~ Beta(a,b)

Hence, (Z2+Z3)/(Z1+[Z2+Z3]) follows a Beta distribution with parameters (2,1), because Z2+Z3 follows a Gamma(2) and Z1 follows a Gamma(1)