For this problem, assume the box contains 6 white balls, 8 red balls, and 5 blue balls, and that we choose two

Question

For this problem, assume the box contains 6 white balls, 8 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?

Answer 1

2 . out of11 ??

Answer 2

If it is given that neither is red, we can just not include the red balls in the calculation.

Yay.

Therefore, if the balls can't be white, then the balls must be blue.

P(2 Blue Balls) = 5/11 * 4/10 because the first ball has a 5 (# of blue balls) out of 11 (# of total balls) chance and the second ball just has 5-1 / 11-1 because there is one less blue ball.

=5*4 / 11*10
=20 / 110
=2/11

The probability that both balls not being white given that neither is red is equal to 2/11, or 22.222%

Answer 3

You are saying that there are 6 white, 8 red, and 5 blue. Added up that is (6+8+5) = 19. Then you said we choose 2 randomly in which neither is white and red meaning you are left with red. You have 5 reds so I would say 5/19 but something tells me I'm wrong.

Answer 4

Balls 8

Answer 5

since neither is red then forget about them.
neither is white means both are blue. theprobability of the first ball being blue is 5/5+6. now we have 10 balls remaining. the probability of the second being blue is 4/10. so the probability of both being blue is 5/11*4/10 =20/110=2/11. though i still don't know why we must multily the probabilities.

Answer 6

The probability of the first one not being either white or red is 5/19. The probability of the next one not being either white or red is 5/18. So the total probability is (5/19)(5/18) = 25/342.

Answer 7

5/11 * 4/10 = 2/11 chance of neither being white

Answer 8

10 out of 143

P(neither red)
= 1 - P(both red)
= 1 - (8/19)(7/18)
= 143/171

P(neither red and neither white)
= P(both are blue)
= (5/19)(4/18)
= 10/171

thus P(neither white | neither red)
= P(neither red and neither white) / P(neither red)
= (10/171)/(143/171)
= 10/143

Answer 9

5/19

Answer 10

1 in 342

Answer 11

I come up with 2/11 like BESTestAn... (guess his method was a little easier)

Sample space consists of BB, BW, WB, WW (B=blue, W=white)

P(BB) = 5/19 * 4/18
P(BW) = 5/19 * 6/18
P(WB) = 6/19 * 5/18
P(WW) = 6/19 * 5/18

So overall probability =
P(BB) / (P(BB) + P(BW) + P(WB) + P(WW)) =
P(BB) / (P(BB) + 3*P(BW)) =
(5*4) / (5*4 + 3*6*5) =
2/11