mmmmmoooooorrrrrrreeeeeeeeeee (1 Viewer)

[McLake]

Originally posted by spice girl


I'm sorry to say that not only is 8C3 NOT obvious, it's also wrong.

For argument's sake, lets rename the 1st E "E1", and the 2nd E as "E2"

Then there would be 8C3 different ways of choosing 3 letters from the now-distinct 8.

However, now remember that in the real question, both E's are the same. So in all the cases where we only chose one "E" we've double-counted ($)(%)(E1) and ($)(%)(E2) have been considered different sets whereas in fact they are the same (since E1=E2)

NB: $, % are arbitrary letters.

So when there aren't 8 distinct letters (such as FREQUENT), there are less than 8C3 ways of choosing three.

There's a similar flaw in method 2 (we've all double-counted every word with only 1 'E')

To backtrack, we need to subtract every selection that has one 'E', i.e. 1 * 3 * 6 * 5 (one way to choose one 'E', 3 different places to put this 'E', 6 different letters to place the 1st non-'E', 5 different letters to place the 2nd non-'E')

318 - 90 = 228 which is my preferred answer.

Dumbarse: sorry, i dun usually show off - i'm just not used to ppl not believing me...

Not to sound like I'm covering my arse but I reverse enginered this question and was worried by the answer.

METHOD III [Correct?]:

PICK 1: 8 letters (2 are the same)

PICK 2: If e is picked, 6 letters left
If e isn't picked, 6 letters left (2 are same)

PICK 3: 6 letters

8*6*6 = 288

does this work?

 

[1234567]

we seem to have a problem in agreeing with answers or soemthing.......
hm.............
well even my teachers can't agree on answers .........

so......does prob have to be this hard in hsc?

 

[McLake]

Heres a lovly one if you don't agree on answers:

2 dice are rolled. A five is always on one of the dice (but not necsarly the same one).

What is the probibility of getting a sum of 8?

Possible Answers:

1/6
2/11 (i think thats the possibility)
I'm sure theres another somwhere.

Anyway, thoughts.

I'm with 1/6.

 

[BlackJack]

The question goes something like:
'On a roll of two dice, you notice one face was a 5. what's the possibility of getting the sum of 8?"

Edit: our class argued over it for the entire period because some people can't agree whether you have to separately define the two dice.
If either can be the 5, the answer is 2/11
If the 5 is fixed, the answer is 1/6

 

[McLake]

Originally posted by BlackJack
The question goes something like:
'On a roll of two dice, you notice one face was a 5. what's the possibility of getting the sum of 8?"

and what are the poss answers BJ?

 

[BlackJack]

That's them. The third one was discredited I think.

 

[McLake]

what about my 8*6*6 for the other question?

 

[BlackJack]

Mr Gospel... no wait, Gosbell.. said he thinks 2/11.
Firstly, the two e's are identical, so first step is seven choices.

If e Then seven(2nd letter)
in this branch:
if e again, six (3rd letter) (1*1*6)
if not e (six of them), six (1*6*6)
if it's not e then six
in this branch,
if e, then there is six more(6*1*6)
if not e, then 5 (6*5*5)

6 + 36 + 36 + 150 = 228
Okay, it works now.

Edit: fix up equations
explanations come later

 

[1234567]

hm.........come back tomrorow.

 

[McLake]

Originally posted by 1234567
hm.........come back tomrorow.

Yes, goodnight.

 

[BlackJack]

Bah, it works... I had to transfer my formal award ideas to Hing just then.

 

[kaseita]

your 8*6*6 escapes my logic.
your basically repeating what your saying for your 2nd pick, in your 3rd pick.
in your 2nd pick, your saying you have 6 letters left, whether you choose e or not
in your 3rd pick, your restating that you have 6 letters to choose from.
so effectively, you've made no 'calculation' in your 2nd pick.

 

[McLake]

Originally posted by kaseita
your 8*6*6 escapes my logic.
your basically repeating what your saying for your 2nd pick, in your 3rd pick.
in your 2nd pick, your saying you have 6 letters left, whether you choose e or not
in your 3rd pick, your restating that you have 6 letters to choose from.
so effectively, you've made no 'calculation' in your 2nd pick.

My illogic loses again.

 

[spice girl]

Originally posted by BlackJack
The question goes something like:
'On a roll of two dice, you notice one face was a 5. what's the possibility of getting the sum of 8?"

Well since you notice that "ONE" face was a 5, that means that the other one isn't. So sample space = 10:

1,5
2,5
3,5
4,5
6,5
5,1
5,2
5,3
5,4
5,6

Two of these give a sum of 8.

Thus probability is 2/10 = 1/5

 

[BlackJack]

Actually... the question didn't say it's just one face... I think it says 'at least one face'.

 

[spice girl]

Originally posted by BlackJack
Actually... the question didn't say it's just one face... I think it says 'at least one face'.

Ok, then there's one more sample space: 5,5

so p = 2/11

If confused over probability, draw a box diagram

 

[p00_p00]

Ok i aint the best math student in the world and i really aint a good ext 1 student but i think all of u ppl have made flaws.

Firstly since it is a word, order does matter, so it should be 8P3 over 2! (if the word is EQUATION)



Well plz tell me if im wrong

 

[spice girl]

You're wrong p00_p00

 

[BlackJack]

Don't you trust spice girl??? You should know better (the word is 'FRQUENT'. 'equation' is on the other thread)
Anyway, spice girl I have a question, just as an aside.
Have you competed in the UNSW schools maths comp (if you've heard of it)? Or the maths olympiads?

 

[p00_p00]

Well if its FREQUENT

shouldnt it still be 8!/2! ??????