# Balls out of a box, with return and adding more balls

A box contains $a$ red balls and $b$ beige balls. We take out a ball at random, return it, and with it placing additional $k$ balls of the other color, e.g. if a red ball was picked first, we return it and additional $k$ beige balls.

I need to calculate the probability to pick a red ball second, event $C_2$.

My attempt: $\newcommand{\set}[1]{\left\{{}#1\right\}}$

The sample space is $\Omega = \set{r_1,\ldots,r_a,c_1,\ldots,c_b}\times\set{r_1,\ldots,r_a,c_1\ldots,c_b,r_{a+1},\ldots,r_{a+k},c_{b+1},\ldots,c_{b+k}}$, where $r$ represents a red ball and $c$ represents a beige ball. We have four basic events that can occur: $A_1 = \set{(r,c)},A_2 = \set{(c,r)}, A_3=\set{(c,c)},A_4=\set{(r,r)}$. Rewriting in terms of subsets of the sample space: $$A_1 = \set{(r_i,c_j) | 1\leq i\leq a,\ 1\leq j\leq b+k} = \set{r_1,\ldots,r_a}\times\set{c_1,\ldots,c_{b+k}}$$ $$A_2 = \set{(c_i,r_j) | 1\leq i\leq b,\ 1\leq j\leq a+k} = \set{c_1,\ldots,c_b}\times\set{r_1,\ldots,r_{a+k}}$$ $$A_3 = \set{(c_i,c_j) | 1\leq i\leq b,\ 1\leq j\leq b} = \set{c_1,\ldots,c_b}^2$$ $$A_4 = \set{(r_i,r_j) | 1\leq i\leq a,\ 1\leq j\leq a} = \set{r_1,\ldots,r_a}^2$$ Now, event $C_2$ is actually $C_2 = A_2\cup A_4$. Since these sets are disjoint, we have that $$P(C_2) = P(A_1) + P(A_4)$$ Since this is a symmetric space, we have that $$P(A_2) = \frac{b(a+k)}{(a+b)(a+b+2k)}$$ $$P(A_4) = \frac{a^2}{(a+b)(a+b+2k)}$$ and therefore $$P(C_2) = \frac{a^2 + ab + ak}{(a+b)(a+b+2k)}$$

A friend of mine made a different calculation, more simple, and reached a similar answer just with $(a+b+k)$ in the denominator. Also, his answer was convincing.

Where am I wrong?

• You state $\Omega=U\times V$. You can do that if you like. However, then probability distribution on this set is not uniform. E.g. $P(\langle r_1,r_{a+1}\rangle)=0$. – drhab Mar 2 '16 at 15:11
• Why is this the case? – Joshhh Mar 2 '16 at 15:17
• If at first ball $r_1$ is drawn then no red balls are added. In spite of that in your construction $k$ red balls are added ($r_{a+1},\dots, r_{a+k}$). Virtually you can do that, but this under the condition that the probability that one of these balls is drawn at the second time equals $0$. – drhab Mar 2 '16 at 15:22
• So what I need to do is to define a piecewise probability function? – Joshhh Mar 2 '16 at 15:26
• There are $(a+b)(a+b+k)$ pairs in $\Omega$ that have equal and positive probability (so $\frac{1}{(a+b)(a+b+k)}$ for each of them). The others have probability $0$. For solving this question you actually do not need to take this road (see the answers). – drhab Mar 2 '16 at 15:31

Not following your sample space breakdown.

To do the problem, note (as you do in your calculation) that there are two paths to victory. using your notation, we write the paths as $A_2$ (first beige then red) and $A_4$ (both of the first two are red). As you point out, the events are disjoint so we just need to compute the two probabilities and add.

$A_2$: probability of that first beige is $\frac b{a+b}$. Probability, then, of the second red is $\frac {a+k}{a+b+k}$. Thus $$P(A_2)=\frac b{a+b} \times \frac {a+k}{a+b+k}$$

$A_4$: probability of that first red is $\frac a{a+b}$. Probability, then, of the second red is $\frac {a}{a+b+k}$. Thus $$P(A_2)=\frac a{a+b} \times \frac {a}{a+b+k}$$

Adding we get $$\frac {b(a+k)+a^2}{(a+b)(a+b+k)}$$

• The sample space is the set of all possible outcomes. So it is the Cartesian of the set of possibilities before taking any ball out and the set of possibilities after taking one ball out. Am I wrong? – Joshhh Mar 2 '16 at 14:54
• The set of possible outcomes has $4$ values: $\{RB,\;BR,\;BB,\;RR\}$ – lulu Mar 2 '16 at 15:02
• Say you number the balls including the not-yet-existing balls (as I think you are doing). That's fine and does give you a larger sample space, but lots of the points in that space have probability $0$. For example, $\{r_1,r_{b+1}\}$ can't occur. – lulu Mar 2 '16 at 15:05

Because you first have $a + b$ balls. Then, no matter what you drew, there are $a + b + k$ balls for the second draw. So the denominator is $(a + b)(a + b + k)$. You correctly calculate the numerator as $a^2 + b(a + k)$.

• But aren't there 2 possible outcomes after taking 1 ball out? Or since I give no significant identity to balls of the same color I shouldn't add $2k$ additional possible outcomes? – Joshhh Mar 2 '16 at 14:57
• No, if your first draw is red you replace that ball back along with $k$ beige ones giving you $a + b + k$ balls for the second draw. If you draw a beige at first you put it back with $k$ red balls, again giving you $a + b + k$ balls for the second draw. – Paul Evans Mar 2 '16 at 15:03

$R_{i}$ denotes the event that a red ball is drawn at the $i$-th draw.

$B_{i}$ denotes the event that a beige ball is drawn at the $i$-th draw.

Then:

$$P\left(R_{2}\right)=P\left(R_{1}\cap R_{2}\right)+P\left(B_{1}\cap R_{2}\right)=P\left(R_{1}\right)P\left(R_{2}\mid R_{1}\right)+P\left(B_{1}\right)P\left(R_{2}\mid B_{1}\right)$$$$=\frac{a}{a+b}\frac{a}{a+b+k}+\frac{b}{a+b}\frac{a+k}{a+b+k}=\frac{a^{2}+b\left(a+k\right)}{\left(a+b\right)\left(a+b+k\right)}$$