Understanding Combinatorics: Probability of Selecting Balls from a Box

[Nikitin]

Hi. OK in a box there are 6 balls, 2 red ones and 4 blue ones. We take 2 balls out of the box without putting any of them back

1) If I wish to know the probability of selecting 2 blue ones, I just do this: (4above2)/(6above2)=6/15 or 4C2/6C2 or (4*3/2)/(6*5/2)2) BUT, if I wish to know the probability 1 blue and 1 red, I must use the first formula here http://en.wikipedia.org/wiki/Hypergeometric_distribution

(4above1)*(2above1)/(6above2)=8/15=p(1 blue and 1 red ball selected)

WHY can't we use the same logic as in the formula in 1) and do this: (4*2/2)/(6*5/2)=4/15 ?? why is just (4*2)/(6*5/2)=8/15 correct?

I mean, why are the number of relevant outcomes "unorganized" or "combinated" in 2) but in 1) they are "organized" and "permuted" ?

Excuse me for my english. I am hoping somebody can please explain this too me?

 

[Nikitin]

so is my english the reason nobody wants to post a reply? If it helps I am talking about "hypergeometric" probability in 2)

 

[mathman]

Instead of mechanically using combinatorial formulas, you are better of with direct calculation.

For the first problem: Prob(first ball chosen is blue)=2/3, prob(second ball chosen is also blue)=3/5. Therefore prob(both balls are blue)=2/5.

For the second problem, there are two mutually exclusive ways of doing it - blue first and red second or red first and blue second. The first has prob 2/3 x 2/5 = 4/15. The second has prob 1/3 x 4/5 = 4/15. Total prob = 8/15.

 

[Nikitin]

aah thank you, i get it now ! :)