Answer:
The probability of having two claims in three days is 0.225.
Step-by-step explanation:
This type of process can be modeled by the Poisson distribution.
This distribution has a rate, that in this case is x: number of claims filled in a one-week period.
The probabilty density function can be written as
[tex]P(y=k)=\frac{(xt)^ke^{-xt}} {k!}[/tex]
In this case, [tex]t =3 \,days =3/7\,weeks= 0.429\,weeks[/tex].
The probablity of having two claims in three days is:
[tex]P(y=2)=\frac{(7*(3/7))^2e^{-7*(3/7)}} {2!}=\frac{3^2*e^{-3}}{2}=\frac{9*0.05}{2}=0.225[/tex]
The probability of having two claims in three days is 0.225.
