Draw a tree diagram to show the sample space of a toss of two dice. Denote the event of an even number on the 1st roll or a sum of 2 on the tree diagram.

There would be a 50/5 chance, of it landing on every side!

The tree diagrams are shown below.

Step-by-step explanation:

If dice is rolled then the possible outcomes are 1,2,3,4,5,6.

After that the second dice is rolled. So, the possible outcomes for second dice are 1,2,3,4,5,6.

Total sample space of a toss of two dice is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

Even number on the 1st roll. It means possible outcomes for first roll are 2,4,6.

A={ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

Sum of 2 rolls is even. The sum of two even numbers is even and sum of two odd numbers is even.

B={ (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6) }

Even number on the 1st roll or sum of 2 rolls is even.

A ∪ B = { (1,1), (1,3), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,3), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,3), (5,5), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }