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We have 6 balls - 2 red, 2 blue, 2 green and 4 different cells .

We throw each ball randomly to one of the four cells.

  1. If in cell#2 there is a red ball, what is the probability that there will be also a green ball in cell#2 ?

  2. What is the probability that at least two balls with the same colour will be in the same cell?

Thanks a bunch for helping.

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closed as off-topic by Michael Rybkin, Jean-Claude Arbaut, max_zorn, Jyrki Lahtonen, ancientmathematician May 12 at 8:12

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    $\begingroup$ What have you tried? For 2, what is the chance the two red balls are in the same bin? $\endgroup$ – Ross Millikan May 11 at 16:08
  • $\begingroup$ For 2 I think its 1/16 because for each ball its 1/4 a chance to be in a cell. For 1 the same? I feel silly to write these answers because I'm not sure. $\endgroup$ – Trolling Killer May 11 at 16:23
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    $\begingroup$ No, it doesn’t matter which cell the first red ball goes into, the second just has to match. For 1 the colors are independent. $\endgroup$ – Ross Millikan May 11 at 16:25
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I am assuming all cells can contain up to 6 balls.

  1. Note that the chance of a particular cell NOT containing a green ball is equal to the chance of both green balls 'missing' that cell; i.e. $(\frac{3}{4})^2 = \frac{9}{16}$. The chance of cell 2 containing a green ball is therefore $\frac{7}{16}$. Red ball doesn't matter.

  2. The chance of the two balls of one colour not to be in the same cell is $\frac{3}{4}$. The chance of this happening for all three colours is $\frac{3}{4}$ cubed. Your desired answer is therefore $ 1 - (\frac{3}{4})^3$.

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  • $\begingroup$ You misread the question. There are four cells in which the balls could be placed. $\endgroup$ – N. F. Taussig May 11 at 16:42
  • $\begingroup$ Yes, I did! Edited with correct numeric solutions. $\endgroup$ – Alexander Geldhof May 11 at 17:51
  • $\begingroup$ In your response to the second question, you calculated the probability that no cell contains two balls of the same colour. Since the question asks for the probability that at least two balls with the same colour will be in the same cell, we need to subtract your answer from $1$. $\endgroup$ – N. F. Taussig May 11 at 19:00
  • $\begingroup$ You are correct. $\endgroup$ – Alexander Geldhof May 11 at 19:01

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