Respuesta :
Answer:
a) Probability: 0.5488
b) Probability: 0.3012
c) Probability: 0.0907
d) Probability: 0.2476
Step-by-step explanation:
a) We can model this as a Poisson process, with parameter r=0.1.
We have to calculate the probability that no claim is made in 6 months.
[tex]P(k)=\frac{e^{-rt}(rt)^k}{k!} \\\\\\P(k=0)=\frac{e^{-0.1*6}(0.1*6)^0}{0!}=e^{-0.6}=0.5488[/tex]
The probability that the dealership will wait at least 6 months until the next claim is 0.5488.
b) In this case, t=12 months.
[tex]P(k)=\frac{e^{-rt}(rt)^k}{k!} \\\\\\P(k=0)=\frac{e^{-0.1*12}(0.1*12)^0}{0!}=e^{-1.2}=0.3012[/tex]
c) In this case, t=24 months
[tex]P(k)=\frac{e^{-rt}(rt)^k}{k!} \\\\\\P(k=0)=\frac{e^{-0.1*24}(0.1*24)^0}{0!}=e^{-2.4}=0.0907[/tex]
d) In this case, we have to multiply the probability that there is no claim in the first 6 months (k=0, t=6) and the probability that there is more than one claim in the next 6 months (k>0, t=6).
As the Poisson process is memoryless, the probabilitiies are independent of past events and can be calculated that way.
[tex]P(k)=\frac{e^{-rt}(rt)^k}{k!} \\\\\\P=P(k=0)*P(k>0)=P(k=0)*(1-P(k=0)\\\\P=(\frac{e^{-0.1*6}(0.1*6)^0}{0!})(1-\frac{e^{-0.1*6}(0.1*6)^0}{0!})=e^{-0.6}*(1-e^{-0.6})\\\\P=0.5488*(1-0.5488)=0.5488*0.4512\\\\P=0.2476[/tex]
