Probability of choosing two 2-digit positive integers with the same units place

Question

Two natural numbers, each of which has 2 digits are chosen randomly.

What is the probability that these two numbers have the same unit digit?

The options were

$\frac{1}{9}$

$\frac{8}{89}$

$\frac{8}{90}$

$\frac{1}{10}$

My attempt

Total number of ways to select two 2-digit numbers = $90 \times 90$

Total pairs of numbers such that their unit place are the same =

unit place zero = $81$ :

$(10, 10), (10, 20), ... (10, 90)$

$(20, 10), (20, 20), ... (20, 90)$

$...$

$(90, 10), (90, 20), ... (90, 90)$

Unit place one = $81$

$(11, 11), (11, 21), ... (11, 91)$

$(21, 11), (21, 21), ... (21, 91)$

$...$

$(91, 11), (91, 21), ... (91, 91)$

. . .

etc for unit places $0 \to 9$

i.e. in total $81 \times 10 = 810$

So req ans = $\frac{810}{8100} = \frac{1}{10}$

IS THIS CORRECT ?

Answer 1

Assuming your sampling is with replacement it is easier to note that the chance of each digit is $\frac 1{10}$. You draw the first number and note the units digit. The chance you get that units digit on the second draw is $\frac 1{10}$, so that is our answer.

Answer 2

Yes, your solution is correct, although your countings need just a bit more formal justification. Here is how you can formally write the solution.

There are $9\times 10$ ways to choose a 2-digit number ($9$ for the tens digit, $10$ for the tens digit). Then the number of ways to choose two 2-digit numbers is $9^2\times 10^2$.

Now, fix a units digit. This can be done in $10$ ways. Now you have the freedom of choosing any two tens digits, which can be done in $9^2$ ways. Thus your probability is $$\frac{9^2\times 10}{9^2\times 10^2}=\frac1{10}$$