Thanks! Help!

Let's say... 1 red, 1 green, 2 blue.

The new question is... What is the probability of selecting 2 blue balls?

First chance... 1 blue ball has chance 2/4, or 1/2.

Second chance... 1 blue ball has chance 1/3. This is because 1 blue ball is left... and only 3 balls are left.

Multiply these two chances... 1/6 probability.

Let's say the the balls are

B₁  B₂  R  G

The number of permuted sets created by choosing any 2 of the 4 balls is  12

Only  1  of these sets contain 2 blue balls....and they can be permuted in two ways

{ B₁, B₂ } or {B₂, B₁}

So

2 / 12   =

1/ 6

Thank you so much!