Math help.

[chaoticbrain]

Lets say you have a certain number of indicators. Each indicator on it's own own has an 80% chance of being accurate, if you have 5 or 7 different indicators say, how likely is it that the majority would be accurate ?

Thanks.

[yogi]

5 * 0.8 = 4

[chaoticbrain]

5 * 0.8 = 4

How does that tell me what I'm asking ?

[chaoticbrain]

Lets say you have a certain number of indicators. Each indicator on it's own own has an 80% chance of being accurate, if you have 5 or 7 different indicators say, how likely is it that the majority would be accurate ?

Thanks.

Overall the majority will be accurate - you have said 80% will be accurate, and 80% is a majority.

Sorry, to clarify, I mean each indicator on it's own is 80% accurate.

Lets say we have 5 indicators that each will say "red" 80% of the time to indicate red, but can also sometimes say "blue" 20% of the time.

What are the chances of 3/5 of them saying "red", what are the chances of all of them saying "blue" ? etc.

[chaoticbrain]

Oh cool, thanks.

That makes sense.

[dancupid]

It's a binomial distribution

p=0.8, 5 indicators
X is number of indicators that are active

for majority active
Pr(X>2)=Pr(X=3) + Pr(X=4) + Pr(X=5)


boils down to: 10(0.8 )^3(0.2 )^2 + 5(0.8 )^4(0.2) + (0.8 )^5
=0.94208





http://en.wikipedia.org/wiki/Binomial_distribution

edit: someone beat me to it

[Haidang1796]

this isnt my level