Annoyinig Geha Probability Q (1 Viewer)

[Veck]

hey I've got a really annoying question that I found in Jeff Geha's book, and since I don't understand what he's up to, I thought I'd post my solution here and see what you guys think..

The question : in a certain box it is known that there are two black marbles and the rest are red. Four marbles are selected at random. The probability that both black marbles are chosen is twice the probability that neither will be chosen. Find the number of marbles

so I let p = 2/n and q = (n-2)/n

so P(2 blacks, 2 reds) = 4C2 (2/n)²((n-2)/n)²
= 24(n-2)^2 / n^4

and P(no blacks)=4C4((n-2)/n)^4
= (n-2)^4 / n^4

so P(no blacks) = 2xP(2 blacks, 2 reds)

so (n-2)^4 / n^4 = 2 x 24(n-2)^2 / n^4
= 48 (n-2)^2 / n^4

cancelling n^4
(n-2)^4 = 48 (n-2)^2

since n is obviously greater than 2, we can divide by (n-2)^2

and we get (n-2)^2 = 48

then we start getting irrational numbers, which obviously make no sense in this context...

hmm.. help anyone?

 

[withoutaface]

(n-2)C2=2*(n-2)C4
(n-2)!/2!(n-4)!=2(n-2)!/4!(n-6)!
1/(n-4)(n-5)=2/12
6=n²-9n+20
0=n²-9n+14
0=(n-2)(n-7)

since n>6
then there are 7 marbles

 

[Veck]

hmm.. not sure how that came about withoutaface

tell me where you got your first line from...

and what did I do wrong..?

 

[withoutaface]

(n-2)C2 are the number of ways you can pick with 2 black, 2 red, since the blacks are already determined, then there are n-2 marbles left select the 2 reds from.

(n-2)C4 is the number of ways you can select 4 reds, since there are only n-2 marbles after the blacks have been excluded

The first line should really read:

(n-2)C2/nC4=2*(n-2)C4/nC4, but the nC4's cancel out so I didn't bother putting them in

 

[Veck]

ahhh ok
so I'm just an idiot
thanks a lot
I'm still a bit hazy though
not sure what I did wrong really...
ahh this is just diggin myself deeper
could you explain what I screwed up?
why can't I use the binominal probability with this..?
hmm.. I hate probability..
thanks