I might be missing something, but you may have a typo --or math error-- when you say there is a "43% chance that even a single person on your plane has COVID".
Shouldn't it be 1 - (1 - 1100/1x10^6)^366 which is about 33% ?
It really depends on the assumptions we're making on prevalence and how long we are going to assume someone is COVID positive.
In the placebo study group, 0.92% of the participants caught COVID. But that was over 8 weeks. So let's assume for the sake of calculation that an infection lasts 1 week we had the infections evenly distributed. That means that in any given week 0.116% of individuals have COVID. That means one in 862 people have COVID in a given week, which give us about a 43% chance that a random set of 366 people have the disease at a given time.
Again, this is a lot of assumptions, but I'm just trying to get a rough quantification that helps us compare vaccinated and unvaccinated probabilities.
I think now I understand the difference between us. It looks like you are calculating the expected number of people on the plane to have Covid (0.43 people), and I am calculating the probability that at least one person on the plane has Covid (between 33% and 35% depending on the exact numbers).
Say the percentage infected in the 8-week study was 80%. Then in any given week 10% have Covid. So, in a plane of 366 people, we would expect 0.1 * 366 = 36.6 people to have Covid. The probability that at least one person had Covid would be 1 – (.9)^366, which is effectively 100%.
My main concern was when you multiplied 366 and 0.116%. That makes sense if you are finding the expected number with Covid, but not with probabilities because P(A U B) = P(A) + P(B) – P(A intersect B).
If the probability of someone having Covid is 0.116% then the probability of someone not having Covid is 99.884%. The probability of two people not having Covid is (99.884%)*(99.884%), and the probability of 366 not having Covid is (99.884%)^366 =65.4%.
By the way, despite my nitpicking, I enjoyed your article. Keep up the good work.
Not to defend the CNN paragraph in the slightest, but it seems to me the mistake they're making is somewhat different then how you put it.
They said "for every million vaccinated people who fly, 100,000 could still become infected." That sounds to me like they're understanding 90% efficacy as something like, "90% of people are entirely immune, 10% of people the vaccine just doesn't work for some reason so are still vulnerable."
Which to my understanding is also wrong, but not ludicrous like "10% of all travelers have Covid." Of course then the problem is that they put it in such a way that it's easy to misunderstand it as "10% of all travelers have Covid," because they foregrounded the big number to make it sound alarming and didn't explain it further.
I might be missing something, but you may have a typo --or math error-- when you say there is a "43% chance that even a single person on your plane has COVID".
Shouldn't it be 1 - (1 - 1100/1x10^6)^366 which is about 33% ?
It really depends on the assumptions we're making on prevalence and how long we are going to assume someone is COVID positive.
In the placebo study group, 0.92% of the participants caught COVID. But that was over 8 weeks. So let's assume for the sake of calculation that an infection lasts 1 week we had the infections evenly distributed. That means that in any given week 0.116% of individuals have COVID. That means one in 862 people have COVID in a given week, which give us about a 43% chance that a random set of 366 people have the disease at a given time.
Again, this is a lot of assumptions, but I'm just trying to get a rough quantification that helps us compare vaccinated and unvaccinated probabilities.
I think now I understand the difference between us. It looks like you are calculating the expected number of people on the plane to have Covid (0.43 people), and I am calculating the probability that at least one person on the plane has Covid (between 33% and 35% depending on the exact numbers).
Say the percentage infected in the 8-week study was 80%. Then in any given week 10% have Covid. So, in a plane of 366 people, we would expect 0.1 * 366 = 36.6 people to have Covid. The probability that at least one person had Covid would be 1 – (.9)^366, which is effectively 100%.
My main concern was when you multiplied 366 and 0.116%. That makes sense if you are finding the expected number with Covid, but not with probabilities because P(A U B) = P(A) + P(B) – P(A intersect B).
If the probability of someone having Covid is 0.116% then the probability of someone not having Covid is 99.884%. The probability of two people not having Covid is (99.884%)*(99.884%), and the probability of 366 not having Covid is (99.884%)^366 =65.4%.
By the way, despite my nitpicking, I enjoyed your article. Keep up the good work.
Not to defend the CNN paragraph in the slightest, but it seems to me the mistake they're making is somewhat different then how you put it.
They said "for every million vaccinated people who fly, 100,000 could still become infected." That sounds to me like they're understanding 90% efficacy as something like, "90% of people are entirely immune, 10% of people the vaccine just doesn't work for some reason so are still vulnerable."
Which to my understanding is also wrong, but not ludicrous like "10% of all travelers have Covid." Of course then the problem is that they put it in such a way that it's easy to misunderstand it as "10% of all travelers have Covid," because they foregrounded the big number to make it sound alarming and didn't explain it further.