Probability

8/15

Neither red nor black means the ball must be green. There are 8 green balls out of 15 total balls. So the probability is 8/15.

1

On a die, every outcome is less than 7. So E is the whole sample space and P(E) = 1. Event F is impossible. Therefore E union F is still the whole sample space, so P(E union F) = 1.

1/6

E = {2,4,6} and F = {6}. So E intersection F = {6}. There is 1 favourable outcome out of 6 total outcomes. Hence P(E intersection F) = 1/6.

E and F are mutually exclusive and exhaustive

E = {2,4,6} and F = {1,3,5}. They have no common outcome, so they are mutually exclusive. Their union is {1,2,3,4,5,6}, which is the sample space, so they are exhaustive too.