3. In the box, there are 30 balls, each with a different natural number from 1 to 30 written on it, meaning that all numbers from 1 to 30 are represented. If two balls are randomly drawn from the box, what is the probability that:

a) the difference between the numbers on the drawn balls will be divisible by 5?

b) the sum of the numbers on the drawn balls will be divisible by 5?
First method:

The total number of all possible pairs of numbers drawn is $\frac{30 \cdot 29}{2}=435$.

$1 \text{ POINT}$

Favorable are those pairs of numbers that give the same remainder when divided by 5.

Among the numbers written on the balls, let's denote:

$A_{0}=\{$ numbers divisible by 5$\}=\{5,10,15,20,25,30\}$

$A_{1}=\{$ numbers that give a remainder of 1 when divided by 5$\}=\{1,6,11,16,21,26\}$

$A_{2}=\{$ numbers that give a remainder of 2 when divided by 5$\}=\{2,7,12,17,22,27\}$

$A_{3}=\{$ numbers that give a remainder of 3 when divided by 5$\}=\{3,8,13,18,23,28\}$

$A_{4}=\{$ numbers that give a remainder of 4 when divided by 5$\}=\{4,9,14,19,24,29\}$

There are 6 numbers in each of these sets.

$1 \text{ POINT}$

Favorable pairs of numbers are those drawn from the same set.

The number of favorable pairs of numbers

from set $A_{0}$ is $\frac{6 \cdot 5}{2}=15$,

from set $A_{1}$ is $\frac{6 \cdot 5}{2}=15$,

from set $A_{2}$ is $\frac{6 \cdot 5}{2}=15$,

from set $A_{3}$ is $\frac{6 \cdot 5}{2}=15$,

from set $A_{4}$ is $\frac{6 \cdot 5}{2}=15$.

$1 \text{ POINT}$

Thus, the total number of favorable outcomes is $5 \cdot 15=75$.

$1 \text{ POINT}$

The desired probability is

$$
p=\frac{75}{435}=\frac{5}{29} .
$$