Finite HELP!!!?

Question

For this problem, assume the box contains 9 white balls, 5 red balls, and 5 blue balls, and that we choose two balls at random from the box.

What is the probability of neither being white given that neither is red?

2/21 was not the correct answer, but thanks for your help

10/171 was not the correct answer either, but thank you for still trying to help!!!

Answer 1

Since all three possibilities (red, white, or blue) have no overlap, we can ignore the red balls, because we know that we are not drawing them. Thus, it becomes a situation of what is the probabilty that neither is white if there are the 14 non-red balls.

The probability the first is not white is 5/14. The probability the second is not white is 4/13 (one of the blue balls is gone). Thus, the probability neither is white given neither is red is 5/14*4/13=10/91

Answer 2

If my understanding is right, no white and red was pick. So...

5+5+9 = 15

5/15*4(one blue ball was picked)/14(one ball was removed from the box) = 2/21(probability of picking up a ball that is neither red nor white..)

Answer 3

Wouldn't the probability that neither is white given that neither is red be the same as the probability of drawing two blue?
(5/19 * 4/18 = 20/342 = 10/171)

Answer 4

you want a finite answer right ?

since given is that no red are drawn you can just ignore the reds ( conditional probability of bayes )

so the answer is then 1 - ( [ 9*8 + 9 + 9] / 14*13 )
1 minus the chance that exactly 2 whites , exactly 1 white are drawn

Answer 5

if neither is red we can forget red balls so probability is 5/14

Answer 6

First, recall the definition of conditional probability:

P( neither white | neither red ) =
P(neither white and neither red)/P(neither red)

I forget if this is called hypergeometric or multinomial, and my books are inaccessible right now. So, someone else will need to finish this!