What is the probability of having $2$ adjacent white balls or $2$ adjacent blue balls in a random arrangement of $3$ red, $2$ white and $2$ blue balls? 

$ 
\textbf{(A)}\ \dfrac{2}{5}
\qquad\textbf{(B)}\ \dfrac{3}{7}
\qquad\textbf{(C)}\ \dfrac{16}{35}
\qquad\textbf{(D)}\ \dfrac{10}{21}
\qquad\textbf{(E)}\ \dfrac{5}{14}
$
1. **Calculate the total number of arrangements:**
   We have 3 red balls (R), 2 white balls (W), and 2 blue balls (B). The total number of arrangements of these 7 balls is given by the multinomial coefficient:
   \[
   \frac{7!}{3!2!2!} = \frac{5040}{6 \cdot 2 \cdot 2} = \frac{5040}{24} = 210
   \]

2. **Calculate the number of arrangements with 2 adjacent white balls:**
   Treat the pair of white balls (WW) as a single unit. Now we have 6 units to arrange: 3 red balls, 1 white pair, and 2 blue balls. The number of arrangements of these 6 units is:
   \[
   \frac{6!}{3!1!2!} = \frac{720}{6 \cdot 1 \cdot 2} = \frac{720}{12} = 60
   \]

3. **Calculate the number of arrangements with 2 adjacent blue balls:**
   Similarly, treat the pair of blue balls (BB) as a single unit. Now we have 6 units to arrange: 3 red balls, 2 white balls, and 1 blue pair. The number of arrangements of these 6 units is:
   \[
   \frac{6!}{3!2!1!} = \frac{720}{6 \cdot 2 \cdot 1} = \frac{720}{12} = 60
   \]

4. **Calculate the number of arrangements with both pairs (WW and BB) adjacent:**
   Treat both pairs (WW and BB) as single units. Now we have 5 units to arrange: 3 red balls, 1 white pair, and 1 blue pair. The number of arrangements of these 5 units is:
   \[
   \frac{5!}{3!1!1!} = \frac{120}{6 \cdot 1 \cdot 1} = \frac{120}{6} = 20
   \]

5. **Use the principle of inclusion-exclusion to find the number of arrangements with either or both pairs adjacent:**
   \[
   \text{Number of favorable arrangements} = 60 (\text{WW}) + 60 (\text{BB}) - 20 (\text{both WW and BB}) = 100
   \]

6. **Calculate the probability:**
   The probability of having 2 adjacent white balls or 2 adjacent blue balls is:
   \[
   \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{100}{210} = \frac{10}{21}
   \]

The final answer is \(\boxed{\frac{10}{21}}\)