Probability question.

Question

There are 2 white balls, 2 red balls, and 3 blue balls in a box. Two balls are selected

at random, and their colors are noted. Find the probability that neither is whie

given that neither is blue.

Answer 1

It is 1/4.

P(no white and no blue)/P(no blue) = P(both red) /P(no blue) = P(both red)/P(one white and one red) =

(2 choose 2)/[(2 choose 1)(2 choose 1)] = 1/4

Answer 2

I'm not seeing any note on replacement, so I can show you the probability for either case.

With Replacement (A ball is selected, recorded, and then put back in the bag before the second ball is drawn).

For this instance the probability of selecting a blue ball is the same on each draw. This means that P(Not Blue) = 4/7. So we simply square this because we don't want a blue ball either time, and that gives us a probability of p=16/49 ≈ 0.3265

Without Replacement (Both balls are drawn at the same time)

Now the probabilities change. The best way I've found to explain this probability is by counting the number of ways I can select balls that are not blue, then divide that by the number of ways that I can select any balls.

Not Blue) We start with 4 balls that are not blue, so that's 4 options for the first ball and 3 remaining for the second ball. Multiplying these together we get 12 ways to select balls that are not blue.

Anything) We start with 7 balls total, so that's 7 options for the first ball and 6 remaining options for the second ball. This makes a total of 42 ways we can select these balls.

Now divide the not blue value by the anything value: p = 12/42 = 2/7 ≈ 0.2857.

Guessing based on the way your question is worded, I would say the without replacement is the way to go, but now you have both available.