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In first box there are 3 white and 7 black balls, in second box there are 4 white and 6 black balls and in third 8 white and 2 black balls. From the first box one ball is taken and placed in second box, then from second box one ball is taken and placed in third box and at the end from the third box one ball is taken and placed in first box. Find probability for every box that they are going to contain the same balls?

I have no clue how to do this. The solutions from my book are, respectively, $\frac{241}{605}$, $\frac{32}{55}$, $\frac{27}{55}$.

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For the second box, the number of white balls is the same if a white ball comes in and a white ball goes out, or if a black ball comes in and a black ball goes out. The chance a white ball comes in is $\frac 3{10}$ and assuming a white ball came in the chance a white ball goes out is $\frac 5{11}$. We do black balls the same way and get the total chance to be $$\frac 3{10}\cdot \frac 5{11}+\frac 7{10}\cdot \frac 7{11}=\frac {64}{110}=\frac {32}{55}$$ The others are similar, except the first is more complicated because you have to condition on all three exchanges.

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  • $\begingroup$ It would be good to write up the rest of the solution as an answer. You will discover if you understand the argument After some time (48 hrs?) you will be able to accept it. $\endgroup$ – Ross Millikan Mar 8 at 14:50
  • $\begingroup$ Okay I think I understood what You wrote so analogue to that I tried to solve it for the third box. So I have two options for the second box: Either I have 5 white and 6 black balls or I have 4 white and 7 black balls (after one ball has been taken from the first box). First option: The chance a white ball came in and out is $\frac{5}{11}$ $\frac{9}{11}$, analogue for the black so in total: $\frac{5}{11}\frac{9}{11}$+$\frac{6}{11}\frac{3}{11}$. Also for the second option: $\frac{7}{11}\frac{3}{11}$+$\frac{4}{11}$ $\frac{8}{11}$. But this is not correct. $\endgroup$ – untitled Mar 8 at 14:59
  • $\begingroup$ You need to compute the chance that you have each of the two mixes because the chance a white ball came in and out of the third box depends on the composition of the second box. You should be multiplying three fractions for each item in the sum, one for each transfer, then have four products to add. $\endgroup$ – Ross Millikan Mar 8 at 15:25
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Number the balls that are taken out virtually and chronologically with $1,2,3$.

For $i=1,2,3$ let $W_i$ denote the event that the $i$-th ball is white and let $B_i$ denote the event that the $i$-th ball is black.

Then the probability that the number of colors in e.g. the second box does not change equals:$$P(W_1\cap W_2)+P(B_1\cap B_2)=P(W_1)P(W_2\mid W_1)+P(B_1)P(B_2\mid B_1)=$$$$\frac3{10}\frac5{11}+\frac7{10}\frac7{11}$$

Now do you get the idea?

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