Lana and Lois each had 15 red and 6 blue marbles. Each placed at least 1 red marble and at least 4 blue marbles into a bag (each had her own bag for this purpose). Each calculated the probability that if she drew out four marbles from her own bag that all four drawn would be blue. It turns out that both the probabilities are the same even though Lana had put more marbles in her bag than Lois had put in hers.
How many of each color had been placed in each bag?
Each woman had to put a different number of blue marbles in her bag. With only 4, 5, or 6 blue marbles as options, listing the possibilities is feasible.
4 blue marbles and x red marbles yields a probability [4!*x!]/[(x+4)!] = 24/[(x+1)*(x+2)*(x+3)*(x+4)]
5 blue marbles and y red marbles yields a probability [5!*(y+1)!]/[(y+5)!] = 120/[(y+2)*(y+3)*(y+4)*(y+5)]
6 blue marbles and z red marbles yields a probability [6!*(z+2)!]/[2!*(z+6)!] = 360/[(z+3)*(z+4)*(z+5)*(z+6)]
The possible numbers of red marbles must follow the inequality x<y<z.
If the first two probabilities are equal then 5*(x+1)*(x+2)*(x+3)*(x+4) = (y+2)*(y+3)*(y+4)*(y+5)
If the first and third probabilities are equal then 15*(x+1)*(x+2)*(x+3)*(x+4) = (z+3)*(z+4)*(z+5)*(z+6)
If the last two probabilities are equal then 3*(y+2)*(y+3)*(y+4)*(y+5) = (z+3)*(z+4)*(z+5)*(z+6)
The last equation has a small integer solution with y=3 and z=4. The next smallest solution is x=11 and z=22 for the second equation, but this already exceeds the 15 red marble limit in the problem. Therefore there is one solution: Lois put 5 blue and 3 red marbles in her bag and Lana put 6 blue and 4 red marbles in her bag.
Solution
