One box contains 3 white and 6 black balls. A second box contains 5 white and 4 black balls. One ball is chosen at random from the 1st box and put into the 2nd box. Then a ball is randomly selected from the 2nd box and put in the 1st box.

(1) What is the probability that the 1st and 2nd selected ball are both white?

So far I have this:

 {3 w and 6 B} box 1
 {5 w and 4 B} box 2

Now if both are white then I calculate the complement like $1- \frac{3}{9} \cdot \frac{5}{10} = \frac{5}{6}$ then I want the opposite probability then $\frac{1}{6}$ of both are whites.

Is that point correct if is are there other perspective to solve this problem?

Thanks, and sorry to don't post my work before.

(2) At the end of the experiment, what is the probability that the configuration of white and black balls in the two boxes is the same as in the beginning?

This question I need to think more I am stuck, but if you can help me with some key advise I will be happy.

  • $\begingroup$ (1) $\frac{1}{3} \cdot \frac{6}{10} + \frac{1}{3} \cdot \frac{5}{10} = \frac{11}{30}$ $\endgroup$ – Gustavo Sep 12 '15 at 2:20
  • 1
    $\begingroup$ Please show your attempt. $\endgroup$ – Seow Fan Chong Sep 12 '15 at 4:57
  • $\begingroup$ I understand the first term , 1/3 of choosing white from the 1st box and 6/10 tfor choose the second white from the 2 box. But I don't understand why you add 1/3 and 5 /10 later on. $\endgroup$ – Electro82 Sep 14 '15 at 18:22

I solved, here is the solution.

1) 1/3 *6/10 = 1/5 =0.2 the probability that the first and second selected ball are both white is 20%.

2) The only way the two boxes end like in the beginning is that we move two white or two black ball in the experiment then

1/3 * 6/10 +6/9 *5/10 = 1/5 + 1/3 =8/15 =0.53

Therefore, there are 53% the experiment en like in the beginning.


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