What is the probability that a random arrangement of the letters in the word 'ARROW' will have both R's next to each other?

$\text{(A) }\frac{1}{10}\qquad\text{(B) }\frac{2}{15}\qquad\text{(C) }\frac{1}{5}\qquad\text{(D) }\frac{3}{10}\qquad\text{(E) }\frac{2}{5}$
1. **Determine the total number of arrangements of the letters in "ARROW":**
   - The word "ARROW" consists of 5 letters where 'R' is repeated twice.
   - The total number of distinct arrangements of the letters can be calculated using the formula for permutations of a multiset:
     \[
     \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
     \]
     where \( n \) is the total number of items, and \( n_1, n_2, \ldots, n_k \) are the frequencies of the distinct items.
     \[
     \text{Total arrangements} = \frac{5!}{2!} = \frac{120}{2} = 60
     \]

2. **Determine the number of arrangements where both 'R's are next to each other:**
   - Treat the two 'R's as a single block. This reduces the problem to arranging the blocks: {RR, A, O, W}.
   - Now, we have 4 blocks to arrange.
   - The number of ways to arrange these 4 blocks is:
     \[
     4! = 24
     \]

3. **Calculate the probability:**
   - The probability is the ratio of the number of favorable arrangements to the total number of arrangements.
     \[
     \text{Probability} = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{24}{60} = \frac{2}{5}
     \]

Conclusion:
\[
\boxed{\frac{2}{5}}
\]