Among $6$ persons $U,V,W,X,Y$ and $Z$, it is known that: $U$ is taller than at-least $3$ persons. $W$ is taller than $V$, who is taller than exactly $1$ person. $X$ is not shorter than $W$. $Y$ is taller than $U$ and $X$ is shorter than $U$. Who is $4^{\text{th}}$ tallest among them?

Let's arrange the $6$ individuals in descending order of their heights ($1^{\text{st}}$ being the tallest, $6^{\text{th}}$ being the shortest):
1. "V is taller than exactly 1 person": This means $V$ is at the $5^{\text{th}}$ position, and there is exactly one person shorter than $V$ at the $6^{\text{th}}$ position. Let's represent this as:

$$\text{\_ > \_ > \_ > \_ > V > \_}$$

2. "W is taller than V": $W$ must be in one of the top four positions.
3. "U is taller than at-least 3 persons": This means there are at least $3$ people shorter than $U$. So $U$ can be at the $1^{\text{st}}$, $2^{\text{nd}}$, or $3^{\text{rd}}$ position.
4. "Y is taller than U and X is shorter than U": This establishes the relation $Y>U>X$. Since $Y$ is taller than $U$, $U$ cannot be the tallest ($1^{\text{st}}$). Thus, $U$ is either $2^{\text{nd}}$ or $3^{\text{rd}}$.
5. "X is not shorter than W": This implies $X\ge W$ (or $X>W$ since heights are unique).

Combining the conditions $Y>U>X>W>V$, we can uniquely map the positions from $1^{\text{st}}$ to $6^{\text{th}}$:

* $1^{\text{st}}$ (Tallest): $Y$
* $2^{\text{nd}}$: $U$
* $3^{\text{rd}}$: $X$
* $4^{\text{th}}$: $W$
* $5^{\text{th}}$: $V$
* $6^{\text{th}}$ (Shortest): $Z$

Let's double-check $U$: $U$ is taller than $4$ people ($X,W,V,Z$), which satisfies "at least 3".
The $4^{\text{th}}$ tallest person is W.