Question

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

• A. $\frac{36}{55}$
• B. $\frac{6}{11}$
• C. $\frac{1}{2}$
• D. $\frac{5}{11}$

Answer 1

Answer: (B)

$\frac{6}{11}$

Answer 2

Total number of solutions $={ }^{10+3-1} C _{3-1}=66$
Favourable number of solutions $={ }^{11} C _{1}+{ }^{9} C _{1}+{ }^{7} C _{1}+{ }^{5} C _{1}+{ }^{3} C _{1}+{ }^{1} C _{1}=36$
$P ( req )=\frac{36}{66}=\frac{6}{11}$