Question

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

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

Answer 1

The correct option is B $\frac{6}{11}$

$x+y+z=10$
Total no. of solutions $={}^{10+3-1}{C}_{3-1}=66$
$\text{Let}z=2n\Rightarrow x+y=10-2n;n\in \{0,1,2,3,4,5\}$
Favourable number of solutions$={}^{(10-2n)+2-1}{C}_{2-1}=11-2n$
$\therefore$Total favourable solutions
$=\sum _{n=0}^5(11-2n)=36$
Required probability
$=\frac{36}{66}=\frac{6}{11}$