Probability of extracting a ball after two balls were swapped

Question

We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls. We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?

A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.

$$P(\text{white ball from box1})=\frac{10}{21}$$ $$P(\text{black ball from box1})=\frac{11}{21}$$ $$P(\text{white ball from box2})= \frac{12+1}{26}\frac{10}{21}+\frac{12}{26}\frac{11}{21}=\frac{13\cdot10+12\cdot11}{26\cdot 21}$$ $$P(\text{black ball from box2})= \frac{13}{26}\frac{10}{21}+\frac{13+1}{26}\frac{11}{21}=\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$

$$P(\text{white ball from box1 after swap})=\frac{10+1}{22}\cdot\frac{13\cdot10+12\cdot11}{26\cdot 21}+\frac{10}{22}\cdot\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$ $$=\frac{2882}{12012}+\frac{2840}{12012}=\frac{5722}{12012}=0.4763$$

Have I done it correctly? I don't know how to verify myself with this kind of problems. Also, is there a trick to find the probability if there were two swaps?

Answer 1

Here’s another approach. [Added: Also, see more on this approach here]

Equivalently, we can do the following.

Reach into the first box and write “S” on one ball (the one to swap). Now choose a ball from the first box. If it is not the “S” ball, that’s your ball. The probability of this happening is $20\over21$, and the ball you choose will be white with probability $10\over21$. If you do choose the “S” ball (you do this with probability $1\over21$), discard it by throwing it into the second box and then choose a ball at random from the second box (now containing an extra ball), and that’s your ball. If you had to do this, the probability of a white result is ${12+{10\over21}\over26}$, because if we add a randomly-chosen ball from box 1 to box 2, the number of white balls in box 2 effectively increases from $12$ to $12+{10\over21}$ and the number of balls in box 2 increases to $26$. Therefore the total probability you want is

$$p = {20\over21}\cdot{10\over21}+{1\over21}\cdot{12+{10\over21}\over26}={2731\over5733}\approx 0.4763649.$$

P.S. I don’t see an easy way to adapt this approach for two swaps.

Answer 2

So... for this, I'll need a lot of names for events that sound similar, so bear with me.

Since you have defined a "swap" to be where a ball is first moved from the first box into the second box, and then a ball is chosen from the second box which now includes the ball just moved and this is moved back to the first box.

Let $b_1$ be the event that a black ball was moved from the first into the second. Let $w_1$ be the event a white ball was moved from the first into the second.

Let $b_2$ be the event that a black ball was moved from the second into the first. Let $w_2$ be the event that a white ball was moved from the second into the first.

Let $W$ be the event that a white ball was selected from the first box after the swap occurred.

We are interested in finding $Pr(W)$

By total probability we have:

$$Pr(W)=Pr(W\mid b_1b_2)Pr(b_2\mid b_1)Pr(b_1)+Pr(W\mid b_1w_2)Pr(w_2\mid b_1)Pr(b_1)\\ +Pr(W\mid w_1b_2)Pr(b_2\mid w_1)Pr(w_1)+Pr(W\mid w_1w_2) Pr(w_2\mid w_1)Pr(w_1)$$

Each of these terms can be found with relatively little effort, though it is tedious to find so many. I'll do one pair of them for now and leave the rest for you.

$Pr(W\mid b_1b_2) = \frac{10}{21}$ and $Pr(b_2\mid b_1)=\frac{14}{26}$ and $Pr(b_1)=\frac{11}{21}$

Answer 3

The probability of ending up with an additional white ball equals:

$$\frac{11}{21} \frac{12}{26}$$

The probability of ending up with an additional black ball equals:

$$\frac{10}{21} \frac{13}{26}$$

We thus find an overall probability of:

$$\frac{11}{21} \frac{12}{26} \frac{11}{21} + \frac{10}{21} \frac{13}{26} \frac{9}{21} + \left(1 - \frac{11 \cdot 12 + 10 \cdot 13}{21 \cdot 26}\right) \frac{10}{21} = \frac{1452 + 1170 + 2840}{11466} = \frac{5462}{11466} \approx 0.4764$$

Answer 4

Although this is already answered, I'll add my two cents. Perhaps someone will find it more intuitive.

Call probability of picking white ball from boxes $p_1$, and $p_2$. Since we're exchanging one ball, the probability will be between $p_1$ and $p_2$, closer to $p_1$, since we have $21$ balls in box 1.

$ p(W) = \frac{20}{21}p_1 + \frac1{21}p_2'$

notice we wrote $p_2'$. The dynamics of the problem will change $p_2$ mid process. But we can apply the same logic here, since 25 balls were in box 2 and now 1 more added.

$p_2' = \frac{25}{26}p_2 + \frac1{26}p_1 $

Combining the two equations will give

$ p(W) = \frac{20}{21}p_1 + \frac1{21}(\frac{25}{26}p_2 + \frac1{26}p_1) $

Here again, the coefficients will tell the story. The final ball will come from box 1 with probability $\frac{20}{21}+\frac1{21}\frac1{26}$, that is either directly or via box 2 detour. Or, with probability $\frac1{21}\frac{25}{26}$ from the second box. Therefore the final probability is the weighted average of the original probabilities with these weights (which themselves are probabilities).

In general let number of balls in box 1 and 2 be equal to $n$ and $m$ respectively. Define $\alpha = \frac{m}{n(m+1)}$. Also, let $p_i(s)$ shows the probability of drawing white in box $i$ after $s$ swaps. Where $p_i=p_i(0)$. Then, we can write

$$ p_1(s+1) = (1-\alpha) p_1(s) + \alpha p_2(s) $$

and

$$ p_2(s+1) = p_2(s) + \frac{n}{m}(p_1(s) - p_1(s+1)) $$

second equation is due to conservation of total white balls. Now you can compute probabilities after any number of swaps from this iterative process.