If five boys and three girls are randomly divided into two four-person teams, what is the probability that all three girls will end up on the same team?

$\text{(A) }\frac{1}{7}\qquad\text{(B) }\frac{2}{7}\qquad\text{(C) }\frac{1}{10}\qquad\text{(D) }\frac{1}{14}\qquad\text{(E) }\frac{1}{28}$
1. **Determine the total number of ways to divide the eight children into two teams of four.**

   The total number of ways to choose 4 children out of 8 to form one team is given by the binomial coefficient:
   \[
   \binom{8}{4} = \frac{8!}{4! \cdot 4!} = 70
   \]
   Since the order of the teams does not matter, we must divide by 2 to avoid counting each division twice:
   \[
   \frac{70}{2} = 35
   \]
   Therefore, there are 35 ways to divide the eight children into two teams of four.

2. **Determine the number of favorable outcomes where all three girls are on the same team.**

   If all three girls are on the same team, we need to choose 1 more child from the remaining 5 boys to complete the team of 4. The number of ways to choose 1 boy from 5 is:
   \[
   \binom{5}{1} = 5
   \]
   The other team will automatically consist of the remaining 4 boys.

3. **Calculate the probability.**

   The probability that all three girls will end up on the same team is the ratio of the number of favorable outcomes to the total number of possible outcomes:
   \[
   \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{5}{35} = \frac{1}{7}
   \]

The final answer is \(\boxed{\frac{1}{7}}\)