A bag contains some marbles, each of which is one of four colors (red, white, blue, and green). At least one of the marbles is red. The composition of the bag is such that if we take four marbles out at random (without replacement), each of the following is equally likely: (1) one marble of each color is chosen, (2) one white, one blue, and two reds are chosen, (3) one blue and three reds are chosen, (4) four reds are chosen. What is the smallest possible number of marbles in the bag?

Mellie May 15, 2015

#1**+11 **

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!

Melody May 17, 2015

#1**+11 **

Best Answer

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!

Melody May 17, 2015

#3**+2 **

Here is my solution.

Let's start by assigning variables.

r is the number of red marbles, w is the number of white marbles, b is the number of blue marbles and g is the number of green marbles.

Moving on, there are r*w*b*g ways to choose one marble of each color.

There are \(wb\binom{r}2\) ways to choose one white, one blue and two reds.

Then, there are \(b\binom{r}3\) to choose one blue and three reds.

Finally, there are \(\binom{r}4\) ways to choose four reds.

The problem tells us that these are all equal, so

\(rwbg=wb\binom{r}2=b\binom{r}3=\binom{r}4\)

The first equality gives us 2g=r-1 which is the same as r=2g+1.

The second equality gives us 3w=r-2 which is the same as r=3w+2.

The final equality gives us 4b=r-3 which is the same as r=4b+3.

Looking at this, we can conclude that r is one less than a multiple of two, three and four.

The smallest number that fits all these categories is 11, and since we want the smallest amount possible, that is $r$.

Now we can just substitute r in the equalities we got.

11=2g+1 gives us g=5.

11=3w+2 gives us w=3.

11=4b+1 gives us b=2.

Therefore, the smallest number of marbles meeting the requirements is 11+5+3+2=21 marbles

Guest May 29, 2016