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I had a biology friend ask me this question:

Suppose I had 50 ml of solution in which exactly five red blood cells that are individually suspended (i.e., not sticking to each other).

If I collect 2 ml of that solution, what is: A) The probability of that 2 ml having no cells B) The probability of that 2 ml having 1 or more cells C) The probability of that 2 ml having exactly 1 or exactly 3 cells

C) was added by me. The biologist doesn't actually care about C :)

I assume the answer to P(B) is 1 - P(A), or P(A) = 1 - P(B) if you prefer. I suspect P(C) is somewhat harder.

Here is what I've tried so far. In response to the comments, let me be clear that I'm not claiming the ideas below are correct. I wanted to show that I have put some effort into this. I'm looking for comments, references, etc. that will lead me to a correct answer. Also, I am assumming that the cells are independently distributed and do not interact. My biology friend was unclear on these points. If these assumptions are not correct, I imagine the problem is considerably more difficult and I will post another question. I'm hoping this these assumptions are a good starting point. Finally, let me be crystal clear that 2 ml of the solution is collected at once! The idea of breaking it up into a 1 ml draw followed by another 1 ml draw was my own idea to help me think about the problem. It may not be necessary or desirable. As an aside, if the cells are uniformly distributed and do not interact, I fail to see how it makes any difference if 2 ml is taken at once or 1 ml is taken followed by another 1 ml. I would like to be enlightened! I reasoned that this is somewhat similar to a card problem without replacement. See for example: Probability of getting 4 Aces

I reasoned that the concentration of cells is 1 per 10 ml or .1 cell per 1 ml. The probability of not drawing any cells in 1 ml is .9 or 90% if you prefer. On the second draw, you only have 49 ml and I think the correct probability for the second draw would be 5/49 ml or .102 for picking 1 or more cells. The inverse of this is .898. Multiply these gives .808 of not picking any cells in 2 ml.

Am I on the right track? Something feels wrong. I'm not sure I've taken into account that 1 ml of solution could have 0 cells, 1 cell, .... 5 cells. I also don't know if my experience with discrete playing cards (and coins) is applicable here. That is, can I arbitrarily break up 50 milliliters into 1 milliliter "compartments"?

Any hints or references would be much appreciated. A name for this type of probability (so I can find more examples!) would also be appreciated.

Finally, my other concern with this whole problem is my initial assumption of ".1 cell per 1ml". I mean I assume that either a cell is in the milliliter or it's not. Not 1/10 of it is there! But perhaps that's a question for my biology friend.

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  • $\begingroup$ You might need to edit the question; do you draw 2ml or 5ml? Are you drawing all at once, or not? If you draw each 1ml at a time, then a hypergeometric distribution will be useful in determining the probability of containing 0,1,2, ..., 5 particles in the total amount drawn. $\endgroup$ – stephematician Sep 1 '17 at 0:27
  • $\begingroup$ "If I collect 2 ml of that solution, what is: A) The probability of that 5 ml ..." --- fix this $\endgroup$ – Glen_b -Reinstate Monica Sep 1 '17 at 2:05
  • $\begingroup$ Glen_b and stephematician, I've edited the question given my rudimentary knowledge of statistics/probability. Thank you for the comments. If there is a more appropriate site (beginning probability?), please let me know. Thanks. $\endgroup$ – Dave Sep 1 '17 at 3:52
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There are a few things that we're going to have to assume, even though you never posed it in the question.First of all, I will assume the cells are non-interacting. That way, each individual cell's probability of being plucked out of a sample is independent of that of any other cell. This effects the result. For example, what if these cells attracted each other? Then you either get all or none in a sample. Secondly, I will assume that they have a uniform distribution in the fluid. Often the chemical potential of cells in fluids dramatically outweighs the gravitational potential inflicted upon them, giving them a uniform distribution, but that's not always the case. (They could be massive cells that are virtually unaffected by the bombardment of H2O molecules.)

FYI, without the above two assumptions, we would just have to throw our hands up and say "not enough information to answer the question". So keep that in mind when you're posing a question.

Because of the independence assumption, this problem is identical to the problem of 5 separate test tubes, each of which have 50 mL and only one cell in them. So imagine that scenario. You draw 5mL out of each tube, and in each draw, you either got a cell or you didn't. What is the probability that none of those five samples yielded a cell? Then, what is the probability that at least one sample yielded a cell? (Your logic on discerning (B) from (A) is perfectly correct, so no need to worry about that.)

Then, what is the probability that either exactly one or exactly three of the five 5mL samples yielded a cell? To answer this question, you'll need the Binomial Distribution.

EDIT: The main reason for the equivalence of one test tube with 5 cells and five test tubes with one cell is because of the independence assumption of the position of each cell in the tube. So if you take a sample from the 5 cell tube, whether you got cell A in the sample has absolutely no bearing on whether you got cell B. (This is unlike draws from a card deck, where the position of aces are not independent.) And each cell has the same probability of being drawn. So you're essentially rolling five dice at once. (More explicitly, you're sampling from the same Bernoulli distribution five times simultaneously.) In the other scenario, where you take 5 separate samples from single-cell tubes, you're just rolling the same five dice, but in sequence rather than all together.

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  • $\begingroup$ Thank you Bridgeburners for the response. Based on your answer and comments, I've edited my question. Yes, I'm assuming the cells do not interact and are uniformly distributed. I'm confused by your stating that the problem is exactly the same as 5 separate test tubes with only one cell. I'm just not seeing how that's the same as on 50ml tube with 5 cells. Can you elaborate? I'm trying to relate it back to playing cards. If I asked what is the probability of drawing 4 aces from a deck, would it be the same as asking what is the probability of drawing 1 ace from each of 4 decks? Thanks! $\endgroup$ – Dave Sep 1 '17 at 3:50
  • $\begingroup$ @Dave The card deck is a bad analogy because the positions of aces are not independent. For example, if the ace of spades takes the nn'th position in the deck then the ace of clubs cannot possibly take the nn'th position in the deck. Think of it this way. If you take 1/52 of the fluid from the test tube, it's possible that you get more than one cell from that sample. If you draw one card from the deck, is it possible that you got more than one ace? I'll edit my answer to try and give an intuition on why your original experiment is equivalent to the 5 test tube experiment. $\endgroup$ – Bridgeburners Sep 1 '17 at 4:32

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