Calculating Probability of 3 Pennies in 30 Boxes Using Poisson Distribution

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The discussion centers on calculating the probability of exactly 3 pennies in box number 1 when 100 pennies are randomly distributed across 30 boxes. The initial attempt used the Poisson distribution, yielding a probability of 0.22021, but the correct answer is 0.22345. It was suggested that a binomial distribution is more appropriate for this scenario, with parameters defined as N (number of pennies), k (number of pennies in box #1), and p (probability of a penny landing in box #1). The user ultimately confirmed understanding after receiving clarification. Accurate probability calculations are crucial in this random distribution problem.
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One hundred pennies are being distributed independently and at random into 30 boxes, labeled 1, 2, ..., 30. What is the probability that there are exactly 3 pennies in box number 1?

I tried using a Poisson distribution f(x) = (e^-λ)*(λ^x)/x! , with λ = 100/30 = 10/3 and x = 3. I got 0.22021 (5 s.f.), but the answer key is 0.22345. Am I using the wrong distribution? Thanks!
 
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For anyone penny, what is the probability that it will end up in box # 1?
Use a binomial distribution...N = number of pennies. k = 3. p = probability of landing in box #1.
 
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RUber said:
For anyone penny, what is the probability that it will end up in box # 1?
Use a binomial distribution...N = number of pennies. k = 3. p = probability of landing in box #1.

Ok I got it! Thanks! :)
 

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