4

Given 3 red balls, 197 black balls, and 5 bins.  The balls are randomly distributed with 40 balls in each bin.  What is the probability that the three red balls are in three separate bins (each of the three bins has one red ball; i.e., no two red balls are in the same bin)?

Different variants of this question has been asked – but somehow I still have trouble computing the correct probability.

One answer I found was:

$$\frac{\left(\begin{array}{l} 5 \\ 1 \end{array}\right)\left(\begin{array}{l} 4 \\ 1 \end{array}\right)\left(\begin{array}{l} 3 \\ 1 \end{array}\right)}{125} = 12/25$$

It only focuses on red balls, and there are 125 ways that 3 red balls can be allocated in bins.

However, another alternative way of computing this seems to be:

$$ \frac{5 \times {3 \choose 1} {199 \choose 39} {3 \choose 1} {199 \choose 39} {3 \choose 1} {199 \choose 39} {200 \choose 40} {200 \choose 40}}{{200 \choose 40}{200 \choose 40}{200 \choose 40}{200 \choose 40}{200 \choose 40}} = 27/25 $$

Obviously it seems that the alternative way is incorrect (it's above 1 as a probability). So my question is: is this alternative way of computing incorrect? How so? How can I fix it?

Scott - Слава Україні
  • 1,015
windweller
  • 223
  • 1
    The question is not clear. You want exact 1 red ball in one of the three bins? – callculus42 Jul 08 '22 at 17:58
  • @callculus42: I think OP wants each of three bins to contain exactly one red ball. – Brian Tung Jul 08 '22 at 17:59
  • 1
    Inferring from the question title and the calculations shown: There are 5 bins. Each of the 200 balls is placed in a bin independently at random, and we want the probability that no two red balls are in the same bin. – Karl Jul 08 '22 at 18:03
  • @Karl Yes, I missed the title. So basically we have to distribute 3 red balls in 5 bins, where one bin contains at most on ball?. In my opinion the probability is $\frac{10}{35}$. – callculus42 Jul 08 '22 at 18:14
  • 3
    The second attempt shows that $40$ balls each are to be put into the $5$ bins. Is this a condition ? – true blue anil Jul 08 '22 at 18:15
  • Yes, exact 1 red ball in 1 bin – windweller Jul 08 '22 at 18:19
  • I forgot to add — each bin will have exactly 40 balls in them – windweller Jul 08 '22 at 18:21
  • @trueblueanil Yes, all bins need to have equal number of balls (40 balls in each). Thanks for bringing it up! – windweller Jul 08 '22 at 18:29
  • @windweller Beside the red balls, you have to distribute 39 black balls in 3 bins and 40 black balls in 2 bins. $39\cdot 3+40\cdot 2=197$. Right? – callculus42 Jul 08 '22 at 18:46
  • @callculus42 Yeah, you are correct. I also think I recognize the error in the first answer now...it's not fixing each bin to have 40 balls. – windweller Jul 08 '22 at 18:49
  • I have posted a simpler solution. – true blue anil Jul 08 '22 at 19:18
  • This is still unclear. One red ball in each of five bins requires five red balls. If you have only three red balls, this is trivially impossible, and the probability is zero. – Scott - Слава Україні Jul 12 '22 at 17:43
  • @Scott-СлаваУкраїні Hi Scott, thanks for pointing that out -- I think my post got edited wrong. 3 red balls, 5 bins. 3 bins each have 1 red ball, the rest 2 bins only have black balls. – windweller Jul 16 '22 at 18:43

3 Answers3

5

Imagine $200$ slots divided into $5$ groups of $40$. By far the simpler computation is just distributing the $3$ red balls one by one into different groups.

