Three randomly chosen non negative integers x , y , z are found to satisfy the equation x + y + z = 10 .…

Let $z=2k,$ where  $k=0, 1, 2, 3, 4, 5$

$\therefore x+y=10-2k$

Number of non negative integral solutions

$\sum _{k=0}^5{ }^{11-2k}{C}_1= \sum 11-2k=36$

Total cases $={ }^{10+3-1}{C}_{3-1}=66$

Required probability $=\frac{36}{66}= \frac{6}{11}$