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# 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.$100 \times {e^{ - 0.01}}$  Verified
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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.