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# Help ASAP please?

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A bag contains some marbles, each of which is one of four colors (red, white, blue, and green). There is at least one of each color. 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?

Please explain thoroughly

Apr 9, 2018
edited by gueesstt  Apr 10, 2018

### 2+0 Answers

#1
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Call the total number in the bag, N

Call the number  of red marbles, R

Call the number white, W

Call the number blue, B

Call the number green, G

4 reds are chosen has  the same probability as  1 blue  and 3 reds

R  *  (R-1) * (R - 2) *   (R - 3)             B      R          R- 1     R - 2

__   ____       _____   ______  =     ___   _____    _____  ____

N    N - 1      N - 2        N - 3             N     N - 1      N  - 2   N - 3

Note that the denominators in each case will be the same  in all cases...so...we can ignore these

So...this simplifies to  R - 3  =  B     (1)

4 reds   has the same probability  as  1  white , 1  blue  and 2 reds

So....subbing (1)  in for B

R (R-1) (R - 2) (R - 3)   =  W (R - 3)  R (R -1)

This simplifies to

R - 2  =   W      (2)

4 reds  has the same probability  as one of each color

Subbing  (1) and (2)  in for B, W   we have

R (R - 1) (R - 2) (R - 3)   =  R (R - 2) (R -3) G

This simplifies to

R - 1   =  G

Since there has to be at least one of each color....let the number of Red  = 4

Let the number of Green  = R - 1  =  3

Let the number of White  = R  - 2  =  2

Let the number of Blue  = R - 3  =  1

So.....the fewest number possible is 10   Apr 10, 2018
#2
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Thank you!

Apr 10, 2018