Let S be the set of outcomes when two distinguishable dice are rolled, let E be the subset of outcomes in which at least one die shows an even number, and let F be the subset of outcomes in which at least one die shows an odd number. List the elements in the given subset. E' ∪ F' (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5), (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) (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) (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)

E'∪F'=[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5], [2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6], [6, 2], [6, 4], [6, 6]

Step-by-step explanation:

Sample Space of the Two Dice is:

[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]

E=At least one die shows an even number.

Sample Space of E

[1, 2],  [1, 4], [1, 6]

[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]

[3, 2] [3, 4],  [3, 6]

[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]

[5, 2],  [5, 4],  [5, 6]

[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]

Complement of E

[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5]

F=At least one die shows an odd number

Sample Space of F

[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]

[2, 1], [2, 3], [2, 5],

[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]

[4, 1], [4, 3], [4, 5]

[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]

[6, 1], [6, 3], [6, 5],

Complement of F

[2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6]

[6, 2], [6, 4], [6, 6]

Therefore:

E'∪F'=[1, 1], [1, 3], [1, 5], [3, 1], [3, 3] [3, 5],[5, 1], [5, 3], [5, 5], [2, 2], [2, 4], [2, 6]

[4, 2],[4, 4], [4, 6], [6, 2], [6, 4], [6, 6]

answer: 59

step-by-step explanation:

yes

step-by-step explanation: