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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.

Kevin
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3 Answers3

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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

enter image description here

$P(A)$, $P(B)$ 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?)

Martín-Blas Pérez Pinilla
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  • I have never seen the backslash notation, what does that mean? – Kevin Jan 13 '15 at 09:50
  • @Ajaxkevi, difference of sets: http://en.wikipedia.org/wiki/Complement_(set_theory)#Relative_complement. – Martín-Blas Pérez Pinilla Jan 13 '15 at 09:52
  • I see. Thanks a lot for your answer, it helped me a lot! – Kevin Jan 13 '15 at 10:22
  • Just to be clear, i am trying to calculate P(F|A ∩ B)? What would I need to be able to continue the calculation. Would P(A|B) help for instance? – Kevin Jan 16 '15 at 14:34
  • @Ajaxkevi, as $P(A|B)=P(A\cap B)/P(B)$, depends only of the relative sizes of $A\cap B$ and $B$ and says nothing of $P(F|A\cap B)$ (check this with the drawing). See the examples of http://en.wikipedia.org/wiki/Bayes%27_theorem. – Martín-Blas Pérez Pinilla Jan 16 '15 at 14:58
  • I am having a hard time to apply these examples on my problem. – Kevin Jan 16 '15 at 15:14
  • @Ajaxkevi, you have six unkowns: the areas of each subrectangle in the drawing. Sum areas=1 is a restriction, $P(F|A)=P(F|B)=0.8$ are two more restrictions. So you need three restrictions more. Can be $P(A)$, $P(B)$ and $P(F|A\setminus B)$, can be another (maybe some $P(\cdots|F)$). – Martín-Blas Pérez Pinilla Jan 16 '15 at 16:09
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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 $1-0.04=0.96$.

Lehs
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0

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

$${w_1w_2/W\over w_1w_w/W+m_1m_2/M}={(4m_1)(4m_2)M\over (4m_1)(4m_2)M+m_1m_2W}={16M\over16M+W}$$

Barry Cipra
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