probability q 2008 paper (1 Viewer)

[jackc91]

If anyone could give me a hand with this question it would be greatly appreciated. Im terrible with probability. Solutions I have seen dont give explanation.

Thanks.

 

[tommykins]

i)
Total amount of balls = n+n+n = 3n

Fix a colour ball the first time -
n/3n first draw = 1/3
(n-1)/(3n-1) second draw because we took out one ball fom the total, and one ball from our wanted colour
(n-2)/(3n-2) third draw, same reasoning above.

Multiply them together to get [n(n-1)(n-2)]/[3n(3n-1)(3n-1)] but we also multiply by 3 as we have 3 different colours to choose from.

same reasoning for the next few

PS. im pretty shit at proability too so this might be wrong, sounds write to me though

 

[kwabon]

when all else fails,

USE A TREE DIAGRAM

am sure you will get it from there.

back to chemistry for me.

(y)

 

[Timothy.Siu]

i)n/(3n-1) x n/(3n-2)
ii) 2n/(3n-1) x n/(3n-2)
just pick one ball at a time.
iii) 2 balls same colour, (n-1)/(3n-1)
then one of another colour, 2n/(3n-2)
so times those together.

iiii)ratio..
n^2 : 2n^2 : 2n^2-2n
as n gets big...
1:2:2

 

[gurmies]

I think the last one is 1:2:6...i'll post up my solution (not sure if it's correct) if nobody else does...

 

[shaon0]

I think the last one is 1:2:6...i'll post up my solution (not sure if it's correct) if nobody else does...

Yeah, you're correct.

 

[lolokay]

i)n/(3n-1) x n/(3n-2)

should be (n-1)/(3n-1) x (n-2)/(3n-2)

iii) 2 balls same colour, (n-1)/(3n-1)
then one of another colour, 2n/(3n-2)
so times those together.

keep in mind that the different colour ball could be picked in any of the 3 draws (the wording is misleading), not just the last one, so you have to multiply this by 3, giving 1:2:6 for iv)

 

[tommykins]

thats so weird.

i dont think i got this question out during ym exam.
hahaha wtf