There are 500 boxes, each containing 1000 ballot papers for election. The chance that a
ballot paper is defective is 0.002. Assuming Poisson distribution, find the number of boxes
containing atmost two defective ballot papers. (Use er2 = 0.1353).
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Answers
Answer:
Here λ=.002×1000=2
Thus probability that a box contain at least one defective ballot =1−P(X=0)=1−e−2=1−0.1353=.8647
Hence number of box out of 500 which contain at least one defective ballot is =.8647×500=432
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338 boxes containing at most two defective ballot papers.
Given : There are 500 boxes, each containing 1000 ballot papers for election . chance that a ballot paper is defective is 0.002.
Assuming Poisson distribution
e⁻² = 0.1353
To Find : the number of boxes containing at most two defective ballot papers.
Solution:
The chance that a ballot paper is defective is 0.002
=> Mean Defective paper in 1000 = 1000 * 0.002 = 2
λ = 2
P(x) = λˣ e^(-λ) / x!
probability of boxes containing atmost two defective ballot papers
= containing 0 , 1 or 2 defective papers
= P(0) + P(1) + P(2)
= 2⁰ e⁻² / 0! + 2¹ e⁻² / 1! + 2²e⁻² / 2!
= e⁻² ( 1 + 2 + 2)
= e⁻² (5)
= 0.1353 ( 5)
= 0.6765
Number of boxes = 500 * 0.6765 = 338.25 ≈ 338
338 boxes containing at most two defective ballot papers.
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