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First box contains 6 Red balls, 4 Green Balls. Second box contains 7 Red Balls and 3 Green Balls. A ball is randomly picked from the first box and then put in the second box. Then a ball is randomly selected from the second box and then put in the first box. (a) What is the probability that a red ball is selected from the first box and a red ball is selected from the second box. (b) At the conclusion of the process, what is the probability that the number of red & green balls in the first box is identical to the numbers at the beginning.

My attempt:

(a)$P(RR)=\frac{6}{10}.\frac{8}{11}=\frac{48}{110} $

(b)The only way the number of R&G balls in the first box will be same if either a red ball was picked from box 1 and then a red ball was picked from box 2 and put back on box 1, or the same process with green ball has to have happened.

$P(RR)+P(GG)=\frac{48}{110}+(\frac{4}{10}.\frac{4}{11})=\frac{64}{110}=\frac{32}{55}$

Am I on the right track? Any help is much appreciated. I apologize for the fact that the title is a bit vague but could not come up with a better title.

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  • $\begingroup$ @ Math__QED. Can you please explain how it is 8/11? I am sorry I am a bit confused.I was thinking 1st box has 10 balls, of which 4 are green. Second box contains 10 balls of which 3 are green. If we pick a green ball from box 1 and put it on box 2, then box 2 will have 11 balls of which 4 are green.Probability of picking a green ball from box 1 and then picking a green ball from box 2 is 4/10*4/11 $\endgroup$
    – nova_star
    Mar 20, 2018 at 19:27
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    $\begingroup$ I'm sorry. Your reasoning is correct. I wasn't thinking clearly. $\endgroup$
    – user370967
    Mar 20, 2018 at 19:32

2 Answers 2

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Your reasoning is correct.

If you want a more rigorous approach, you can do the following for (a):

Let $R_1$ be the event that the ball chosen from the first box is red.

Let $R_2$ be the event that the ball chosen from the second box is red.

Then, $$\mathbb{P}(R_1 \cap R_2) = \mathbb{P}(R_2 \mid R_1) \mathbb{P}(R_1) = 8/11*6/10 $$

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  • $\begingroup$ Thanks. I was looking for a more rigorous way to do this! $\endgroup$
    – nova_star
    Mar 20, 2018 at 20:19
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Yes, your reasoning is correct for both $(a)$ and $(b)$

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