[REQUEST] 1998 2U James Ruse Trial Worked Solutions OR Help me with my question (1 Viewer)

[mirakon]

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

True. What I know is that we know there are 24 ways of choosing the first red ball, therefore the probability is 24/x

we know after this happens, the probability of the next red ball is 1/2 so it becomes 12/x

The problem is according to the answer given x=34. I'm simply trying to see how we get that value lol

i think the issue lies in the perception that one ball is chosen, then the other. But what if they are chosen at the same time? Does this make a difference?

 

[michaeljennings]

Look i got no idea how to solve this problem so I asked Descartes for his help. I pmed him, hopefully he logs on tonight or something

 

[HumanXJT]

Heres my attepmt:

Hopefully someone can confirm it before the exams. XD

 

[barbernator]

far out this is annoying me

 

[CecilyMare]

we aren't going to get this question in the 2u exam. it's too weird.

 

[jaafar1]

Heres my attepmt:

Hopefully someone can confirm it before the exams. XD

this is correct! but to get a red for the first and second pick it will be
4/7 x 3/6 = 2/7

 

[asianese]

that is an intense INTENSE diagram humanxjt! How'd you do it?

Btw barbernator dw too much about it!

 

[HumanXJT]

that is an intense INTENSE diagram humanxjt! How'd you do it?

Lol, made it with Photoshop.

 

[mirakon]

Ok, here is my next attempt.

1. First we have to figure out how many combinations of two apples can be picked (regardless of whether its red or not). NOTE: There is no "first pick" or "second pick", he picks out two apples at random so they are presumably at the same time..

So he can pick 7 different apples in one hand, leaving 6 possible apples for his other hand meaning there are 7x6= 42 different combinations.

2. However, let the apples be

r1
r2
r3
r4
g1
g2
g3

Now if he picks r1g1 and g1r1 they are the same combination of two apples, its a repitition

so counting the repetitions you should get 8.

meaning you have 42-8= 34 combinations

3. Now for the combinations where both apples are red, given it is known that at least one is red

there are 4 red apples and one of them is defintely picked, that leaves 6 other apples, including

so the probability that he picks at least one red apple is 4x6=24

4. Of the second apple, 3 of the 6 remaining apples are red ie. 1/2. Therefore the combinations of two red apples are 24x(1/2)= 12

5. Therefore probability equals valid combinations/total combinations- 12/34= 6/17

 

[cutemouse]

Lol conditional probability... and yes I think the given answer is wrong too...

I get 1/3 using 2 different ways

Method 1 (conditional probability method):

P(at least one red) = 1- (3/7)*(1/3) = 6/7

P(both red) = 4/7 * 1/3 = 4/21

So P(both red | at least one red) = (4/21)/(6/7) = 1/3

Method 2 (combinations):

No of ways at least one red = Total number of ways - No. of ways both green = 7.6 - 3.2 = 36

No of ways both red = 4.3 = 12

So Prob = 12/36 = 1/3