MedVision ad

Question 9(a) (iv) HSC 1997 (1 Viewer)

_deloso

Member
Joined
Jan 25, 2011
Messages
470
Location
Andromeda Galaxy
Gender
Male
HSC
2011

MrBrightside

Brightest Member
Joined
Jan 18, 2010
Messages
2,032
Gender
Undisclosed
HSC
N/A
There is 5 possible combination for getting at least one red: RR,RB,RW,BR and WR
Now if the red ball is dropped, the combination of RR is the only combination where the second ball is red, therefore, there is a 1/5 probability that the other ball is red.
But what are the probabilities of getting each colour?

I drew the tree diagram like it said and for RR e.g. it would be (2/4*1/3), how can you get 1/5?

do you just say there are 5 outcomes therefore 1/5 chance? (have no logical reasoning behind this)

Initially I did P(1 - Chance of at least 1 R) but it doesn't make sense and it's very hard to explain from my point of view.

I know RR, RB, RW, BR and WR are the possibilities of having 1 hidden R after 1 has falling from his hand.

where do I go from here?
 

krnofdrg

Mq Law Student :)
Joined
Mar 8, 2010
Messages
1,672
Location
Strathfield
Gender
Male
HSC
2012
Uni Grad
2017
If one of the selected balls is red, there are 5 possible outcomes only one of which will yield another red:

RR, RB , RW , BR , WR

Therefore, P(HIDDEN RED)= 1/5

hope this helps!
 

MrBrightside

Brightest Member
Joined
Jan 18, 2010
Messages
2,032
Gender
Undisclosed
HSC
N/A
If one of the selected balls is red, there are 5 possible outcomes only one of which will yield another red:

RR, RB , RW , BR , WR

Therefore, P(HIDDEN RED)= 1/5

hope this helps!
Ahhh I think I get it now, it said the first ball he dropped was red, so that means only RR will contain the other red ball. and there's 1/5 chance it will be RR. Thanks, I'm a bit rusty on probability at times.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top