Probability question, please help

[mmmt20]

Given numbers from 1 to 10. Two numbers are to be selected from these 10 numbers and they can be same also. What is the probability that at least one of them is even?
A. 1/2
B. 3/4
C. 1/4

[cramya]

odd->1,3,5,7,9
p(odd) = 5/10
=1/2
even->2,4,6,8,10
p(even) = 5/10
=1/2

p(atleast 1 is even) = 1 -p(none of them are even)
= 1- 1/2*1/2
= 1 -1/4
= 3/4

I am taking that this problem is with replacement since it says the numbers can be the same also.

[mmmt20]

Thanks Cramya, but I still did not understand

p(atleast 1 is even) = 1 -p(none of them are even)
= 1- 1/2*1/2
= 1 -1/4
= 3/4