Respuesta :
Answer:
We have
[tex]p(T)=\int_{0}^{T}\lambda e^{-\lambda t}dt[/tex]
a)
Thus probability that the bulb lasts 3 months equals
[tex]p(3)=\int_{0}^{3}0.2 e^{-0.2 t}dt\\\\p(3)=0.45[/tex]
b)
It is a case of conditional probability
Let A be the event that the bulb will last another 3 months provided it has lasted for 2 months
'B' be the event that that the light bulb lasts for 2 months
thus we have
[tex]p(A|B)=\frac{p(A\cap B)}{p(B)}[/tex]
[tex]p(B)=\int_{0}^{2}0.2 e^{-0.2 t}dt\\\\p(B)=0.33[/tex]
also
[tex]p(A\cap B)[/tex]=[tex]0.45\times 0.33=0.148[/tex]
[tex]\therefore p(A|B)=\frac{0.148}{0.33}\\\\p(A|B)=0.448[/tex]
c)
On an average the expectation value of the life of the bulb will be given by mean of the distribution
[tex]mean=\beta =\frac{1}{\lambda }\\\\\therefore mean=\frac{1}{\frac{1}{5}}=5[/tex]
thus on an average the bulb will last for 5 months.
