Question

Suppose that the amount of time (in months) that a light bulb lasts before it burns out has an exponential distribution with parameter λ = 1/5. a) What is the probability that it lasts at least three months? b) If it has already lasted two months, what is the probability that it will last at least three more months? c) On average, how many months will the light bulb last?

Answer 1

We have

`p(T)=\int_(0)^(T)\lambda e^(-\lambda t)dt`

a)

Thus probability that the bulb lasts 3 months equals

`p(3)=\int_(0)^(3)0.2 e^(-0.2 t)dt\\\\p(3)=0.45`

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

`p(A|B)=(p(A\cap B))/(p(B))`

`p(B)=\int_(0)^(2)0.2 e^(-0.2 t)dt\\\\p(B)=0.33`

also

`p(A\cap B)`=
`0.45* 0.33=0.148`

`\therefore p(A|B)=(0.148)/(0.33)\\\\p(A|B)=0.448`

c)

On an average the expectation value of the life of the bulb will be given by mean of the distribution

`mean=\beta =(1)/(\lambda )\\\\\therefore mean=(1)/((1)/(5))=5`

thus on an average the bulb will last for 5 months.