Question

At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per minute. Processing time for parts has an exponential distribution with mean of 2 minutes. What is the probability that a random part arrival finds that there are already 8 parts in the system (in machine + in queue)?

• A. 0.0173
• B. 0.0576
• C. 0.082
• D. 0.0247

Answer 1

The correct option is A 0.0173

$\text{Arrival rate (Poisson distribution)}$

$\lambda =0.35\text{parts/minute}=0.35\times 60\text{= 21 parts/hour}$

$\text{Service rate (exponential distribution)}$

$\mu =\frac{60}{2}=30\text{parts/hour}$

$\rho =\frac{\lambda }{\mu }=\frac{21}{30}=\frac{7}{10}=0.7$

$\text{Probability of having exactly n-customers in the system,}$

$P_n={\rho }^n{P}_0={\rho }^n(1-\rho )$

$\therefore \text{Probability of having exactly 8-customers in the system,}$

$P_8={\rho }^8(1-\rho )=\left(0.7{)}^8\right(1-0.7)$

$=\left(0.7{)}^8\right(0.3=0.01729=0.0173)$

$\text{Points to Remember :}$

• Parts arriving according to a Poisson process means Arrival is random.
• Service rate is an exponential distribution that means some parts take less time for service and others take more time for service.