Suppose you are trying to model the lifespan of a bulb. You want to model this in such a way such that the lifespan is memoryless. i.e.,  the extra time it will take to malfunction from this moment onwards does not depend on how long the bulb has already been working for. You also know that on average the bulb takes 10 months to malfunction.

What is your density of the lifespan of this bulb?

What is the chance that the bulb will last longer than 60 months?

To model the lifespan of a bulb with the memoryless property, we use the **exponential distribution**. The exponential distribution is commonly used to model the time until an event occurs, such as the failure of a component, and has the key property of being memoryless.

### Part 1: Finding the Density of the Lifespan

An exponential distribution is defined by a single parameter, the **rate parameter** \(\lambda\), which is the reciprocal of the mean. If the average lifespan of the bulb is 10 months, we have:

\[
\text{Mean} = \frac{1}{\lambda} = 10 \, \text{months}
\]

Solving for \(\lambda\):

\[
\lambda = \frac{1}{10} = 0.1
\]

The probability density function (PDF) of an exponential distribution with rate parameter \(\lambda\) is given by:

\[
f(t) = \lambda e^{-\lambda t}, \quad t \geq 0
\]

Substituting \(\lambda = 0.1\):

\[
f(t) = 0.1 e^{-0.1 t}, \quad t \geq 0
\]

Thus, the **density of the lifespan of the bulb** is:

\[
f(t) = 0.1 e^{-0.1 t}, \quad t \geq 0
\]

### Part 2: Chance the Bulb Will Last Longer Than 60 Months

The probability that the bulb will last longer than a certain time \(t\) in an exponential distribution can be calculated using the survival function, which is:

\[
P(T > t) = e^{-\lambda t}
\]

For \(t = 60\) months:

\[
P(T > 60) = e^{-0.1 \times 60} = e^{-6}
\]

Calculating \(e^{-6}\):

\[
P(T > 60) \approx 0.00247875
\]

So, the **chance that the bulb will last longer than 60 months** is approximately:

\[
P(T > 60) \approx 0.0025 \text{ or } 0.25\%
\]

### Summary

1. **Density of the lifespan** of the bulb: 
   \[
   f(t) = 0.1 e^{-0.1 t}, \quad t \geq 0
   \]

2. **Probability the bulb lasts longer than 60 months**: 
   \[
   P(T > 60) \approx 0.25\%
   \]

Would you like further details on exponential distributions or more examples? Here are some related questions:

1. How would you calculate the expected value of a bulb's lifespan given a different rate parameter?
2. What is the cumulative distribution function (CDF) for the exponential distribution?
3. Can we model multiple bulbs’ lifespans using the exponential distribution? How would that work?
4. What is the relationship between the exponential distribution and the Poisson process?
5. How would you find the median lifespan for this bulb?

**Tip:** The exponential distribution is widely used in queuing theory, reliability analysis, and survival studies due to its memoryless property.

Explore how to model the lifespan of a bulb using the exponential distribution with memoryless property. Calculate the density of the lifespan and determine the probability that the bulb will last longer than 60 months. Understand key concepts like the probability density function and survival function. Ideal for advanced high school students or anyone interested in probability and statistical modeling.

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