Bailey will roll a number cube and flip a coin for a probability experiment. The faces of the number cube are labeled 1 through 6. The coin can land on heads or tails. If Bailey rolls the number cube once and flips the coin once, which list contains only the outcomes in which the number cube lands on a number greater than 4? A 5, Heads 5, Tails 6, Heads 6, Tails B 5, Heads 6, Tails C 1, Heads 1, Tails 2, Heads 2, Tails 3, Heads 3, Tails 4, Heads 4, Tails 5, Heads 5, Tails 6, Heads 6, Tails D 4, Heads 4, Tails 5, Heads 5, Tails 6, Heads 6, Tails

A permutation is an act of arranging the objects or elements in order. Combinations are the way of selecting objects or elements from a group of objects or collections, in such a way the order of the objects does not matter.

Given

The faces of the number cube are labeled 1 through 6.

The coin can land on heads or tails.

To find

The number cube lands on a number greater than 4.

Let, H be the head and T be the Tail.

The total sample will be

Total sample = 6×2

Total sample = 12

All the samples are shown.

(1, H)(2, H)(3, H)(4, H)(5, H)(6, H)

(1, T) (2, T) (3, T)(4, T) (5, T)(6, T)

The number cube lands on a number greater than 4 will be 4. The combination is (5, T)(5, H)(6, T)(6, T).

Thus, the correct option is A.

More about the Permutation and the combination link is given below.

https://brainly.com/question/11732255