Problem 7 - A First Course in Probability

Question

A person has 100 light bulbs whose lifetimes are independent exponentials with mean 5 hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, approximate the probability that there is still a working bulb after 525 hours.

Step-by-Step Solution

Answer

The probability that there is still a working bulb after 525 hours is approximately \(1 - (1 - e^{-105})^{100}\).

Step 1: Calculate the probability that a single bulb lasts at least 525 hours

First, we need to determine the probability that a single light bulb, with an exponential lifetime with mean 5 hours, lasts at least 525 hours. The probability density function (pdf) for an exponential distribution is given by: \(f(x) = \frac{1}{\mu} e^{-x/\mu}\), where \(\mu\) is the mean of the distribution, which in this case is 5 hours. We are interested in the probability that a single bulb lasts at least 525 hours, which can be calculated by finding the complement of the probability that the bulb fails before 525 hours. Thus, we need to calculate the following: \(P(X \geq 525) = 1 - P(X < 525)\), where \(X\) is the random variable representing the lifetime of a single bulb. To find \(P(X < 525)\), we need to integrate the pdf from 0 to 525: \(P(X < 525) = \int_{0}^{525} \frac{1}{5}e^{-x/5} dx\).

Step 2: Evaluate the integral

To evaluate the integral, we use the substitution method. Let: \(u = -\frac{x}{5}\), so \(du = -\frac{1}{5} dx\). The integral becomes: \(-5 \int_{u(0)}^{u(525)} e^{u} du\). The definite integral of \(e^{u}\) is just \(e^{u}\), so we get: \(-5 [e^{u(525)} - e^{u(0)}]\). Now we substitute back: \(-5 [e^{-\frac{525}{5}} - e^{-\frac{0}{5}}]\).

Step 3: Calculate the probability for a single bulb

Now we can calculate the probability: \(P(X < 525) = -5 [e^{-\frac{525}{5}} - e^{-\frac{0}{5}}] = 1 - e^{-105}\). Thus, the probability that a single bulb lasts at least 525 hours is: \(P(X \geq 525) = 1 - (1 - e^{-105}) = e^{-105}\).

Step 4: Calculate the probability that none of the 100 bulbs last 525 hours

Since the lifetimes of the bulbs are independent, the probability that all of the 100 bulbs fail before 525 hours is the product of their individual probabilities: \(P(\text{all bulbs fail before 525 hours}) = [P(X < 525)]^{100} = (1 - e^{-105})^{100}\).

Step 5: Find the probability that there is still a working bulb after 525 hours

Finally, we can use the complement rule to find the probability that there is still a working bulb after 525 hours: \(P(\text{at least one bulb still working after 525 hours}) = 1 - P(\text{all bulbs fail before 525 hours}) = 1 - (1 - e^{-105})^{100}\). This is the probability that there is still a working bulb after 525 hours.