You're in the first 2 boxes, so 75% chance to see green first (total space), which causes the GG box to be weighted more.
({GG} / {GG, GR})
/
({GG,GG,GR} / {GG,GG,GR,RG})
=
(1/2) / (3/4) = 2/3
Decent reduction excluding the RR box though and a proof that it's irrelevant.
Close. It's either 100% green for the first green ball, 100% green for the second green ball (both in the first box) and 0% for the last green ball. Add those probabilities up and you get 200%, divided by three balls is an average chance of 66% or two thirds.
No, if you pull a green ball then there is only one ball left in the box. If the original box was (GG) then the remaining ball will be G. If the original box was (GR), then the remaining ball will be R. Since the (RR) box is irrelevant, there are only two possibilities left. One is G and one is R, therefore, 50% chance for G.
If the first ball is green the there are only 2 boxes out of the 3 which have green balls then it's a 50/50 chance.
No, because it's more likely that you pulled G from GG in the first place rather than GR
anybody saying 50% is trolling or a brainlet
That's not how probabilities work
...
>Aryan toothpaste answer
>Brainlet shitskin answer
Thanks Yas Forums
Holy shit Zoomers don't know 7th grade math.