Three randomly chosen non-negative integers $x$, $y$ and $z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is

$\text{Given: } x,y,z \ge 0 \text{ integers with } x+y+z=10$ $x,\ y,\ z \text{ chosen at random}$ $\text{To find: } P(z \text{ is even})$ $\text{Options: } \dfrac{36}{55},\ \ \dfrac{6}{11},\ \ \dfrac{1}{2},\ \ \dfrac{5}{11}$ $1.\ \text{total solutions of } x+y+z=10$ $2.\ \text{count solutions with } z \text{ even} \qquad 3.\ P=\dfrac{\text{favourable}}{\text{total}}$ $x,y,z\ge 0,\quad x+y+z=10$ $\text{No. of solutions }=\dbinom{10+3-1}{3-1}=\dbinom{12}{2}$ $\dbinom{12}{2}=\dfrac{12\cdot 11}{2}=66 \ \Rightarrow\ \text{total outcomes}=66$ $\text{For fixed } z:\ x+y=10-z \ \Rightarrow\ (11-z) \text{ solutions}$ $z \text{ even}:\ z=0,2,4,6,8,10$ $z=0:11,\qquad z=2:9,\qquad z=4:7$ $z=6:5,\qquad z=8:3,\qquad z=10:1$ $11+9+7+5+3+1=36$ $\text{Favourable outcomes}=36$ $P(z \text{ even})=\dfrac{36}{66}$ $\dfrac{36}{66}=\dfrac{36\div 6}{66\div 6}=\dfrac{6}{11}$ $\boxed{P(z \text{ even})=\dfrac{6}{11}}\quad \text{option (2)}$