Answer:
5/11.
Step-by-step explanation:
The number of ways 2 balls can be selected = 12P2
= 12!/10!
= 12*11
= 132.
There are 6 even numbers and 6 odd numbers in the bag.
If both numbers drawn are even there are 6P2 = 30 ways that sum is even.
If the 2 numbers are odd there are also 30 ways that sum is even.
Any other combination wil be odd.
So the required probabilit = 2(3) / 132
= 60/132
= 5/11.
