Answer:
a) 0.13533
b) 0.36788
Step-by-step explanation:
Let X be the random variable that measures the lifetime of a bulb.
If the random variable X is exponentially distributed and X has an average value of 1000 days, then its probability density function is
[tex] \bf f(x)=\frac{1}{1000}e^{-x/1000}\;(x\geq 0)[/tex]
and its cumulative distribution function (CDF) is
[tex] \bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/1000}[/tex]
(a) Find the probability that the light bulb will function for more than 2000 days
[tex]\large P(X>2000)=1-P(X\leq 2000)=\\\\=1-(1-e^{-2000/1000})=e^{-2}=0.135335[/tex]
(b) Find the probability that the light bulb will function for more than 2000 days, given that it is still functional after 500 days
Let A and B be the events,
A = “The bulb will last at least 2000 days”
B = “The bulb will last at least 500 days”
We want to find P(A | B).
By definition P(A | B) = P(A∩B)P(B)
but B⊂A, so A∩B = B and
[tex]\large P(A | B) = P(B)P(B) = (P(B))^2[/tex]
We have
[tex]\large P(B)=P(X \geq 500)=1-P(X\leq 500)=1-(1-e^{-500/1000})=e^{-0.5}=0.60653[/tex]
hence,
[tex]\large P(A | B)=(P(B))^2=(0.60653)^2=0.36788[/tex]
