Marbles Probabilities

[George0305]

Hello! I can`t find the solution to a problem from a certain point. The problem:

From a box containing n=white marbles, m=black marbles and p=red marbles, we get a marble at a time, check its color and place it back in the box until we get 'x' iterations of white marbles.
s=the number of marbles retrieved. We need to calculate the probability L={the number of black marbles retrieved is bigger than the red ones}.

Until now, i replaced n, m and p with values - 2, 2, 10 (from a similar problem) so i could get a rough formula(but i couldn`t). I found that 1/7 is the chance to get a white ball, and 1/49 is the chance to get 2 white balls one after the other(unnecessary i think) and this is where i`m stuck at.

Thank you!

 

[Steven G]

Hello! I can`t find the solution to a problem from a certain point. The problem:

From a box containing 'n' white marbles, 'm' black marbles and 'p' red marbles, we get a marble at a time with replacement until we get 'x' iterations of white balls.
s represents the number of white balls retrieved. We need to calculate the probability L={the number of black marbles retrieved is bigger than the red ones}.

Until now, i replaced n, m and p with values - 2, 2, 10 (from a similar problem) so i could get a rough formula(but i couldn`t). I found that 1/7 is the chance to get a white ball, and 1/49 is the chance to get 2 white balls one after the other(unnecessary i think) and this is where i`m stuck at.

Thank you!

What does - "we get a marble at a time with replacement until we get 'x' iterations of white balls" - mean??
In your small example (2,2,10) can you list all ways that there are more black marbles than red ones? Can you find P(L)? Do you see a pattern?

 

[Denis]

Until now, i replaced n, m and p with values - 2, 2, 10 (from a similar problem) so i could get a rough formula(but i couldn`t).

WHAT's that suppose to mean

 

[JeffM]

Hello! I can`t find the solution to a problem from a certain point. The problem:

From a box containing 'n' white marbles, 'm' black marbles and 'p' red marbles, we get a marble at a time with replacement until we get 'x' iterations of white balls.
s represents the number of white balls retrieved. We need to calculate the probability L={the number of black marbles retrieved is bigger than the red ones}.

Until now, i replaced n, m and p with values - 2, 2, 10 (from a similar problem) so i could get a rough formula(but i couldn`t). I found that 1/7 is the chance to get a white ball, and 1/49 is the chance to get 2 white balls one after the other(unnecessary i think) and this is where i`m stuck at.

Thank you!

It truly helps to put things into notation AND TO ASSIGN NAMES TO THINGS.

\(\displaystyle x = \text {number of white marbles retrieved in } t \text {trials.}\)

\(\displaystyle y = \text {number of black balls retrieved in } t \text {trials.}\)

\(\displaystyle z = \text {number of black balls retrieved in } t \text {trials.}\)

\(\displaystyle \text { Probability of white marble in a trial} = \dfrac{1}{n}.\)

\(\displaystyle \text { Probability of black marble in a trial} = \dfrac{1}{m}.\)

\(\displaystyle \text { Probability of red marble in a trial} = \dfrac{1}{p}.\)

I think you are looking for

\(\displaystyle \text {Given } t \ge x \ge 1, \text { find P(} y > z).\)

I'd start by trying things.

\(\displaystyle x = t \implies \text {P(}y > z \text {)} = 0.\)

\(\displaystyle x + 1 = t \implies \text {P(}y > z \text {)} = \dfrac{1}{m}.\)

\(\displaystyle x + 2 = t \implies \text {P(}y > z \text {)} = \left ( \dfrac{1}{m} \right )^2.\)

\(\displaystyle x + 3 = t \implies \text {P(}y > z \text {)} = 3 * \left \{ \left( \dfrac{1}{m} \right )^2 * \dfrac{1}{p} \right \}+ \left ( \dfrac{1}{m} \right)^3.\)

If I understand what you are doing, you want a conditional probability so this is not the end.

 

[George0305]

What does - "we get a marble at a time with replacement until we get 'x' iterations of white balls" - mean??

We extract a marble from the box, check its color, and place it back. We repeat doing this until 2 white balls have been extracted from the box.

WHAT's that suppose to mean

n=2, m=2, p=10 and i tried working that out instead of n,m and p without values.

It truly helps to put things into notation AND TO ASSIGN NAMES TO THINGS.

\(\displaystyle x = \text {number of white marbles retrieved in } t \text {trials.}\)

\(\displaystyle y = \text {number of black balls retrieved in } t \text {trials.}\)

\(\displaystyle z = \text {number of black balls retrieved in } t \text {trials.}\)

\(\displaystyle \text { Probability of white marble in a trial} = \dfrac{1}{n}.\)

\(\displaystyle \text { Probability of black marble in a trial} = \dfrac{1}{m}.\)

\(\displaystyle \text { Probability of red marble in a trial} = \dfrac{1}{p}.\)

I think you are looking for

\(\displaystyle \text {Given } t \ge x \ge 1, \text { find P(} y > z).\)

I'd start by trying things.

\(\displaystyle x = t \implies \text {P(}y > z \text {)} = 0.\)

\(\displaystyle x + 1 = t \implies \text {P(}y > z \text {)} = \dfrac{1}{m}.\)

\(\displaystyle x + 2 = t \implies \text {P(}y > z \text {)} = \left ( \dfrac{1}{m} \right )^2.\)

\(\displaystyle x + 3 = t \implies \text {P(}y > z \text {)} = 3 * \left \{ \left( \dfrac{1}{m} \right )^2 * \dfrac{1}{p} \right \}+ \left ( \dfrac{1}{m} \right)^3.\)

If I understand what you are doing, you want a conditional probability so this is not the end.

Yes, i`m pretty sure it implies a conditional probability, but i think that`s a good start.

I also edited the problem to make it less confusing.

Thank you and sorry for the poor description of the problem.