I have come across the following probability problem. I've used the binomial formula to come up with an answer, but I'm wondering if this is the correct approach, rather than modelling the question with, say, a Poisson process.
Question: You are running a site with 1 million customers, one hundred of which use an Android phone. Everyday, the website is visited independently and randomly by 10,000 people. On how many days of the year would you would expect to your site to have three or more visitors using an Android mobile?
I used the following logic:
- 1,000,000 customers, 100 Android phones. --> $P($Android$) = 0.0001$
- $P(\geq 3 \text{ Android})$ = 1 - $P(0 \text{ Android})$ - $P(1 \text{ Android})$ - $P(2 \text{ Android})$
- $P(0 \text{ Android})$ = $(1 - 0.0001)^{10,000}$ = 0.3679
- $P(1 \text{ Android})$ = $\binom{10000}{1} (1 - 0.0001)^{9,999} (0.0001)^1$ = 0.3679
- $P(2 \text{ Android})$ = $\binom{10000}{2} (1 - 0.0001)^{9,998} (0.0001)^2$ = 0.1839
This gives me a $P(\geq 3 \text{ Android}) = 0.0803$. Multiplying this probability by the number of days in the year gives me ~29.3 days.
Is this logic correct, and is there an equivalent way that this could be derived using a Poisson process?
Thanks in advance!
Same result using a poisson distribution. Each day you have a $Poi(1)$ thus the probability you nasty calculated with a binomial is simply
$$\mathbb{P}[X\geq3]=1-e^{-1}-e^{-1}-\frac{e^{-1}}{2}=1-\frac{5}{2}e^{-1}\approx0.0803$$