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6 sides to a die.

1 and 2 are factors of 2

The other 4 sides are not

2/6

1/3

The probability of not rolling factors of 2 on first die is 0.5.

The numbers on a die are 1, 2, 3, 4, 5, 6.

Among them the factors of 2 are: 2, 4, 6.

So the events of not rolling factors of 2 are = 1, 3, 5.

n(not rolling factors of 2) = 3.

n(all cases) = 6.

So the probability of not rolling factors of 2 on first die is = [tex]\frac{3}{6}=\frac{1}{2}=0.5[/tex].

Learn more: https://brainly.com/question/13769924

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