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If there are $72$ white and $24$ red balls to be arranged in $4$ boxes equally (i.e. $24$ balls in each box) what is the probability that all $24$ red balls end up in the same box.

(If this is a simple question please forgive me, I've lost touch with these kind of problems)

N. F. Taussig
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Priyatam bolisetty
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  • There are only four arrangements where all $24$ red balls are together... do you see why? Next, can you calculate the total number of possible arrangements, $N$? In which case the probability is $4/N$. – mjqxxxx Jan 21 '16 at 04:26

3 Answers3

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An important point to remember is that arrangements using the stars and bars formula are not equiprobable. A recent answer explaining this can be seen here

Favorable ways are obviously 4, and total ways = $\binom{96}{24}\;$, thus $Pr = \dfrac{4}{\binom{96}{24}}$

Further explanation

A lot of comments are being received, so following the dictum "There's more than one way to skin a cat", here's a colorful explanation that may appeal.

Imagine a line of the 96 balls randomly lined up, and divided by | into 4 boxes
$\color{red}\bullet\circ\circ\circ\color{red}\bullet\circ\circ\circ\circ\circ\circ\circ\color{red}\bullet\circ\circ\circ\circ\circ\circ\circ\color{red}\bullet\circ\circ\circ | \circ ......$

The red balls could be placed in the line in $\binom{96}{24}$ equiprobable ways of which only 4 would have all the red balls in the same box.

Community
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true blue anil
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  • I think this does not take into account the capacity of the box – Alex Jan 21 '16 at 10:18
  • You could work out a smaller analogous problem: distribute 2 red cards and 6 black cards to 4 people. What is the probability that one person gets both the red cards ? – true blue anil Jan 21 '16 at 10:18
  • I still get $\binom{5}{3}$ in the denominator – Alex Jan 21 '16 at 10:22
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    @Alex: I don't understand. Once you distribute the red balls, the white ones automatically get distributed. You might like to enumerate the problem in my previous comment. – true blue anil Jan 21 '16 at 10:22
  • @Alex: I corrected a typo, the denominator is obviously $\binom{96}{24}$ – true blue anil Jan 21 '16 at 10:26
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    @Alex: A full expression would be $\frac{\binom41\binom{24}{24}\binom{72}{72}}{\binom{96}{24}\binom{72}{72}}$ – true blue anil Jan 21 '16 at 10:31
  • 'Once you distribute the red balls, the white ones automatically get distributed' - right, and that's how I got $\binom{27}{3}$, like @Jack's wasted life – Alex Jan 21 '16 at 10:43
  • Since the capacity of the box is 24, all we need to look at is how to distribute red balls, which can be done in $\binom{24+3}{3}$ ways. – Alex Jan 21 '16 at 10:44
  • I'm also doubtful about this solution, using $\binom {96} {24}$ aren't we assuming that we place the 24 red balls first? Shouldn't we account for the cases in wich we put one or more white balls first? – N74 Jan 21 '16 at 10:57
  • @Alex: The whole point is that $\binom{27}{3}$ are not equiprobable. Pl. consult the answers referred to. – true blue anil Jan 21 '16 at 11:42
  • @N74: Pl. see the full expression in another comment of mine. Both red and white balls have been distributed simultaneously, – true blue anil Jan 21 '16 at 11:45
  • Perhaps I don't understad something, but $\binom{96}{24}$ is the total number of ways to allocate 24 identical balls in 96 slots. How does this account for the fact that we have 4 boxes? – Alex Jan 21 '16 at 11:52
  • I have simplified the answer, and added a diagrammatic explanation which should clear all doubts. – true blue anil Jan 21 '16 at 16:46
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You can choose $24$ balls for the $1$st box in $\binom{96}{24}$, for the $2$nd box in $\binom{72}{24}$, for the $3$rd box in $\binom{48}{24}$, for the $4$-th box in $\binom{24}{24}=1$ way. If we have to put $24$ red balls in one box we can choose the box to contain them in $4$ ways and fill up the remaining boxes in $\binom{72}{24}\binom{48}{24}\binom{24}{24}$ ways. So the required probability is $$ {4\binom{72}{24}\binom{48}{24}\binom{24}{24}\over\binom{96}{24}\binom{72}{24}\binom{48}{24}\binom{24}{24}}={4\over\binom{96}{24}} $$ like the post of true blue anil says.

Jack's wasted life
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  • Yes, but I simplified computations by focusing only on red balls, since the white ones will automatically occupy whatever slots are left. Maybe that is what has been raising so many queries ! – true blue anil Jan 21 '16 at 13:22
  • I need my steps :) – Jack's wasted life Jan 21 '16 at 13:25
  • See my colorful "stepless" PS, maybe that appeals :) – true blue anil Jan 21 '16 at 13:49
  • @trueblueanil Ah now I understand your simplification, a minor correction 3rd arrangement should be rw-ww-rw-ww and your PS is a sight for sore eyes. +1 – Jack's wasted life Jan 21 '16 at 14:24
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    +1 This counting considers the balls are distinguishable, while true blue anil's answer consider them undistinguishable (except for colour). Obviously, both ways must give the same result, but the counting is different. I slightly prefer this way. – leonbloy Jan 21 '16 at 15:13
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The value is the same of the methods by true blue anil and Jack's wasted life but I feel this method easier to understand:

First of let's recognise that placing 24 red balls in 4 buckets is (in this case) the same that placing 24 red balls and 72 white balls in 4 buckets. So I'm evaluating this probability.

We have 4 bucket of capacity 24, so 96 places to place our red balls into.

The first ball can be placed everywhere so its probability is 1 (or 96/96 if you prefer). The second ball, given the position of the first, has only 23 favorable places out of the 95 remaining. The third 22 out of 94 and so on.

So the final probability is: $$ \frac{23}{95}\times\frac{22}{94}\times\frac{21}{93}\times\cdots\frac{1}{73}=\frac{1}{\binom{95}{23}} $$

As $96/24=4$: $$ \frac{1}{\binom{95}{23}}=\frac{4}{\binom{95}{23}\frac{96}{24}}=\frac{4}{\binom{96}{24}} $$ Same value, different ways to reach it.

N74
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