A question about Poisson process such that... I got the following problem:

Suppose that instances of some event occur in accordance with a Poisson process having a rate of 24 instances an hour
Suppose we take a time-interval of length 1 hour and we partition it into 4 sub-intervals each of length $\frac{1}{4}$ hour.
If it is known that in this time interval (the one with length 1 hour) exactly 20 instances already occurred.
What is the probability that at every $\frac{1}{4}$ hour (according to the partition we defined above) exactly 5 instances occur (from the 20 instances that already occurred). 

I got stuck for at least 2 hours and I don't know how to proceed (I tried to use conditional probability but I got wrong answer).
Tanks on any hint/help.
 A: The number $N(1/4)$ of events that occur in a time period of $1/4$ of an hour is Poisson distributed with parameter $λ=24/4=6$. But the conditional distribution of $N(1/4)$ given that the number $N(1)$ of occurances in $1$ hour is equal to $20$ is binomially distributed with parameters $n=20$ and $p=1/4$. Using the properties of a Poisson process that occurances in different time intervals are independent and that the distribution of the occurances depends only on the length of the interval and not on the starting time, this can be shown as follows $$\begin{align*}P(N(1/4)=x \mid N(1)=20)&=\frac{P(N(1/4)=x), N(1)=20)}{P(N(1)=20)}\\[0.2cm]&=\frac{P(N(1/4)=x), N(1-1/4)=20-x)}{P(N(1)=20)}\\[0.2cm]&=\frac{P(N(1/4)=x))P(N(3/4)=20-x)}{P(N(1)=20)}=\frac{e^{-6}\frac{6^x}{x!}e^{-18}\frac{18^{20-x}}{(20-x)!}}{e^{-24}\frac{24^{20}}{20!}}\\[0.2cm]&=\frac{\not e^{-6-18}20!6^x18^{20-x}}{\not e^{-24}24^x24^{20-x}x!(20-x)!}=\frac{20!}{x!(20-x)!}\left(\frac{6}{24}\right)^x\left(\frac{18}{24}\right)^{n-x}\\[0.2cm]&=\dbinom{20}{x}\left(\frac{1}{4}\right)^{x}\left(\frac{3}{4}\right)^{n-x}\end{align*}$$ Similarly you can show that the number of events that occur in each interval is multinomially distributed with parameters $n=20$ and $p_1=p_2=p_3=p_4=1/4$. Thus the required probability is equal to $$P(X_1=5, X_2=5, X_3=5, X_4=5)=\frac{20!}{5!5!5!5!}(1/4)^5(1/4)^5(1/4)^5(1/4)^5=\frac{20!}{5!^44^{20}}$$
