Respuesta :
Answer:
a) [tex]P(X=0) = 0.1^0 \frac{e^{-0.1}}{0!}=0.905[/tex]
b) [tex]P(X\geq 1) = 1-P(X <1) = 1-P(X=0) = 1-0.905=0.095[/tex]
c) [tex]P(X=0)=0.2^0 \frac{e^{-0.2}}{0!}=0.819[/tex]
Step-by-step explanation:
Definitions and concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
And the parameter [tex]\lambda[/tex] represent the average ocurrence rate per unit of time.
On this case the mean is given by [tex]\lambda =0.1[/tex]
Part a: Find the probability that two next inspected disk will have no missing pulse
For this case we want this probability:
[tex]P(X=0) = 0.1^0 \frac{e^{-0.1}}{0!}=0.905[/tex]
Part b: Find the probability that the next inspected disk wkl have more than one missing pulse
For this case we want this probability:
[tex]P(X\geq 1) = 1-P(X <1) = 1-P(X=0) = 1-0.905=0.095[/tex]
Part c: Find the probabihty that neither of the next two inspected disks will contain any missing pulse.
For this case the new mean would be [tex]\lambda = 0.1x2 = 0.2[/tex] since we have now two inspected disks, and we want this probability:
[tex]P(X=0)=0.2^0 \frac{e^{-0.2}}{0!}=0.819[/tex]
