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The first basket contains 2 black and 2 white balls, while the second basket contains 2 black and 1 white balls. A randomly selected ball was replaced from first basket to the second one. Further balls in the second basket were properly mixed and one of them is returned back to the first basket. Find the probability that set of black and white balls is the same after all these manipulations.

I've figured out that we basically need to find the sum of probabilities of the next two events: take black and return black & take white and return white. But when I started to calculate the p of the first event I've run into a problem:

$0.5 (\text{because $2$ black balls out of $4$}) \cdot 2 (\text{since we can choose $1$ among $2$ black balls}) \cdot 0.75 (\text{to choose black again in the second basket}) \cdot 3 (\text{$3$ different black balls})$

But it exceeds $1$, what am I doing wrong? :(

N. F. Taussig
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Alice
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    Please edit your post to make it readable. – lulu Oct 26 '17 at 20:05
  • I can't read what you wrote, but the computation looks easy. There are only two possible paths: you draw white both times or you draw black both times. Just work out the probability of each and add. – lulu Oct 26 '17 at 20:09
  • Why did you erase half an hour ago your previous post on 4 and 7 with two dies ? It is not a good practice. – Jean Marie Oct 26 '17 at 20:14
  • @lulu Alice is calculating the probability for drawing black both times, but over counting by multiplying by 2 and 3, which is redundant since "choosing the ball" is already considered by the fractions. It should be just $0.5\cdot0.75$. – Graham Kemp Oct 26 '17 at 20:16
  • @GrahamKemp Ah, thank you. There was an earlier formatting problem that rendered the computation illegible. – lulu Oct 26 '17 at 20:18
  • @JeanMarie I've found a similar problem on math exachange with sum of 5 and sum of 7 and figured it out by myself. I believe that developing a dialog about unrelated to the particular problem stuff in the comments isn't a good practice either. https://math.stackexchange.com/questions/1828255/probability-sum-of-5-before-sum-of-7 – Alice Oct 27 '17 at 04:37
  • Thanks for your answer. IMHO, it is better to leave the question as it is and specify in a comment that you have found an answer either on a certain site that has to be cited or by yourself (in this last case, you can even give an answer to your question as ... an answer, or include the answer at the bottom of your question. – Jean Marie Oct 27 '17 at 05:58

2 Answers2

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The probability of a black ball being moved equals $\frac{2}{4}$, and the probability of a black ball being moved back then equals $\frac{3}{4}$. The probability of a white ball being moved equals $\frac{2}{4}$, and the probability of a white ball being moved back then equals $\frac{2}{4}$. We thus find:

$$P(orig) = \frac{2}{4} \cdot \frac{3}{4} + \frac{2}{4} \cdot \frac{2}{4} = \frac{10}{16} = \frac{5}{8}$$

jvdhooft
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Taking black and having white return happens with probability $.5$(picked black 2 of 4)*$.25$(picked white 1 of 4) = $1/8$.

the probability that you pick white and return black is $.5$(picked white 2 of four)*$.5$(picked black 2 of four) = $1/4$,

The probability of either case happening then is $1/8+1/4=3/8$. The probability that you want which is the probability that this doesn't happen is $1-3/8=5/8$

shai horowitz
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