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?

supermanaccz Nov 8, 2017

#1**+2 **

Let R = No. Red, B = No. Blue, W = No. White , G = No. Green and T = Total

And we have that

R / T * (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

B/T * ( R ) /[ T - 1] * (R - 1) / [T - 2] * (R - 2) / [ T --3] ⇒

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

(R - 3) = B

And

R / T * (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

W/T * (R -3)/ [T -1] * R/ [ T - 2] * (R -1) / [T - 3] ⇒

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

(R - 2) = W

And

R / T * (R -1)/[T - 1] * (R - 2)/[T- 2] * (R -3)/ [T - 3] =

R /T * (R -3)/[ T - 1] * (R -2) / [ T - 2] * G / [T - 3] ⇒

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

(R -1) = G

So

R = R

B = R - 3

W = R - 2

G = R - 1

Let the number of blue = 1

Let the number of white = 2

Let the number of green = 3

Let the number of red = 4

Probability of drawing 4 red =

[ 4 * 3 * 2 * 1 ] / [10 * 9 * 8 * 7 ] = 24 / 5040

Probability of one blue, three reds =

[ 1 * 4 * 3 * 2] / [ 10 * 9 * 8 * 7] = 24 / 5040

Probability of one white, one blue, two reds =

[ 2 * 1 * 4 * 3 ] / [ 10 * 9 * 8 * 7] = 24 / 5040

Probability of one of each color =

[1 * 2 * 3 * 4 ] / [10 * 9 * 8 * 7 ] = 24 / 5040

So...the minimum number is 10

CPhill Nov 8, 2017