I can't really find a math/stat subforum, and this isn't homework.

My question is, if something has a 16% of happening, what's the chance it'll happen if it repeats 5 times?
I came up with x = chance, n = repeat .... x ^ (1/n)
(.16)^(1/5) = 69%

I can't find this on google, but my answer seems very plausible to me. Is this how you're supposed calculate it?

This post was edited by Foxic on Apr 12 2013 02:07pm

Lol
P(A) + P(B) - P(A)P(B)

79.99%

Been awhile since I had stats but about 99% sure here

This post was edited by 0n35 on Apr 12 2013 03:02pm

I don't think that's right, your answer is the same as 16% + 16% + 16% + 16% + 16%. You can't assume it'll happen for sure after 7 times (16 x 7) in your case, it's just a high chance

Are you asking?:
If you have an outcome from an event that has a 16% chance of happening, what is the probability you will get this outcome 5 times if you preform 5 events?
That is (0.16)^5


Or are you asking:
If you have an outcome from an event that has a 16% chance of happening, what is the probability you will get this outcome 1 or more times, if you preform 5 events?
In this case we do the calculation "backwards":
The chance you will NOT get a success on one attempt is 1 - 0.16 = 0.84
The chance you will NOT get a success in 5 attempts is (0.84)^5
The chance you will get a success (one or more) in 5 attempts is 1 - (0.84)^5

This post was edited by Azrad on Apr 12 2013 09:13pm

The latter. Thanks, that makes sense