Question

Patients arrive randomly and independently at a Doctor's room from 8 AM at an average rate of one in 5 minutes. The waiting room can accommodate 12 persons. The probability that the room will be full when the doctor arrives at 9AM is

• A. $\frac{e^{-12}(12{)}^{12}}{12!}$
• B. $\sum _{x=0}^{11}\frac{e^{-12}(12{)}^x}{x!}$
• C. $1-\sum _{x=0}^{11}\frac{e^{-12}(12{)}^x}{x!}$
• D. $1-\sum _{x=0}^{\infty }\frac{e^{-12}(12{)}^x}{x!}$

Answer 1

The correct option is C $1-\sum _{x=0}^{11}\frac{e^{-12}(12{)}^x}{x!}$

Random Arrival process is a Poisson process
Given by
$P\left(k\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$
$given,\lambda =1/5mi{n}^{-1}=12s^{-1}$
Room is not full, if No.of patients $<12$
Hence
$P$(room is full) $=1-\left(P\right(1)+....+P(11\left)\right)$
$=1-\sum _{x=0}^{11}\frac{e^{-12}(12{)}^x}{x!}$