A random variable is constructed through the following procedure. First you through a fair six-sided die once. If the outcome of this throw is $6$, you pick a real number at random between $0$ and $1$. If the outcome of the throw is not $6$, you pick a real number at random between $1$ and $5$.
The real number you picked up in this way is a random variable $X$, with density function $f.$ Compute $f(4.5).$ Give your answer using three decimal digits or give an exact answer.
MY WORKING:
As far as I have understood this problem is related to Binomial Distribution, with $Pr(S)=\frac{1}{6}$ and $Pr(F)=\frac{5}{6}$. But I don't understand how to compute $f(4.5)$ value and what is significance of picking real number randomly. Please any one help me. It will be appreciated. Thanks
$f(4.5) = f(4.5 | \text{throw 6})\Pr(\text{throw 6}) + f(4.5|\text{Not throw 6})\Pr(\text{Not throw 6}) = 0 \times \frac{1}{6} + \frac{1}{5 - 1} \times \frac{5}{6} = \frac{5}{24}$
Here $f(4.5 | \text{throw 6}) = 0$, because when you throw a 6 then you pick a number between $0$ and $1$ and since $4.5 \not\in [0, 1]$ therefore the $0$.
And $f(4.5 | \text{Not throw 6}) = \frac{1}{5 - 1}$, because when you dont throw a 6, then you pick a number uniformly at random between $1$ and $5$ which has $f(x) = \frac{1}{5 - 1}, \forall x \in [1, 5]$.