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cpaavola
04-30-2010, 11:30 AM
For instance, let's say a person is a truck driver and there's an:
80% chance of they smoke
40% chance of they are overweight
60% chance of they are male
If we grab one person at random who is a truck driver, what's the chance they smoke, are overweight AND are male? Or, what's the chance they smoke and are male but are not overweight?

Thanks!

vinux
04-30-2010, 12:41 PM
Hint:
If you are assuming the qualities are indepenent .. Use the property of probability of independent events.

P(A and B)= p(A)*P(B).

cpaavola
04-30-2010, 01:05 PM
So, the probability of al three occuring is:
(80 X 40 X 60) / (100 X 100 X 100) = 19.2%

Is that correct? Sorry, I'm not a stats guy, but I like those who are!

statsguy
05-02-2010, 11:18 PM
So, the probability of al three occuring is:
(80 X 40 X 60) / (100 X 100 X 100) = 19.2%
P(A and B and C) = (.80)(.40)(.60)
P(A and not B and C) = (.80)(1-.40)(.60) = (.80)(.60)(.60)

For independent events A and B, the probability of their intersection is the product of the individuals probabilities. Don't sum probabilities in this case; that only applies for events that are disjoint (which they are not here, clearly.)
Is that correct? Sorry, I'm not a stats guy (I'm statsguy), but I like those who are!

Hope this helps!

cpaavola
05-03-2010, 11:15 AM
Hope this helps!

Hahaha... your name suits you! Thanks man. :wave: