Melody

I am not technical at all I am sorry.

I was using Firefox but I had problems with it (different from yours) and I returned to Google Chrome.  :/

Thanks for answering that MathsGod1

Well I will give you 100% for your initiative but that site made it look really hard.

It is not hard!

14.5% = 14.5/100 or  0.145

of   means multiply

so

14.5% of 215

= 14.5/100    x     215            See it is really simple.

$${\frac{{\mathtt{14.5}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{215}} = {\frac{{\mathtt{1\,247}}}{{\mathtt{40}}}} = {\mathtt{31.175}}$$

I wish you all the luck in the world Zegroes :)

That is how I apprached it and Heureka too I think.

I tried to use the table fuction on the forum for layout.  But it does not work anywhere good enough and it looks like a mess.

I expect Heureka's and mine are very similar but mine is probably riddled with errors. :)

It does not pay to rush MG

You said you were going to  -4 but then you added 4 instead!

Plus you used the latex for cancelling when there was no cancelling to do!

You need to edit this one  - or redo it and delete the original.   :)

You welcome Mellie.

I have given this question my best shot :)

Thanks Nauseated - I have added this to our LaTex thread  :)

BOXES FROM NAUSEATED      18/5/15

Poor CPhill    

Let the number of reds be a where a is at least 1

Let the number of whites be b

Let the number of blues be c

Let the number of greens be d

The probabilities below are all "Order does not count."

P(WRBG)=P(WBRR)=P(BRRR)=P(RRRR)

The denominators of these are all going to be the same so I am going to look at the numerators only

4!*abcd = (4!/2!)bca(a-1)=(4!/3!)ca(a-1)(a-2)=a(a-1)(a-2)(a-3)

24abcd = 12bca(a-1)=4ca(a-1)(a-2)=a(a-1)(a-2)(a-3)

24bcd = 12bc(a-1)=4c(a-1)(a-2)=(a-1)(a-2)(a-3)

let

24bcd                (1)

12bc(a-1)          (2)

4c(a-1)(a-2)      (3)

(a-1)(a-2)(a-3)      (4)

24bcd=12bc(a-1)      (1)=(2)

2d=a-1

$${\mathtt{d}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}$$

4c(a-1)(a-2)=(a-1)(a-2)(a-3)         (3)=(4)

4c=a-3

$${\mathtt{c}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}$$

12bc(a-1)  = 4c(a-1)(a-2)             (2)=(3)

3b = (a-2)

$${\mathtt{b}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}$$

SO WE HAVE

$${\mathtt{a}}$$                $${\mathtt{b}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}$$                  $${\mathtt{c}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}$$                         $${\mathtt{d}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}$$

NOW all these must be whole numbers.

a has to be odd   (from d)

a maybe 7, 11    ( looking at c and checking with b)    a=11 works

a=11,     b=3,   c=2,    5

Minimum number of marbles is   11+3+2+5 = 21

I have not checked that these numbers work Mellie - you will have to do that!