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This question is linked to this question. I was told to create another question instead if I have additional variation of this setup that I need to ask :) Sorry for spamming.

Question: Given 3 red balls, 197 black balls. What is the probability of 2 red ball in 1 bin and 1 red ball in another bin. Each bin needs to have 40 balls in them.

What is the probability of this placement happening?

windweller
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    I think you can use the approach I demonstrated in the solution I provided to your other question to get the probability of $$\frac{{5 \choose 2}2!\times\frac{3!}{2!}\times\frac{197!}{38!\cdot 39!\cdot(40!)^3}}{\frac{200!}{(40!)^5}}$$ – user429040 Jul 08 '22 at 19:02
  • @user429040 you answer is correct!! The worked out exact number is 0.4751 and it's very close to my simulated result. Let me re-read your explanation and try to understand it. Thank you very much! (Feel free to post this as an answer and I'll mark it the correct one) – windweller Jul 08 '22 at 19:35

3 Answers3

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Choose a bin to contain two red balls: $5$ options.

Choose two balls in that bin: $\binom{40}{2}$ options.

Choose a bin to contain the other red ball: $4$ options.

Choose one ball in that bin: $40$ options.

$$\frac{5\times\binom{40}{2}\times 4\times 40}{\binom{200}{3}}=\frac{1040}{2189}\approx 0.4751$$

Daniel Mathias
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  • This math seems so easy to follow! I try to extend it to a new question: exactly 1 red ball for each bin (for 3 bins), but seems it wouldn't work...I wrote: $$\frac{5 \times {40 \choose 1} \times 4 \times {40 \choose 1} \times 3 \times {40 \choose 1}}{{200 \choose 3}} \approx 2.7$$ and this is incorrect... – windweller Jul 08 '22 at 20:22
  • @windweller You need $\binom 53$ choosing three bins. Then choose a ball for each bin. – Daniel Mathias Jul 08 '22 at 21:52
  • Ahhh got it. Thank you! – windweller Jul 08 '22 at 22:37
  • Can you explain why the denominator is (200 choose 3)? It seems like we are choosing 3 red balls out of 200 balls... – windweller Jul 08 '22 at 23:51
  • For your original answer, you have 5 x 4, how can I understand this in terms of (n choose k)? One answer I thought of was: we care about bin with 2 balls and bin with 1 ball, while 3 bins each with 1 ball is interchangeable... – windweller Jul 09 '22 at 00:03
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Using the idea of placing red balls turn by turn, as in the linked question, but with one more twist, if we start by first choosing the duo to be together in $\binom32= 3$ ways,

thus $Pr = 3\times\frac{39}{199}\frac{160}{198} = \frac{1040}{2189}$

true blue anil
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  • Hi, thank you for answering this! I don't think it's correct...the correct answer is what the comment suggested (which is close to 0.4751) – windweller Jul 08 '22 at 19:33
  • A factor of $3$ was needed. – true blue anil Jul 08 '22 at 20:15
  • Hi true blue anil, I really appreciate you answering both questions!! I'm giving you the correct answer mark here :) – windweller Jul 08 '22 at 20:23
  • Thanks, this approach came to me when a questioner asked for an alternative intuition. You might like to see because a variety of approaches of different degrees of complexity were posted. https://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace/1345436#1345436 – true blue anil Jul 08 '22 at 20:41
  • Yeah, that is so much appreciated. I do need to understand the different degrees of complexity and different variants of this setup! – windweller Jul 08 '22 at 23:39
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My approach is:

$$\frac{(5 \times 4) \times \left[\binom{3}{2} \times \binom{197}{38}\right] \times \left[\binom{1}{1} \times \binom{159}{39}\right]}{\binom{200}{40} \times \binom{160}{40}} \tag1 $$

$$\times \frac{\binom{120}{40} \times \binom{80}{40} \times \binom{40}{40}}{\binom{120}{40} \times \binom{80}{40} \times \binom{40}{40}}. \tag2 $$

Combining the denominators of (1) and (2) above gives the number of equally likely ways that the $(200)$ balls can be distributed into the $5$ bins. For convenience, I am assuming that each of the $(200)$ balls is distinguishable, and that each of the $(5)$ bins is distinguishable.

So the combined numerators of (1) and (2) must be computed in a manner consistent with how the denominator was computed.

In the combined numerator, the $(5 \times 4)$ factor expresses the number of distinct ways that one bin can be selected to receive two red balls, and then a second bin can be selected to receive one red ball.

Then, using hypergeometric distribution, the rest of the numerator and denominator in (1) above is explained. With (2) above cancelling out, (1) above may be simplified as follows:

  • $\displaystyle \frac{\binom{3}{2} \times \binom{197}{38}}{\binom{200}{40}} = \frac{3!}{2!} \times \frac{(197!)}{(159!)\times (38!)} \times \frac{(160!) \times (40!)}{(200!)}$

    $\displaystyle = \frac{3!}{2!} \times \frac{197!}{200!} \times \frac{160!}{159!} \times \frac{40!}{38!}$

    $\displaystyle = \frac{3 \times 160 \times 40 \times 39}{200 \times 199 \times 198} = \frac{16 \times 13}{199 \times 11}.$

  • $\displaystyle \frac{\binom{159}{39}}{\binom{160}{40}} = \frac{159!}{(39!) \times (120!)} \times \frac{(40!) \times (120!)}{160!} = \frac{40}{160}= \frac{1}{4}.$

So, the final computation is

$$(5 \times 4) \times \frac{16 \times 13}{199 \times 11} \times \frac{1}{4} = \frac{80 \times 13}{199 \times 11} = \frac{1040}{2189}.$$

user2661923
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