Statistics & Probability

Is the value n, and p, a random variable, i read some where if this is the case that 2 random variables can not be calculated but may be of track to your question, im a poker player and only know so much

In the binomial distribution,
mean = n*p
variance = n*p*(1-p)

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

You know the mean and variance, so solve these equations for n and p.

guys you seem you know a lot about variance ,would you please help me with my question two posts below,

mean=np=18
variance=npq=3.52
npq/np=0.195=q
p=1-q
=1-.195=.80
np=18
n(.80)=18
n=22.50

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