Balls in a box shaking experiment

[rmjmu507]

A box contains V balls of which N$_{g}$ are green, N$_{r}$ are red and V-N$_{g}$-N$_{r}$=N$_{b}$ are blue.
Now somebody shakes the box vigorously, brings it to rest and then observes the
arrangement of the balls in the box. Suppose this is repeated many times so that
probabilities of different ball configurations can be defined as frequencies of occurrence.

What is the probability, P(G, R) that there are G green balls and R red balls in the
left half of the box? (Assume G+R<V/2)

I believe this probability is equal to $\Gamma$(G, R)/$\Gamma$(total)...

In this case, I think $\Gamma$(G, R) would be:

$\Gamma$(G, R) = (($\frac{(N_g)!}{(N_g)!(V-N_r-N_b)!}$$)^2)((\frac{(N_r)!}{(N_r)!(V-N_g-N_b)!}$)^2)(($\frac{V}{2}$!)^2)

And $\Gamma$(total) is V!.

Does this look right? In the first expression, the first term represents the number of ways to select which green balls are on the left hand side times the number of ways to select which red balls are on the left hand side, the second term represents the number of ways to select which red balls are on the left hand side times the number of ways to select which green balls are on the left hand side, and the third term represents the number of ways to arrange V/2 balls on the LHS times the number of ways to arrange V/2 balls on the RHS.

Am I on the right track?

 

[mXSCNT]

You need to clarify. Do you assume that each ball is equally likely to end up on the left or the right half of the box, and that the final positions of the balls are independent (which might NOT be a reasonable assumption since if part of the box gets too crowded, it may force balls into the other part!)?

 

[rmjmu507]

every arrangement of the balls is equally likely to occur. in other words, every arrangement will occur with equal frequency.

and yes, independence is assumed. the balls are identical except for their color

 

[haruspex]

A box contains V balls of which N$_{g}$ are green, N$_{r}$ are red and V-N$_{g}$-N$_{r}$=N$_{b}$ are blue.
Now somebody shakes the box vigorously, brings it to rest and then observes the
arrangement of the balls in the box. Suppose this is repeated many times so that
probabilities of different ball configurations can be defined as frequencies of occurrence.

What is the probability, P(G, R) that there are G green balls and R red balls in the
left half of the box? (Assume G+R<V/2)

I believe this probability is equal to $\Gamma$(G, R)/$\Gamma$(total)...

In this case, I think $\Gamma$(G, R) would be:

$\Gamma$(G, R) = (($\frac{(N_g)!}{(N_g)!(V-N_r-N_b)!}$$)^2)((\frac{(N_r)!}{(N_r)!(V-N_g-N_b)!}$)^2)(($\frac{V}{2}$!)^2)

And $\Gamma$(total) is V!.

Does this look right? In the first expression, the first term represents the number of ways to select which green balls are on the left hand side times the number of ways to select which red balls are on the left hand side, the second term represents the number of ways to select which red balls are on the left hand side times the number of ways to select which green balls are on the left hand side, and the third term represents the number of ways to arrange V/2 balls on the LHS times the number of ways to arrange V/2 balls on the RHS.

Am I on the right track?

None of your below-the-line terms look right. What happened to G and R?
I'd expect to see terms like (^N_g C _G), etc.

 

[blue_raver22]

Yes, you are on the right track. Your calculation for \Gamma(G, R) is correct. The first two terms represent the number of ways to select G green balls and R red balls from a total of N_{g} and N_{r} balls respectively, while the last term represents the number of ways to arrange these balls on the left half of the box. The total number of arrangements is represented by \Gamma(total) which is V!. Therefore, the probability of having G green balls and R red balls in the left half of the box is \Gamma(G, R)/\Gamma(total). This approach is valid assuming the balls are randomly distributed within the box after each shake.