Probability of Green Balls – Final Episode

Five types of ball-picking problems are introduced here. Welcome to another episode of arrangements (and more confusion). A bag contains 4 green balls and 4 red balls. You like to pick four balls at random, in which:

One green ball

Probability to choose one green AND 3 reds = [4C1 x 4C3] / [8C4]. Remember, the multiplication in the numerator happened because of AND.

At least one green ball

At least one green = P([1 green AND rest red] OR [2 green AND rest red] OR [3 green AND rest red] OR [4 green]). Applying multiplication for AND and summation for OR,

Probability to choose at least one green = [4C1 x 4C3 + 4C2 x 4C2 + 4C3 x 4C1 + 4C4] / [8C4]

At most one green ball

At most one green = P([0 green AND rest red] OR [1 green AND rest red].
Probability to choose at most one green = [4C4 + 4C1 x 4C3] / [8C4]

No green ball

This one is easy; just pick red balls, 4C4/8C4

One green or two greens

What is the probability to pick one green AND three reds or two greens AND two reds?

[4C1 x 4C3 + 4C2 x 4C2] / [8C4]