The first red ball can go anywhere, and $Pr = \frac{160}{199}\frac{120}{198} = \frac{3200}{6567}$

true blue anil
  • 41,295
  • I appreciate it!! Can you maybe explain this line of logic a bit more? I do agree it's a lot simpler and beautiful -- but I'm a bit rusty on combinatorics :( – windweller Jul 08 '22 at 19:23
  • Let's just name the red balls $A,B,C$, so in more formal terms, we are computing P(Place $A$)(Place $B|A$ placed)P(Place $C|A,B$ placed), and since they have to go to different groups, and to start with $A$ can be placed anywhere (ie $\frac{200}{200}$ ), you should understand. The first wrong answerer used the same idea, but with wrong figures. – true blue anil Jul 08 '22 at 19:36
4

Let's enumerate all of the balls as follows: $$R_1R_2R_3B_1B_2B_3\dots B_{196}B_{197}$$

Using this enumeration, we see that there is a one to one correspondence between placing the $200$ balls into the $5$ bins and two hundred digit numbers whose digits are sampled from $\{1,2,3,4,5\}$ with replacement. For example, the two hundred digit number $$143555\dots55$$ signifies that we put $R_1$ into bin $1$, $R_2$ into bin $4$, $R_3$ into bin $3$, and the remaining $197$ black balls into the fifth bin.

The numbers of ways to distribute the $200$ balls into the $5$ bins in such a way that the the red balls all belong to different bins is precisely the number of two hundred digit numbers whose first three digits are all distinct. The corresponding probability is $$\frac{{5 \choose 3}3!\times 5^{197}}{5^{200}}=\frac{12}{25}$$

If we're distributing the balls in such a way that each bin contains $40$ balls, then this becomes $$\frac{{5 \choose 3}3!\times\frac{197!}{(39!)^3 (40!)^2}}{\frac{200!}{(40!)^5}}=\frac{3200}{6567}$$

user429040
  • 535
  • @windweller I don't know if it is right. But if you think the answer is right then accept ($\checkmark $) the answer. For the next (extended) question open a new question. – callculus42 Jul 08 '22 at 18:52
  • @callculus42 That's a great comment! Thank you. I didn't want to spam people with multiple questions with the same setup – windweller Jul 08 '22 at 18:54
  • @callculus42 I ran some simulation on this and the result is very close to 0.487 – windweller Jul 08 '22 at 18:55
  • @windweller I don't say that the answer is not right. I just haven't checked it. If your simulation support you calculation, it is even better. – callculus42 Jul 08 '22 at 18:57
  • @callculus42 The new question is here: https://math.stackexchange.com/questions/4488913/balls-in-bins-3-red-balls-for-5-bins-2-balls-in-1-bin-1-ball-in-another – windweller Jul 08 '22 at 18:58
  • @windweller Good luck with this question. Tip: Any attempt is always welcomed for this forum. – callculus42 Jul 08 '22 at 19:02
0

The alternative way is attempting to calculate ways to place the 3 red balls and 39 others into three boxes, and 40 balls into the two other boxes, when selecting from the 200 balls ... each time.

The error is in not decrementing the source count for the balls. The process is done without replacement.

$$\small{\dfrac{\dbinom 53\, \dbinom{3}{1}\dbinom{197}{39}\,\dbinom{2}{1}\dbinom{158}{39}\,\dbinom{1}{1}\dbinom{119}{39}\,\dbinom{0}{0}\dbinom{80}{40}\,\dbinom{0}{0}\dbinom{40}{40}}{\dbinom{200}{40}\dbinom{160}{40}\dbinom{120}{40}\dbinom{80}{40}\dbinom{40}{40}}\\=\dfrac{\dbinom 53\dbinom{3}{1,1,1,0,0}\dbinom{197}{39,39,39,40,40}}{\dbinom{200}{40,40,40,40,40}}\\=\dfrac{5!}{3!2!}\dfrac{3!}{1!^30!^2}\dfrac{197!}{39!^340!^2}\dfrac{40!^5}{200!}\\=\dfrac{3200}{6567}}$$

Another approach. Count ways to place the three balls among five groups of forty places when selecting three among two hundred places. Which produces the same result.

$$\dfrac{\dbinom 53~\dbinom{40}{1}\dbinom{40}1\dbinom{40}1\dbinom{40}0\dbinom{40}0}{\dbinom{200}3}=\dfrac{3200}{6567}$$

Graham Kemp
  • 129,094