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\% = \dfrac{{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) $ = \dfrac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}} $
Therefore number of boxes with no defective bottles (x=0) =
 $
  100 \times p(x = 0) \\
   = 100 \times \dfrac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}...........(x = 0) \\
   = 100 \times \dfrac{{{e^{ - 0.5}}({{0.5}^0})}}{{0!}} \\
   = 100 \times {e^{ - 0.5}} \\
$
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.