JEE 2017 Probability

Here, This was JEE Adv 2017 Problem .

$x+y+z=10$ .

So total number of non-negetive solutions are $\dbinom{10+2}{2} \implies \dbinom{12}{2} \implies\boxed{ 66}$ .

Now we see that, $z$ is even for $0 , 2, 4, 6, 8 ,10$ .

For triplets (x,y,0) $\implies x+y=10$

So total solutions = $\boxed{11}$ .

For triplets of the form (x,y,2) $\implies x+y=8$

So total solutions $\implies \boxed{9}$

For triplets of the form (x,y,4) $\implies x+y=6$

So total solutions $\implies \boxed{7}$

For triplets of the form (x,y,6) $\implies x+y=4$

So total solutions $\implies \boxed{5}$

For triplets of the form (x,y,8) $\implies x+y=2$

So total solutions $\implies \boxed{3}$

For triplets of the form (x,y,10) $\implies x+y=0$

So total solutions $\implies \boxed{1}$

So total no. of even z's are $(11+9+7+5+3+1) \implies \boxed{36}$

So probability that z is even = $\dfrac{36}{66} \implies \dfrac{6}{11}$