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Re:I'm giving up on PGP by caf on Tuesday December 06, 2016 @06:06PM (#13119062)
You are mistaken. If probability of each of the 5 men being colorblind is independent (so eg. they're not related etc), then there's a 41% chance that at least one of them is (1 - 0.9⁵).

(There's a 33% chance that exactly one of them is: 0.1 × 0.9⁴ × ⁵C₁).

Re:I'm giving up on PGP by DigitalJack on Tuesday December 06, 2016 @07:18PM (#13119452)
Thanks. It's been too long since I've studied statistics. Might be time to watch some khan academy.

Oh, I see. Assuming probability that a man is not colorblind is 0.9, then you found the probability of none of them being colorblind and subtracted that amount from 1 to find the other side.

For anyone reading this that was bothered by my use of a specific gender, it's because colorblindness occurrence is significantly higher in men than women.

Re:I'm giving up on PGP by mkesper on Wednesday December 07, 2016 @08:15AM (#13122262)
That's not soo much lower, though.
Re:I'm giving up on PGP by jballanc on Tuesday December 06, 2016 @06:54PM (#13119331)
I think a Poison distribution would be more appropriate here, so the probability of 1 man in 5 being colorblind, assuming 10% of the population on average is colorblind, is:

    e^-0.5*sum((i)->(0.5^i/factorial(i)), 1:5) = ~39%
Re:I'm giving up on PGP by softawre on Tuesday December 06, 2016 @07:48PM (#13119610)
> Poison distribution

Why? This is basic statistics (% of population).

Re:I'm giving up on PGP by caf on Tuesday December 06, 2016 @10:17PM (#13120181)
I don't believe so - it's n instances of a bi-valued random variable, IID. It's exactly the case the binomial distribution covers.
Re:I'm giving up on PGP by mnw21cam on Wednesday December 07, 2016 @04:21AM (#13121409)
Exactly. You have five chances at an event, and an event probability per chance, which is exactly what the binomial distribution is for.

The Poisson distribution is what you use when the effective number of chances is "large". The Poisson distribution is effectively a special case of the binomial distribution, where the number of chances is infinity, the probability per chance is infinitesimal, and the product of the two is the expected number of events.