# Number of arrangements of colored balls into distinct bins

I have 6 colored balls :2 red,2 yellow,2 blue.

The balls are thrown into 4 distinct bins independently , where each bin can hold more than one ball.

I'm trying to calculate the probability that in the second bin there is a red ball, and also calculating the probability of having a red ball in the first bin when there is already a red ball in the second bin.

Here is what I have tried I think the the total arrangements is the same thing as arranging 6 stars and 2 bars which is $$\frac{8!}{2!*6!}=28$$, then the arrangements in the bins where there is a red ball in second bin are $$\frac{7!}{2!*5!}=21$$. Then $$Pr[$$second bin there is a red ball$$]=\frac{21}{28}=3/4$$.

is this correct for the first probability?

• Not following your calculation. The other colors have nothing to do with the problem, you are only asking about the red ones. Therefore, you are just placing two balls in the urns. – lulu May 8 at 13:26
• All balls are being placed for example : R,R,YB,YB. is one arrangement – user3133165 May 8 at 13:31
• Yes, the other balls are being placed but they are irrelevant....their placement has nothing to do with the problem. The probability that the second bin contains a red ball is $1-\left( \frac 34\right)^2=\frac 7{16}$ – lulu May 8 at 15:57

The placement of the non-red balls is irrelevant...the problem is decided by the placement of the red ones alone.

To do the first part, it is easier to work from the complimentary event. The probability that a specified red ball goes to a bin other than the second is $$\frac 34$$. Thus the answer is $$1-\left( \frac 34\right)^2=\boxed {\frac 7{16}}$$

To do the second one, it might be helpful to simply list the possible placements of the two red balls. Given that the second bin must contain a red ball, there are only four cases: $$(1,1,0,0)\quad (0,2,0,0)\quad (0,1,1,0)\quad (0,1,0,1)$$

Where writing, say, $$(1,1,0,0)$$ means that the first two bins each have one red ball and the second two each have none.

Note that the probability of getting $$(0,2,0,0)$$ is $$\frac 1{16}$$ since we must have both red balls going to the second bin. The probability of getting, say, $$(1,1,0,0)$$ is $$\frac 18$$ since we can get this configuration in two ways (either the first red ball goes in the first bin and the second red ball goes in the second bin, or the first red ball goes in the second bin and the second red ball goes in the first bin). This gives another way to see $$\frac 7{16}$$ as the result for the first problem, since $$\frac 18+\frac 1{16}+\frac 18+\frac 18=\boxed {\frac 7{16}}$$

We see from this that the conditional probability that the first bin has a red ball (conditioned on the fact that the second bin also has one) is $$\frac {1/8}{7/16}=\boxed {\frac 27}$$

Side note: if you prefer to work with equi-probable scenarios (not a bad idea) then you need to indicate the placement of $$r_1, r_2$$ (the two red balls) separately. Thus the scenario $$(1,1,0,0)$$, say, becomes two scenarios, $$(r_1,r_2,0,0)$$ and $$(r_2,r_1,0,0)$$. You wind up with seven scenarios in which the second bin contains at least one red ball, each of which having probability $$\frac 1{16}$$.