Chapter 7, question 11: A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?
Given that each bulb has an exponential lifetime with a mean of 1000 hours, let \(\lambda_i\) denote the rate parameter for the \(i\)-th bulb. For 100 bulbs, we have \(\lambda_i = \frac{1}{1000}\).
Then, the total rate at which the bulbs burn out is \(\sum_{i=1}^{100} \lambda_i = 100 \times \frac{1}{1000} = \frac{1}{10}\).
Now, considering the probability density function (PDF) of the minimum of exponential random variables, denoted by \(X_{\text{min}}\), it is given by \(f_{X_{\text{min}}}(x) = \lambda e^{-\lambda x}\), where \(\lambda\) is the rate parameter.
Therefore, for the first bulb to burn out, the PDF becomes \(f_{X_{\text{min}}}(x) = \frac{1}{10} e^{-\frac{1}{10} x}\).
Now, to determine the probability that a specific bulb, say bulb \(k\), is the first to burn out, we utilize the conditional probability:
\[ P(k \, | \, X_k = \text{min}(X_1, \ldots, X_{100})) = \frac{\lambda_k}{\sum_{i=1}^{100} \lambda_i} \]
Substituting the values, we get:
\[ P(k \, | \, X_k = \text{min}(X_1, \ldots, X_{100})) = \frac{\frac{1}{1000}}{\frac{1}{10}} = \frac{1}{10} \]
This implies that regardless of the specific bulb, the probability that any bulb is the first to burn out is \(\frac{1}{10}\).
Here is the r code:
bulbs=100
lambdaBulb=1/1000
sumLambda=(bulbs*lambdaBulb)
(minE = 1/sumLambda)## [1] 10