Topic: Math help. (Read 376 times)
this isnt my level
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
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
Oh cool, thanks.
That makes sense.
That makes sense.
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.
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.
5 * 0.8 = 4
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.
Thanks.
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