How Do You Solve a Poisson Distribution Problem Where 3P(X=1)=P(X=2)?

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SUMMARY

The discussion focuses on solving a Poisson distribution problem defined by the equation 3P(X=1) = P(X=2). The probability mass function (pdf) for a Poisson distribution is given by P(n) = e^(-κ)κ^n/n!. Through analysis, it is determined that the value of the constant κ is 6. Consequently, the pdf of X can be expressed, and P(X=4) can be calculated using this value.

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tiger2380
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Can someone help me with this question:
If X has a Poisson distribution so that 3P(X=1)=P(X=2)
find the pdf of X, and P(X=4)?
 
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"pdf," I assume means Poisson Distribution Function.

Well, the function is [tex]P(n)=\frac{e^{-\kappa}\kappa^n}{n!}[/tex]

So since we know values of n, the important thing is to find the particular value of the constant [tex]kappa, \kappa[/tex]
 
Last edited:
It looks like k=6.
 

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