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]