A bag contains 6 red, 7 white and 5 blue marbles, how many ways can four marbles be picked where there are an equal number of red and white marbles? Is it 315 ways?

Answer 1

Are you looking for the probability of drawing 2 reds and 2 whites when drawing any 4 marbles ?

you could have 4!/(2!2!) ways or 6 ways for it to happen:
RRWW
WWRR
RWWR
RWRW
WRWR
WRRW

prob of (RWRW) = (6/18)*(7/17)*(5/16)*(6/15)
= (6*7*5*6)/73440

prob(WRRW) = (7/18)*(6/17)(*(5/16)*(6/15)
= (7*6*5*6)/73440
notice the calculations are the same, so

prob(of what you want) = 6* (1260/73440)
= 7/68

How did you interpret this?
How did you end up with 315 ?

Answer 2

I am looking for the number of ways you can draw 4 marbles to have an equal number of red and white marbles after the draw. I was using combination probability set up as 6C2*7C2 = 315 but this doesn't include the probability of the 5 blue marbles. I'm not sure how to include that in the setup.

Answer 3

Ok
Let's assume that all the marbles are distinguishable , that is,
I can tell any red apart from the other reds, etc

Now lets "choose" 2 reds and 2 whites
= C(18,2)*C(16,2) = 18360

Remember in those the order does not matter, and we can tell reds apart and whites apart
so we could arrange them in 4! ways
But in reality we have 2 alike of the reds and another 2 alike of the whites.

Final number = 18360(4!/(2!2!)) = 18360(6)
= 110160

Answer 4

Thank you!