text: user1771: In Shapiro et al., when discussing about loss of molecules as source of error in single-cell sequencing, it is written that:
  
  Another source of error is losses, which can be severe. The detection limit of published protocols is $5$–$10$ molecules of mRNA. If, as seems likely, the limit of detection is primarily determined by losses during sample preparation, this would indicate that $80$–$90\%$ of mRNA was lost. Or, to put it the other way around, a $90\%$ loss leads to an approximately $50\%$ chance of failing to detect a gene that is expressed at a level of seven mRNA molecules (from the binomial distribution).
  
  How is this probability computed using the binomial distribution? I thought that $90\%$ loss corresponds to $5$ detected molecules, and I assume that $k=7$ for the binomial calculation, but I am unable to go further.
  
  user77: A 90% loss can be rephrased as a 10% chance of detecting anything. So what we want to find is the probability of detecting 0 molecules, when we start with 7 and have 10% probability of success. Once can do that in R as follows:
  > pbinom(0, 7, 0.1)
  0.4782969
  
  So ~50%, as they stated. I suspect that part of the confusion arises from the fact that the detection limit is due to lowly-expressed genes/transcript being heavily affected by this loss. So the probability of detecting a single molecule out of 4 original molecules (assuming 90% mRNA loss) is 34%, for 3 molecules it's 27%, for 2 it's 19% and for 1 it's 10%. I think the threshold of 5 is mentioned more because it's a nice round number than there's anything particularly different in detecting a gene with 4 vs. 5 molecules (34 vs. 41% probability).
  
original_text: 1771: <p>In <a href="https://www.nature.com/articles/nrg3542" rel="noreferrer">Shapiro et al.</a>, when discussing about loss of molecules as source of error in single-cell sequencing, it is written that:</p>
  
  <blockquote>
    <p>Another source of error is losses, which can be severe. The detection limit of published protocols is $5$–$10$ molecules of mRNA. If, as seems likely, the limit of detection is primarily determined by losses during sample preparation, this would indicate that $80$–$90\%$ of mRNA was lost. Or, to put it the other way around, a $90\%$ loss leads to an approximately $50\%$ chance of failing to detect a gene that is expressed at a level of seven mRNA molecules (from the binomial distribution).</p>
  </blockquote>
  
  <p>How is this probability computed using the binomial distribution? I thought that $90\%$ loss corresponds to $5$ detected molecules, and I assume that $k=7$ for the binomial calculation, but I am unable to go further.</p>
   77: <p>A 90% loss can be rephrased as a 10% chance of detecting anything. So what we want to find is the probability of detecting 0 molecules, when we start with 7 and have 10% probability of success. Once can do that in R as follows:</p>
  
  <pre><code>&gt; pbinom(0, 7, 0.1)
  0.4782969
  </code></pre>
  
  <p>So ~50%, as they stated. I suspect that part of the confusion arises from the fact that the detection limit is due to lowly-expressed genes/transcript being heavily affected by this loss. So the probability of detecting a single molecule out of 4 original molecules (assuming 90% mRNA loss) is 34%, for 3 molecules it's 27%, for 2 it's 19% and for 1 it's 10%. I think the threshold of 5 is mentioned more because it's a nice round number than there's anything particularly different in detecting a gene with 4 vs. 5 molecules (34 vs. 41% probability).</p>
  