1. Let $A = \{\text{multiples of }4\}, \quad B = \{\text{multiples of }6\}, \quad C = \{\text{multiples of }7\}$ among the integers from $1$ to $50$.

2. We need: $P(A\cup B\cup C)=\frac{n(A\cup B\cup C)}{50}$

3. Use the inclusion-exclusion principle: $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$

4. Count each set:

   • Multiples of $4$ up to $50$: $n(A)=\left\lfloor \frac{50}{4} \right\rfloor=12$
   • Multiples of $6$ up to $50$: $n(B)=\left\lfloor \frac{50}{6} \right\rfloor=8$
   • Multiples of $7$ up to $50$: $n(C)=\left\lfloor \frac{50}{7} \right\rfloor=7$

5. Count pairwise intersections using LCM:

   • $A\cap B$ are multiples of $\operatorname{lcm}(4,6)=12$ $n(A\cap B)=\left\lfloor \frac{50}{12} \right\rfloor=4$
   • $B\cap C$ are multiples of $\operatorname{lcm}(6,7)=42$ $n(B\cap C)=\left\lfloor \frac{50}{42} \right\rfloor=1$
   • $C\cap A$ are multiples of $\operatorname{lcm}(7,4)=28$ $n(C\cap A)=\left\lfloor \frac{50}{28} \right\rfloor=1$

6. Count the triple intersection:

   • Multiples of $\operatorname{lcm}(4,6,7)=84$ $n(A\cap B\cap C)=\left\lfloor \frac{50}{84} \right\rfloor=0$

7. Substitute into inclusion-exclusion: $n(A\cup B\cup C)=12+8+7-4-1-1+0=21$

8. Therefore, the required probability is $\frac{21}{50}$

9. Compare with options:

   • A: $\frac{8}{25}$
   • B: $\frac{9}{50}$
   • C: $\frac{14}{25}$
   • D: $\frac{21}{50}$

Hence, the correct option is D.