Current Events > If you were in a room with 124 random Americans, odds are 1 would have corona.

(+) Yeah! (+)

I wrote this up in a convo with a really good friend who is having a wedding but is justifiably nervous:

There are 328.2 million americans
There are 3.285 million total cases
There are 1.453 million recovered cases

So currently active cases should be around 1.832 million.

So that's a chance of 0.005582 that any random american currently has it. Or in percentage: 0.5582%

So moving from there you can do the math of given a group of randomly sampled americans, how many would you need to get over 50% that at least 1 had it.

The odds of 1 least event happening is 1 minus the odds that it never happens, so for odds of it not happening with 1 person is 0.99441803778
The odds of it not happening at all for X people is 0.99441803778^x

If we want to solve for when that hits 50% we do 0.99441803778^x=.5, do logs on the sides and get x=ln(.5)/ln(0.99441803778)=123.829361041

So by 124 random americans in a room you have a 50% chance at least 1 has it.


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Interesting deduction.

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Also worthy of note is that the total cases could be higher, but I don't think it would affect the odds in any significant manner. Unless testing is waaaay below the desired.

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LordFarquad1312 posted...

Interesting deduction.

My cousin is having a 150 person wedding next week in wisconsin

we rsvpd but are now strongly considering not going

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