[REQUEST] 1998 2U James Ruse Trial Worked Solutions OR Help me with my question (2 Viewers)

[qawe]

Can someone please post up the worked solution to this paper, or help me with the following question:

Q2cii) A box contains 4 red and 3 green apples. Peter took 2 apples at random. If it is known that at least one of the apples is red, find the probability that they are both red.

Answer: 6/17

So if someone can explain the answer or post up the worked solutions that would be great.

 

[asianese]

If one is red, consider ALL the other possible combinations.
So if we have 4 red, there are 3 left. 3 also for the green apples.
RR RR RR RG RG RG
RR RR RR (different way) GR GR GR
(other Gs) GR GR GR GR GR

So the only way RR can occur is 6/17. It's quite a tricky question...you just have to count all the possibilities...

 

[jaafar1]

isnt it, using a tree diagram the probability of get the first apple is 4/7, the second is 3/6
so u do 4/7 x 3/6

 

[qawe]

If one is red, consider ALL the other possible combinations.
So if we have 4 red, there are 3 left. 3 also for the green apples.
RR RR RR RG RG RG
RR RR RR (different way) GR GR GR
(other Gs) GR GR GR GR GR

So the only way RR can occur is 6/17. It's quite a tricky question...you just have to count all the possibilities...

this doesn't make sense to me, how on earth do you get five GR's in the last line!

 

[qawe]

can anyone do this... or can someone with maths experience please confirm the solutions are simply wrong

 

[michaeljennings]

can anyone do this... or can someone with maths experience please confirm the solutions are simply wrong

TBH i believe the answers are wrong I think the answer is 1/2

If we assume the one apple picked is a red one (like the question says) that leaves 6 apples in total remaining with 3 red and 3 green left. hence to pick a red apple you have 3 possibilities out of 6 which would be 1/2

 

[Jaymay]

^But theres a difference if he's looking at the second apple picked, compared to the first apple picked... right?!

 

[michaeljennings]

^But theres a difference if he's looking at the second apple picked, compared to the first apple picked... right?!

I dont think so but rarely are worked solutions wrong so maybe someone could show how its done

 

[Jaymay]

Yeah i got it... But the explanations gonna be shit...

No i didn't.... got 7/16 lol soz guys

 

[mirakon]

Guys, it is known one of the apples is picked is red AFTER THEY ARE RANDOMLY PICKED. It's not like he picked out a red apple deliberately and there are 3 red and 3 green left

 

[michaeljennings]

Guys, it is known one of the apples is picked is red AFTER THEY ARE RANDOMLY PICKED. It's not like he picked out a red apple deliberately and there are 3 red and 3 green left

Yes so either his first draw was a red apple or his 2nd draw was a red apple and its easiest to look at the red apple being the first one draw and then finding the probability for the 2nd one drawn. Thats my reasoning but it would be best if someone could show how the answer is 6/17

 

[yomammasfatass]

Guys, it is known one of the apples is picked is red AFTER THEY ARE RANDOMLY PICKED. It's not like he picked out a red apple deliberately and there are 3 red and 3 green left

But if it is KNOWN that at least one is red, then the probability of one the balls being red is 1. Therefore the chance of the other one being red is 3 in 6, i.e 1/2

 

[mirakon]

Think of the fact that 7!/(3!x4!)= 35 (3 apples are repetitions of green, 4 are repetitions of red) of these 24 consist of red being chosen.

Now one of the possibilities is excluded (I cannot figure out why atm, if someone has an idea why it will help bring this together)

That means the probability of one red being chosen is 24/34

Now we have 3 red and 3 green remaining. Probability of red being chosen is 1/2

24/34x1/2= 6/17

 

[mirakon]

But if it is KNOWN that at least one is red, then the probability of one the balls being red is 1. Therefore the chance of the other one being red is 3 in 6, i.e 1/2

yes but in doing so you're disregarding the fact that the apple known to be red was nevertheless randomly picked.

 

[michaeljennings]

Think of the fact that 7!/(3!x4!)= 35 (3 apples are repetitions of green, 4 are repetitions of red) of these 24 consist of red being chosen.

Now one of the possibilities is excluded (I cannot figure out why atm, if someone has an idea why it will help bring this together)

That means the probability of one red being chosen is 24/34

Now we have 3 red and 3 green remaining. Probability of red being chosen is 1/2

24/34x1/2= 6/17

Where does this come from?

 

[yomammasfatass]

yes but in doing so you're disregarding the fact that the apple known to be red was nevertheless randomly picked.

yeah but if something is known, then there is no probability involved in finding out what it is right? cos u know that its red, regardless of being randomly picked

 

[mirakon]

Where does this come from?

counting lol, there's probably a more efficient method though...

 

[michaeljennings]

I dont understand your 7! business though. We arent trying to discover how many arrangments of 7 apples there are

 

[yomammasfatass]

mirakon, are u sure ur not doing perms? because isnt 7!/(3!x4!) considering the other 5 balls, when we are only concerned with the picking of 2, rather than the ordering of all 7

 

[aphorae]

Ughhhh I am getting 6/18. Can someone take a look at this and tell me the flaw in my method? Well, two methods (which are virtually the same thing rofl)

http://i55.tinypic.com/2eelc3q.jpg
http://i51.tinypic.com/2vmyma8.jpg

edit:
lol i just realised i accidentally wrote the wrong possibilities in the second method. ignore the left side, calculations still turn out the same