Respuesta :
The given Poisson Distribution when solved gives these answers
(a)
[tex]P(X\geq 1) =1-e^{-\lambda }[/tex]
[tex]P(X\geq 2) =1-e^{-\lambda }-\lambda e^{-\lambda }[/tex]
(b)
[tex]P(X=k|X\geq 1) =\frac{\lambda ^{k}}{k!(e^{\lambda }-1)}[/tex]
What is Poisson Distribution?
Poisson Distribution is a distribution that shows how many times an event is likely to occur in a given period.
What do we mean by Probability Mass Function?
A probability mass function (PMF) is a function over the sample space of a discrete random variable X which gives the probability that X is equal to a certain value. Let X be a discrete random variable on a sample space S . Then the PMF f(x) is defined as. f(x)=P[X=x].
How do we solve the given question?
We are informed that X is the number of purchases that Fred will make on the online website.
We are given the PMF of X is
[tex]P(X=k)=(\frac{\lambda ^{k}}{k!})e^{-\lambda}[/tex] for k=1, 2, 3, .....
Now, we know that this is a Poisson Distribution with parameter λ.
(a) We are asked to find P(X≥1) and P(X≥2)
[tex]P(X\geq 1) = 1 - (P(X\geq 1))^{c} = 1 - P(X=0) = 1 - \frac{\lambda ^{0}}{0!}e^{-\lambda}=1-e^{-\lambda }[/tex]
[tex]P(X\geq 2) = 1 - (P(X\geq 2))^{c} = 1 - P(X=0) - P(X=1))\\ =1 - \frac{\lambda ^{0}}{0!}e^{-\lambda} - \frac{\lambda ^{1}}{1!}e^{-\lambda} \\=1-e^{-\lambda }-\lambda e^{-\lambda }[/tex]
(b) We want to find when the customer is making one or more purchases, that is X≥1,
∴ We need to find P(X=k | X≥1).
When k=0 we get P(X=k | X≥1) = 0.
For other values of k, we use basic probability theory,
[tex]P(X=k|X\geq 1) = \frac{P(X=k,X\geq 1)}{P(X\geq 1)}= \frac{P(X=k)}{P(X\geq 1)}=\frac{ \frac{\lambda ^{k}}{k!}e^{-k}}{1-e^{-\lambda }} = \frac{\lambda ^{k}}{k!(e^{\lambda }-1)}[/tex]
Learn more about Poisson Distribution at
https://brainly.com/question/7879375
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