Question

4 red,4 white,4 blue
at random take out 2 , say 1 red 1 white do not replace.
what is probarbility of now picking a)blue
b)red c)white
so red=1/12 or 1/4
white=1/12 or 1/4
now only 10 left
red 3/10,white3/10, blue 4/10

Answer 1

no
After taking out any two, you could have the following cases:

R W B

4 4 2
4 2 4
2 4 4
3 3 4
3 4 3
4 3 3

let's take the 3 3 4 situation
prob of that having happened
= C(4,1)xC(4,1)xC(4,0)/C(12,2) = 16/66 = 8/33
The other two cases of choosing 2 different colours = 8/33 as well
for the case of 4 4 2
prob of that happening = C(4,0)xC(4,2)xC(4,2)/C(12,2) = 6/66 = 1/11
(check by adding them up, you should get 1
8/33 + 8/33 +8/33 + 1/11 + 1/11 + 1/11 = 1 , Yeahh! )

a) You will then have to find the prob(blue) for each of the 6 cases.
I will continue with with our particular case of 3 3 4
The prob of that event is 8/33 and it contains 4 blues
and the prob of getting one blue = 4/10

So for that particular event to happen = (8/33)(4/10) = 16/165
Repeat for each of the remaining cases, then add them up

Just a lot of calculations .....