Chances of someone being of a certain gender at websites I have 2 of websites and I know the chances of a visitor being a female or male.
Let's say I have 2 website where the chance of a new visitor being a female is 80%.
If the visitor comes on website 1 I know the chance of that visitor being female is 80%.
But what if that visitor comes on both websites. Is the chance still 80%? Or am I more certain that visitor is a female? If so what is the equation I should use?
The website are not dependent of each other.
 A: General case without independence suppositions:
$$A=\text{visit to website 1},$$
$$B=\text{visit to website 2},$$
$$F=\text{visitor is female},$$
$$0.8P(A)=P(A)P(F|A)=P(F\cap A)=P(A\setminus B)P(F|A\setminus B)+P(A\cap B)P(F|A\cap B),$$
$$0.8P(B)=P(B)P(F|B)=P(F\cap B)=P(B\setminus A)P(F|B\setminus A)+P(A\cap B)P(F|A\cap B).$$
Now, $P(A)$, $P(B)$, $P(A\cap B)$ are free parameters (with $P(A)+P(B)\ge 1$, $P(A\cap B)>0$,...), $P(A\setminus B)=P(A)-P(A\cap B)$, $P(B\setminus A)=P(B)-P(A\cap B)$ and we have a system of two equations with three unknowns: $P(F|A\setminus B)$, $P(F|B\setminus A)$, $P(F|A\cap B)$, i.e.,  we have a lineal relationship between the three unknowns.
If $P(A|F)$... are known we can use Bayes (maybe in a future edition).
EDIT: an illustrative diagram

$P(A)$, $P(B)$ are areas and also lenghts (why?) $P(F|A\setminus B)$, $P(F|B\setminus A)$, $P(F|A\cap B)$ are quotients of areas and also lenghts (why?)
A: Suppose there are $n$ sites with the same probability for female. If visiting these sites really are independent events and a person that visit one of the sites are making totally independent choices, then the probability that the visitor is female increase for each first time visit on one of the sites.
The probability for being a female is then $1-0.2^n$ where $n$ is the number of first time visits of some of these sites.
So in your case, the probability should be $1-0.04=0.96$.
A: I get the answer
$${16M\over 16M+W}$$
where $M$ and $W$ are the total number of Men and Women who either do or don't visit the two websites.
Here's my thinking.  Suppose $w_1$ women and $m_1$ men plan to visit website $1$, and likewise for $w_2$ and $m_2$.  The $80\%$ hypothesis is that $w_1=4m_1$ and $w_2=4m_2$.  Independence of website visits implies the numbers of women and men who visit both sides are (approximately) $w_1w_2/W$ and $m_1m_2/M$.  So among this class, the fraction that are women is
$${w_1w_2/W\over w_1w_w/W+m_1m_2/M}={(4m_1)(4m_2)M\over (4m_1)(4m_2)M+m_1m_2W}={16M\over16M+W}$$
