I want to find $\mathbb{V}\mathrm{ar}[Y]$, where $Y=[\min(X,4)]$ and $$f(x)=\frac{1}{5}, 0<x<5.$$ So what I have done is to use this formulae: $$\mathbb{V}\mathrm{ar}[Y]=\mathbb{E}_I[\mathbb{V}\mathrm{ar}[X\mid I]]+\mathbb{V}\mathrm{ar}_I[\mathbb{E}[X\mid I]],$$ and $I$ is an indicator function on whether the values of $x$ exceed $4.$ So, I have substituted all values: $$\mathbb{V}\mathrm{ar}[Y]=\mathbb{V}\mathrm{ar}_I[2,4]+\mathbb{E}_I[1.333333,0]$$ and we got $$\mathbb{V}\mathrm{ar}[Y]=4.3.$$
May I know which step I have gone wrong, thank you
It's hard to tell where you went wrong since there is not much of your work shown. Here are the values I get
$$ \text{Var}(Y|X\le4)=\frac43\text{ with probability }\frac45 $$ $$ \text{Var}(Y|X\gt4)=0\text{ with probability }\frac15 $$ $$ \text{E}_I\!\left(\text{Var}(Y|I)\right)=\frac43\cdot\frac45+0\cdot\frac15=\boxed{\frac{16}{15}} $$
$$ \text{E}(Y|X\le4)=2\text{ with probability }\frac45 $$ $$ \text{E}(Y|X\gt4)=4\text{ with probability }\frac15 $$ $$ \text{Var}_I\!\left(\text{E}(Y|I)\right)=\left(2^2\cdot\frac45+4^2\cdot\frac15\right)-\left(2\cdot\frac45+4\cdot\frac15\right)^2=\boxed{\frac{16}{25}} $$
$$ \text{Var}(Y)=\frac{16}{15}+\frac{16}{25}=\boxed{\frac{128}{75}} $$