In a working office there are $2$ workers who take care of clients . All the clients go to the first worker and
if they needed additinal help they go to the second worker .
we know that $40$% of the clients also go to the second worker. the number of clients going to the first worker is poisonic random variable with mean $40 ~\frac{Clients}{Hour}$.
if we know that $4$ clients went to the second worker in $15$ min what is the probability that $2$ of them went to the second worker in the first $10$ min and the remaining $2$ went in the additional $5$ min .
Answer Attempt :
Let Y be Random variable of the number of clients going to worker $2$ in $15$ min.
Let X be Random variable of the number of clients going to worker $1$ in $15$ min.
We know that $40$% of the clients who went to worker $1$ , go to worker $2$ as well .
So :
$E(Y) = E(0.4X) = 0.4E(X) = 0.4 * 10 = 4$ ( change unit $40$ per hour to $10$ per $15$ min)
$P(Y=4) = \frac{e^{-4}~(4)^{4}}{4!}$
Let $Z_{10}$ be Random variable of the number of clients going to worker $2$ in the first $10$ min.
Let $Z_{5}$ be Random variable of the number of clients going to worker $2$ in the additional $5$ min.
so :
$E(Y) = E(Z_{10}+Z_{5})= E(Z_{10}) + E(Z_{5}) = 4$
$E(Z_{10}) = \frac{40}{15}$
$E(Z_{5}) = \frac{20}{15}$
the question asks for : $P(Z_{10} = 2~And~Z_{5} = 2 | Y=4) = \frac{P(Z_{10})P(Z_{5})}{P(Y=4)}$
is this right ? my answer is $0.2962$ and the Final answer is $0.2954$ so i dont know if thats wrong
Essential is that the number of clients of the second worker is also poissonic.
If $N_{t}$ denotes the number of clients of the second worker in time periond $\left[0,t\right]$ then - if we take $5$ minutes as time unit - we must find $P\left(N_{2}=2\mid N_{3}=4\right)$ where $N_{t}$ denotes a homogeneous Poisson-proces.
We can calculate the rate of that process ($0.40\times40=16$ clients per hour so $\frac{4}{3}$ clients per $5$ minutes) but actually we do not need that because for every rate $\lambda$ we get the same outcome:$$P\left(N_{2}=2\mid N_{3}=4\right)=\frac{P\left(N_{2}=2,N_{3}-N_{2}=2\right)}{P\left(N_{3}=4\right)}=\frac{e^{-2\lambda}\frac{\left(2\lambda\right)^{2}}{2!}e^{-\lambda}\frac{\lambda^{2}}{2!}}{e^{-3\lambda}\frac{\left(3\lambda\right)^{4}}{4!}}=\frac{4!}{3^{4}}\approx0.296296$$