1. ## probability question-marbles.

there are a certain number of marbles in a bag, some are green, some red and some blue. I pick 3 blue marbles in a row. i observe that there was a 1 in 3417 chance that this would occur. if only my first two marbles where blue there would be a 1 in 204 chance.

-what is the smallest number of marbles that can be in my bag?
-how many blue marbles are there?
-how many possibllities are there for numbers of marbles in my bag?

i really need any help i can get with this, been working on it for hours

2. Let the number of blue marbles be b.
Let the total number of marbles be n.
P(3 blue marbles in a row) =b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417
P(2 blue marbles in a row) =b/n * (b-1)/(n-1) = 1/204

If we replace the second relation into the first one, we get:

1/204 * (b-2)/(n-2) = 1/3417

Hence: 3417 * (b-2) = 204 * (n-2)

Finding the least number of marbles in the bag boils down to finding the LCM(204, 3417) = 2 * 2 * 3 * 17 * 67 = 13668

So: 3417 * 4 = 204 * 67

One possible solution: b-2 = 4 and n-2 = 67

b=6 and n= 69. When replaced back in the initial condition b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417 it does not work

Therefore we have to go for multiples of 4 and 67

The next possible solution: b-2 = 2*4 and n-2 = 2*67
b=10 and n=136 - which is the correct one.

For your third question, just try to play woth multiples of 4 and 67.

3. ## in response

Originally Posted by Ithaka
Let the number of blue marbles be b.
Let the total number of marbles be n.
P(3 blue marbles in a row) =b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417
P(2 blue marbles in a row) =b/n * (b-1)/(n-1) = 1/204

If we replace the second relation into the first one, we get:

1/204 * (b-2)/(n-2) = 1/3417

Hence: 3417 * (b-2) = 204 * (n-2)

Finding the least number of marbles in the bag boils down to finding the LCM(204, 3417) = 2 * 2 * 3 * 17 * 67 = 13668

So: 3417 * 4 = 204 * 67

One possible solution: b-2 = 4 and n-2 = 67

b=6 and n= 69. When replaced back in the initial condition b/n*(b-1)/(n-1) * (b-2)/(n-2)=1/3417 it does not work

Therefore we have to go for multiples of 4 and 67

The next possible solution: b-2 = 2*4 and n-2 = 2*67
b=10 and n=136 - which is the correct one.

For your third question, just try to play woth multiples of 4 and 67.
Thanks very much, i found out that by simultaneosly solving (b-2)/4=(n-2)/67 and b/n * (b-1)/(n-1) = 1/204 you get the quadratic equation (for b):

175b^2 - 22207b - 216648 = 0

the other solution is -1593/175.