Answer:
The probability that the server is busy is P=0.56.
Step-by-step explanation:
We have a quieing theory problem with M/M/1
In queing theory, the probability of the server being busy can be expressed as:
[tex]P=\frac{\lambda}{\lambda+\mu}[/tex]
being μ: the time between services and λ: the time between customers arrival.
Then we can calculate:
[tex]P=\frac{\lambda}{\lambda+\mu}=\frac{\frac{1}{28} }{\frac{1}{28}+\frac{1}{35} } =\frac{0.036}{0.036+0.029}=0.56[/tex]
The probability that the server is busy is P=0.56.
