Two non-negative integers x and y are chosen at random with replacement. The probability that $x^2+y^2$ is divisible by 10, is

$\frac{9}{50}$

By division algorithm, we have $x = 10x_1+a_1$ and $y = 10y_1 +b_1 $, where $x_1, y_1 , a_1, b_1 $ are integers such that $0 ≤ a_i ≤ 9 $ and $ 0 ≤ b_1 ≤ 9.$

$∴ x^2 + y^2 = (10x_1 +a_1)^2 + (10y_1 + b_1)^2 $

$= 100(x_1^2 +y_1^2)+ 20 (a_1 x_1 +b_1y_1) + (a_1^2 +b_1^2)$

It is evident from this expression that $x^2 + y^2 $ will be divisible by 10 iff $a_1^2+b_1^2$ is divisible by 10.

 Now, there are 10 choices each for $a_1$ and $b_1$, so that there are 10 × 10 = 100 ways of choosing them.

Now, $a_1^2+b_1^2$ will be divisible by 10 in each of the following cases:

(0, 0), (1, 3), (1, 7), (2, 4), (2, 6), (3, 1), (3, 9), (4, 2), (4, 8),

(5,5), (6, 2), (6, 8), (7, 1), (7, 9), (8, 4), (8, 6), (9, 3), (9,7)

∴ Favourable number of elementary events = 18.

Hence, required probability $=\frac{18}{100}=\frac{9}{50}$