# Why Do We Calculate the Probability of Phone Failure After Two Years?

• IntegrateMe
In summary, the probability that a client will not have to replace their phone before the company gives them a new one is given by the integral \int_2^\infty 1-e^{-λt}dt, which represents the probability of the phone lasting at least two years before failing. This is because the density function p(t) describes the time t in years that a phone will fail, so the interval T>2 corresponds to the probability that the phone will fail after two years.
IntegrateMe
The phones offered by a cell phone company have some chance of failure after they are activated. Suppose that the density function p(t) describing the time t in years that one of their phones will fail is

$$p(t) = 1-e^{-λt}$$ for t ≥ 0, and 0 otherwise.

The cell phone company offers its clients a replacement phone after two years if they sign a new contract. What is the probability that the client will not have to replace his phone before the company will give him a new one?

I tried solving the problem as follows:

$$\int_{-\infty}^2 1-e^{-λt}dt$$ which would end up becoming $$\int_0^2 1-e^{-λt}dt$$ since the function is 0 for everything t < 0.

However, the solution says that the answer is actually $$\int_2^\infty 1-e^{-λt}dt$$

I'm having trouble understanding why. If we're trying to find the probability that the client will not have to replace his phone before two years, why is the solution finding the probability that the phone will be defective after 2 years?

Your density function is for the "T = time to failure". You want it to last at least two years so you want P(T>2).

OK, but if the function describes when the phones *will* fail, wouldn't the interval T>2 describe the probability the phone will fail after 2 years, so we actually need to subtract 1 from this value?

IntegrateMe said:
OK, but if the function describes when the phones *will* fail, wouldn't the interval T>2 describe the probability the phone will fail after 2 years, so we actually need to subtract 1 from this value?

No. You want the phone to last two years so they don't have to replace it before then. That means you want it to fail sometime after 2 years. In other words, the probability of it lasting at least two years (which is what you want) is the same as the probability that the time of failure is greater than 2.

## 1. What is a probability density function (PDF)?

A probability density function (PDF) is a mathematical function used to describe the likelihood of a continuous random variable taking on a particular value. It is a function that maps the probability of a given outcome to each possible outcome.

## 2. How is a probability density function different from a probability distribution function?

A probability density function (PDF) is the continuous version of a probability distribution function (PDF). While a PDF is used to describe the probability of discrete outcomes, a PDF is used for continuous outcomes.

## 3. How is a probability density function related to the concept of probability?

A probability density function (PDF) is used to describe the likelihood of a continuous random variable taking on a specific value. It is an integral of the probability distribution function (PDF) and is used to calculate the probability of a continuous variable falling within a certain range of values.

## 4. What is the area under a probability density function (PDF) curve?

The area under a probability density function (PDF) curve is always equal to 1. This represents the total probability of all possible outcomes.

## 5. How is a probability density function (PDF) used in statistical analysis?

A probability density function (PDF) is used in statistical analysis to model and analyze data that follows a continuous distribution. It can be used to calculate probabilities, determine the characteristics of a distribution, and make predictions about future outcomes.

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