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A box has 10 red balls and 5 black balls. A ball is selected from the box. If the ball is red, it is returned to the box. If the ball is black, it and 2 additional black balls are added to the box. Find the probability that a second ball from the box is red? Find the probability that a second ball from the box is black?

I understand this question has been asked before, but I am having difficulty understanding the process behind it and why my initial process was incorrect. Also, this question can be found in Chapter 1, 21 in Introduction to Probability Theory.

My Process:

$$P(R_2)=P(R_2|R_1)+P(R_2|B_1)$$ $$=\frac{r}{b+r}+\frac{r}{b+r+2}$$ $$=\frac{10}{15}+\frac{10}{17}$$ $$=1.25...$$

This is clearly wrong. When I looked at the example in the text provide, they state $$P(R_2)=P(R_1\cap R_2)+P(B_1\cap R_2)$$ which is how I eventually got to the correct answer. They concluded this because they assume $B_1$ and $R_1$ are mutually disjoint events, which I understand because they are the first event in the sequence. But, why was my first idea incorrect?

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  • $\begingroup$ What is color of two additional balls.? $\endgroup$
    – Rishi
    Commented Oct 1, 2019 at 16:00
  • $\begingroup$ Black. My apologies I missed a vital part of the question. $\endgroup$ Commented Oct 1, 2019 at 16:02
  • $\begingroup$ Why is your first idea incorrect? How should we answer that? Why should it have been correct in the first place? Nothing should have suggested that it was a correct thing to do. In general $Pr(A) \neq Pr(A\mid B)+Pr(A\mid B^c)$. Rather, what is correct is $Pr(A)=Pr(A\cap B)+Pr(A\cap B^c)$ as this is a special case of the law of total probability. This continues to expand as $Pr(A)=Pr(B)Pr(A\mid B)+Pr(B^c)Pr(A\mid B^c)$. You effectively forgot to condition your probabilities on the event of having actually entered into that case. $\endgroup$
    – JMoravitz
    Commented Oct 1, 2019 at 16:06
  • $\begingroup$ "They assume $B_1$ and $R_1$ are mutually disjoint events" They don't have to assume this, it is clearly true. That is because when drawing a ball there is a clear single color that the ball is at a time. It is impossible for the single ball that we draw to simultaneously be black and red. And this isn't true just for the first draw... $R_k$ and $B_k$ are just as disjoint. $\endgroup$
    – JMoravitz
    Commented Oct 1, 2019 at 16:08
  • $\begingroup$ I was thinking of your third equation but I forgot to condition the probabilities as you pointed out. That is why I was incorrect, but almost had the right idea. No need to be mean. $\endgroup$ Commented Oct 1, 2019 at 16:11

1 Answer 1

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There are two cases

  1. In first draw ball is red and again it's red , probability to it is $$\frac{r}{r+b}.\frac{r}{r+b}$$

  2. In first draw ball is black and again it's red , probability to it is $$\frac{b}{r+b}.\frac{r}{r+b+2}$$

so required probabilty is $$P(R)= \frac{r}{r+b}.\frac{r}{r+b}+\frac{b}{r+b}.\frac{r}{r+b+2}$$

Similarly

$$P(B)=\frac{r}{r+b}.\frac{b}{r+b}+\frac{b}{r+b}.\frac{b+2}{r+b+2}$$

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