(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean μ = 1600, find the probability that both of the lamp's bulbs fail within 1800 hours. (Round your answer to four decimal places.)(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 1800 hours. (Round your answer to four decimal places.)

a) 0.4560

Step-by-step explanation:

a) Failure of Bulbs has an Exponential Density Function with a mean(\mu) = 1600 hours

Cumulative distribution function of Exponential Distribution is given by:   1-e^{-\λ *x}

We want the probability of both bulbs failing before 1800 hours

This means P_{1st bulb} (X<1800) and P_{2nd bulb} (X<1800) =??

So, we want area to the left of value 1800 on the distribution curve

Here, λ = 1 / 600

P(X<1800) = 1-e^{-λ *x}

              =1-e^-{ 1 / 1600 *1800}

              =0.6753

This is the probability that one bulb will fail within 1800 hours.

Probability of both bulbs is simply, 0.6753 * 0.6753 = 0.4560

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