Question

The manufacturer who produces medicines bottles finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 100 boxes from the producer of bottles. Using Poisson distribution, the number of boxes with no defective bottle is ;
A.$100\times e^{-0.1}$
B.$100\times e^{-0.5}$
C.$100\times e^{-0.05}$
D.$100\times e^{-0.01}$

Answer 1

Hint: The Poisson distribution is the discrete probability distribution of the number of events occurring in a given time period, given the average number of times the event occurs over that time period.

Complete step by step solution :
According to the question:
Probability of getting a defective bottle (p) = $0.1\mathrm{\%}={\displaystyle \frac{0.1}{100}}=0.001$
And number of bottles in box (n) = 500
Therefore according to the poisson distribution
$\lambda =np=500\times 0.001=0.5$
And number of boxes (N) = 100
And from poisson distribution we know that
P(x) $={\displaystyle \frac{e^{-\lambda }{\lambda }^x}{x!}}$
Therefore number of boxes with no defective bottles (x=0) =
 $$\begin{gathered} 100\times p(x=0) \\ =100\times {\displaystyle \frac{e^{-\lambda }{\lambda }^x}{x!}}...........(x=0) \\ =100\times {\displaystyle \frac{e^{-0.5}({0.5}^0)}{0!}} \\ =100\times e^{-0.5} \end{gathered}$$
Hence the required number of boxes with 0 defective bottles = $100\times e^{-0.5}$

Note:In poisson distribution $\lambda$= np here $\lambda$ is the mean of the given distribution and n is the number of units.