Here is a fun and challenging probability question.

1) A box contains some Blue balls and 9 Red balls. If the probability that two balls randomly removed without replacement from the box are Red is 6/11, what is the total probability that the third ball randomly removed without replacement of the balls drawn is a blue ball?

The first step in answering this question is to find how many blue balls there are by solving for what we are given in the problem. We are given that the probability that two balls randomly removed without replacement from the box are red is 6/11, so then lets use this info to find x, the number of blue balls. We make an equation:

Let x be the number of blue balls
9/(9+x)*8/(8+x)=6/11 <--because there are first 9 balls, then 8 red balls.
Solving for x (skipping a few steps of basic algebra) we get x =3, as 9/12*8/11=6/11

Therefore we know that there are 3 balls. Now, because there are 3 blue balls left, and 7 red balls left, the odds of getting a blue ball next is 3/10. The next and final step is to multiply 6/11 by the probability of choosing a blue ball, or 6/11*3/10 = 9/55, as the question asks for total probability, given that the 1st 2 balls picked, without replacement, are red balls.
Good job!