Statistical Problem

Rev. Green | 10:15 Mon 25th Jul 2022 | Science

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If one of a group of n items is distinguishable, then it can be found in (n+1)/2 attempts, on average, by examining a random item from the group, provided that the same item is not chosen twice. How many attempts are needed, on average, if items are examined at random without checking whether they have been examined previously? (This is only for interest - I've never studied statistics. Don't put great effort into answering.)

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ScienceNoob

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If there are N items and you are looking for precisely one of them, with replacement, then you want to work out the sum to infinity of 1/N + 2*(1/N)*(N-1)/N + 3*(1/N)*[(N-1)/N]^2 + 4*(1/N)*[(N-1)/N]^3 ... etc. Write as (1/N)*Sum(i=1,infinity) i [(N-1)/N]^(i-1) This is the same as N* d/dN{Sum(i=1,infinity) [(N-1)/N]^(i)} which is the same as N* d/dN...

11:06 Mon 25th Jul 2022

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Peter Pedant

this comes up in another guise
very rare occurrences of side effects
If you estimate that a covid vaccine side effect is 1 in 10 000
how many do you have to follow up in order to be reasonably sure it is much less than that ?

3n - - 30 000

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