Question

The lifetime of a lightbulb can be modeled with an exponential random variable with an expected lifetime of 1000 days. (a) Find the probability that the lightbulb will function for more than 2000 days. (b) Find the probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days

Answer 1

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

$\bf f(x)=\frac{1}{1000}e^{-x/1000}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/1000}$

(a) Find the probability that the light bulb will function for more than 2000 days

$\large P(X>2000)=1-P(X\leq 2000)=\\\\=1-(1-e^{-2000/1000})=e^{-2}=0.135335$

(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  

$\large P(A | B) = P(B)P(B) = (P(B))^2$

We have  

$\large P(B)=P(X \geq 500)=1-P(X\leq 500)=1-(1-e^{-500/1000})=e^{-0.5}=0.60653$

hence,

$\large P(A | B)=(P(B))^2=(0.60653)^2=0.36788$