Consider a Poisson distribution with #μ=3#. What is #P(x≥2)#?

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Answer 1 · 2017-01-13T09:28:36

The answer is #=0.8009#

The Poisson distribution is

#P(x)=(e^-mu mu^x)/(x!)#

#mu=3#

#3! =3*2*1#

#P(x>=2)=1-P(1)-P(0)#

#P(1)=(e^-3 *3^1)/(1!)=3/e^3#

#P(0)=e^-3=1/e^3#

Therefore,

#P(x>=2)=1-P(1)-P(0)=1-0.0498-0.1494=0.8009#

Answer 2 · 2017-01-13T09:41:41

#P(x>=2)=0.800852#

In a Poisson probability distribution, if mean value of success is #mu#,

the probability of getting #x# successes is given by

#P(x)=(e^(-mu)mu^x)/(x!)#

Now #P(x>=2)# means #1-P(x=0)-P(x=1)#

Here #mu=3# and #e^(-mu)=e^(-3)=0.049787#

and hence, desired probability is

#1-(e^(-3)3^0)/(0!)-(e^(-3)3^1)/(1!)#

= #1-e^(-3)xx(1+3)#

= #1-0.049787xx4#

= #1-0.199148#

= #0.800852#