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.

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)}

it is definatly the 3rd one. very much looks like it, for a good chance, the third one would be best for you.

8(4)^8=72^feet to say that the total can conclude to 8(4) which as you know would mean 8 x 4. that equals 32w. 32^ could make a 72 answer as well. but your coefficent will remain w so your answer is 72w.

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