I got this Poisson distribution problem.
There's a clothes shop, and average number of customers per 1 hour are 5 men , 10 women, independent each other, in Poisson distribution. If for 30 minutes there had been totally 10 customers, what's the probability of women customer more than 2?
- This is how I solved. Poisson average of man customers for 30 minute = m = 2.5
women = w = 5
all = a = 7.5
then the solution is p(w>2 | a=10)
=1- [ p(w>2 ∩ a=10)/p(a=10)]
= 1- { [p(w=0)*p(m=10) + p(w=1) * p(m=9)]/p(a=10)
(solve with Poisson formula)
=0.9996
I'm not quite sure if this is right,
because not sure if p(w=1 ∩ a=10) can be interpreted right into
p(w=1) * p(m=9)
. Is this approach right? or am I missing something Thank you verymuch
Your approach is generally correct, but where you have $w\gt2$ the second time it should be $w\le2$ instead, and you're missing the $w=2$ case in the sum below that.
However, a somewhat simpler approach would be: Conditional on the total number of customers, each customer is independently a man with probability $\frac13$ and a woman with probability $\frac23$, so the probability for more than $2$ women is
$$ 1-\sum_{w=0}^2\binom{10}w\left(\frac23\right)^w\left(\frac13\right)^{10-w}=\frac{19616}{19683}\approx0.9966\;. $$
See Finding expectation of a Poisson variable when condition has been made.