Question

A lamp has two bulbs each of a type with an average lifetime of 4 hours The probability density function for the lifetime of a bulb is f(t) = 1/4 e⁻ᵗ/⁴ t≥ 0, t greater than equal to 0. What is the probability that both of the bulbs will fail within 6 hours

Answer 1

To find the probability that both bulbs will fail within 6 hours, we integrate the pdf over 0 to 6 hours for one bulb, and then square the result to account for both bulbs failing independently.

Step-by-step explanation:

The student asks what the probability is that both bulbs will fail within 6 hours given that the average lifetime of the bulbs is 4 hours and the lifetime is modeled by an exponential distribution with probability density function (pdf) f(t) = 1/4 e−t/4 for t ≥ 0.

To find this probability, we need to calculate the probability that one bulb will fail within 6 hours and then square that result, since the bulbs fail independently.

The probability that one bulb will fail within 6 hours is given by the integral of the pdf from 0 to 6, which is ∫06 (1/4 e−t/4) dt. This integral evaluates to 1 − e−6/4, which can be approximated with a calculator. Subsequently, the probability that both bulbs fail within 6 hours is the square of this result.

Given that the average lifetime of a bulb is 4 hours, we can calculate that the probability of a bulb failing within 6 hours is:

P(T < 6) = 1 - e^(-6/4)

P(T < 6) = 1 - e^(-3/2) ≈ 0.7769

Since the bulbs are independent, we can multiply the probabilities together to get the probability that both bulbs will fail within 6 hours:

P(both bulbs fail within 6 hours) = (0.7769)*(0.7769) ≈ 0.6041