We are given that both will be reaching the park between $4:00$ and $5:00$.
Probability that one friend arrives between $4:00$ and $4:50 = \dfrac{5}{6}$
Probability that one friend arrives between $4:00$ and $4:50$ and meets the other arriving in the next $10$ minutes $=\dfrac{5}{6}\times \dfrac{1}{6}\times {2} =\dfrac{10}{36} =\dfrac{5}{18}.$
(For any time of arrival between $4:00$ and $4:50,$ we have a $10$ minute interval possible for the second friend to arrive, and $2$ cases as for choosing which friend arrives first)
Probability that both friend arrives between $4:50$ and $5:00 =\dfrac{1}{6}\times \dfrac{1}{6} =\dfrac{1}{36}.$
This covers all possibility of a meet. So, required probability of non-meet
$=1 - \left( \dfrac{5}{18} + \dfrac{1}{36}\right)$
$= 1 - \dfrac{11}{36}$
$=\dfrac{25}{36}.$