Answer:
[tex]P(T>0.1h)=5(e^{-2})[/tex]
Step-by-step explanation:
We are going to solve this exercise using count-time duality.
The event [tex]T>0.1h[/tex] means that the time when third mail arrives is later than 0.1 h
This fact means that in the previously interval there have come two or fewer emails.
Let be X the random variable : ''The number of mails that arrived during the first 0.1 h''
Using count-time duality ⇒ [tex]P(T>0.1h)=P(X\leq 2)[/tex]
X ~ Po (20. 0.1)
X ~ Po (2)
Now we calculate [tex]P(X\leq 2)[/tex]
The probability function for X is :
[tex]P(X=x)=\frac{m^{x}e^{-m}}{x!}[/tex]
Where m = λ
[tex]P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)[/tex]
[tex]P(X\leq 2)=\frac{2^{0}e^{-2}}{0!}+\frac{2^{1}e^{-2}}{1!}+\frac{2^{2}e^{-2}}{2!}[/tex]
[tex]P(X\leq 2)=e^{-2}+2e^{-2}+2e^{-2}=5(e^{-2})[/tex]
Finally, [tex]P(T>0.1h)=5e^{-2}[/tex]
