# Probability of pipetting X number of cells in a solution

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.

• 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. – stephematician Sep 1 '17 at 0:27
• "If I collect 2 ml of that solution, what is: A) The probability of that 5 ml ..." --- fix this – Glen_b -Reinstate Monica Sep 1 '17 at 2:05
• 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. – Dave Sep 1 '17 at 3:52