Part A.) A lamp has two bulbs of a type with an average lifetime of 1200 hours.

Question

Part A.) A lamp has two bulbs of a type with an average lifetime of 1200 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean ? = 1200, find the probability that both of the lamp's bulbs fail within 1200 hours. (Round your answer to four decimal places.)

Part B.) Another lamp has just one bulb of the same type as in part (a). If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1200 hours. (Round your answer to four decimal places.)

Explanation / Answer

F(t) = 1 - exp(-t/1200)

Thus, P(both fail within 1200 hours) = (1 - exp(-1200/1200))(1 - exp(-1200/1200)) =

(1-exp(-1))^2 = 1 - 2 exp(-1) + exp(-2) =0.399576400893728

B) This can be considered a truncated Poisson process with parameter 1200/1200 = 1, as it terminates once there are 2 failures.

Then, P(2 + failures) in the Poisson process = 1 - P(0) - P(1) = 1 - exp(-1) - exp(-1) = 1 - 2 exp(-1) =

0.264241117657115

We can calculate this in another way;

we can find P(x+y <= 1) where f(x,y) = exp(-x-y)

Thus, we integrate this for x from 0 to 1 - y and y from 0 to 1

Then, the integral with respect to x gives us exp(-y)(1 - exp(y-1)) = exp(-y) - exp(-1) and the integral of this is -exp(-y) - yexp(-1) evaluated from 0 to 1 = 1 - exp(-1) - exp(-1) = 1 - 2 exp(-1)

This matches the solution using the Poisson result.

If you want to reason out why the probability that both fail in B is less, think this way; in A), we have 2 lights burning until there is a failure, so during this time, there is twice as high a rate of failure; then, once one light fails, we have the same failure rate that we always have in B.

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