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I thought there should be 33 in the space but its wrong
12C3 = 12! / (9! 3!) = 12 * 11 * 10 / (3 * 2 * 1) = 220 different combinations of three balls among the 12 balls in the box. This is the number of different possible 3-ball draws. There are 6C3 = 6! (3! 3!) = 20 combinations of 3 balls that are all red. There are 4C3 = 4 combinations of 3 balls that are all blue. There are no combinations of 3 balls that are all green. So the probability of drawing 3 balls the same color, without replacement, is (20 + 4) / 220 = 24/220 = 6/55 = about 10.9%
there are no red balls in this problem
?? Let me re-do. This was my problem, I've had this before.
I did this awhile ago, mine was just with red instead of white, so I'm not sure.
so there are 6?
nono so w can be drawn either 1s 2nd 3rd... or 6th
for each of these 6 cases u can have different combinations of balls before
right so how do i find out how many outcomes there are
I don't know ;(
okay so take a random case see if we can figure out a pattern because i dont like this method it looks too long
lets say for w showing up on the 4th try _ _ _ W that means we have 3 places for ,2 red and 3 green
no red just yellow
that means we either have 3 green,2, green or 1 green in there
just 3 possiblites
but i dont wanna count like this, theres gonna be a permutation adn combination way of getting 3
we have 3 spots, 3 Green,2 Yellow* to fill those 3 spots
would it be 35?
how did u get that
well I originally made a tree chart and came up with 33
ugh thats too annoying
Ik, is there a faster way to do it?
thats what im trying to think about
I think i missed a W on my chart so I think it 34
Got it :D