Answer:

a. S = { 2, 4, 6, 4, 5, 6 } n(S) = 6

b. A = { 2, 4, 6 }, n(A) = 3

B = { 4, 5, 6 }, n(B) = 3

c. A (union) B = { 2, 4, 5, 6 } , n(A (union) B) = 4

d. To solve for the ratio, we use the probability addition rule:

P(A union B) = P(A) + P(B) - P(A intersection B)

4 and 5 exists in A and B therefore there are 2/6 chance that you'll get 4 or 5 in a die.

P(A union B) = 3/6 + 3/6 - 2/6

P(A union B) = 4/6

P(A union B) = 2/3

P(A union B) = 2:3

≈ 66.6666667%

≈ 67%

A simpler way is to find the number of terms in the union set and divide it to the 6. (since there are 6 faces in a die.)

P(A union B) = 4/6

P(A union B) = 2/3

P(A union B) = 2:3

≈ 66.66666667%

≈ 67%

You can choose either two to put in your paper. Since their answers are the same. Hope this helps. ^^