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3 bottles per case are underfilled on average. What is the chance that at least 4 underfilled bottles will be contained in a random case?

Using the formula: $1-\left( \dfrac{3^0 e^{-3} }{ 0!} + \dfrac{3^1 e^{-3}}{1!} + \dfrac{3^2 e^{-3}}{2!} + \dfrac{3^3 e^{-3}}{3!} \right)$, I obtained $0.3528$

However, my answer was marked as incorrect. How do you solve this correctly?

Michael Hardy
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Florencio
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1 Answers1

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My guess is you were told something about the number of bottles in each case and you should have used the binomial distribution rather than the Poisson distribution.

Michael Hardy
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  • No more information is given in the problem. Just 3 underfilled bottles per case,average, then you randomly select one case wanting to know the probability mentioned. – Florencio Aug 28 '15 at 14:39
  • To elaborate on Michael Hardy's answer, if you were told the total number in the case, then you condition that the number of underfilled bottles is no larger than that total number. This process (conditioning on the total number) can be thought of as inverse to the standard "infinite size with fixed rate" limit which reduces a binomial distribution to its limiting a Poisson distribution. – josh Aug 28 '15 at 14:41
  • Perhaps you were to assume a total number $N$. A true Poisson distribution gives a non-zero probability to an arbitrarily high number of underfilled bottles. The case would need to be infinite if Poisson were appropriate. – josh Aug 28 '15 at 14:42