2 Unit probablity (1 Viewer)

[crazylilmonkee]

how? wat?
explain!

 

[...]

ooo...

working?

 

[Affinity]

that's the fun part.. given the answer is 1/5, try to make sense out of it

my math teacher caught me off guard today with this..

 

[crazylilmonkee]

u dont get it either do u?
grrrrrr that shits me, its really gonna kill me...
i'll just have to work it out wont i.....

 

[Affinity]

hint:

do it the old fashion way.. list down the possible sample space

at first I thought it was same as (i) but.. there's a subtle difference, see if you can spot it

 

[Fosweb]

Got it... (using the oh duh method...)

There are 5 ways you can get combinations with a red ball in it:

RR, BR, RB, WR, RW. (like - 2*RW, 2*RB cause there are 2 r's...)

The only way you can drop the red ball, and then have a red ball left is with RR, so 1/5...

 

[...]

oh fork..

i treat those 2 red marbles differently..thats y i got it as 6...

*slams head*

 

[underthesun]

Bleh, case bashing gives 1/5

Label the two red marbles r1 and r2, these are possible:

r1r2 -> one red, another one red
r1b -> one red
r1w -> one red
r2r1 -> one red, another one red
r2b -> one red
r2w -> one red
br1 -> one red
br2 -> one red
bw
wr1 -> one red
wr2 -> one red
wb

2/10, or 1/5. classic...

 

[crazylilmonkee]

ooooooooooooooohhhhhhh i get it now
and im screwed

 

[Harimau]

... I think we were just thinking too complex...

 

[underthesun]

It seems to defy logic however, that the answer isn't 1/3. Suppose one ball is taken, and it is red. The chance that the other ball that will be chosen is red is 1/3, or i'm missing something as I always do

 

[Fosweb]

You are missing the fact that you KNOW that one of the balls taken WAS red already, which rules out the case of having a BW together (like - this question is not dealing with the probabilty of getting the red ball out of the BAG, it is dealing with getting ANOTHER red ball from your hand, which you have already picked.) So there are only 5 ways you can get the balls from your hand. Kindof logical.

 

[ND]

Originally posted by underthesun
It seems to defy logic however, that the answer isn't 1/3. Suppose one ball is taken, and it is red. The chance that the other ball that will be chosen is red is 1/3, or i'm missing something as I always do

Yeh but you have to realise that there are two ways of choosing both RB and RW, whereas there is obviously only one way of choosing RR.

 

[Affinity]

it's related to the 'monty hall' question, where you have 3 doors one with the prize behind it, and you choose 1 of them initially, then the host will show you the empty one out of the remaining 2 and you are given a chance to switch, should you switch or not?

both showing red ball and the empty door are not 'random' events and thats why things turned out this way I suppose

 

[Fosweb]

God... I remember our teacher copying us pages of notes on that in about year 9/10... Wonder where they are now.

Didnt they have goats behind the losing doors? And dont you have a greater probability of winning if you DO actually change your choice after he opens one door? (Because you have a 1/2 change of losing now as opposed to a 2/3 chance of losing if you dont change?)

 

[Affinity]

yes you do have a greater chance if you change