Problem 32 - A First Course in Probability

Question

The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter \(\lambda=\frac{1}{2} .\) What is (a) the probability that a repair time exceeds 2 hours? (b) the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours?

Step-by-Step Solution

Answer

The probability that a repair time exceeds 2 hours is approximately \(0.368\). The conditional probability that a repair takes at least 10 hours, given its duration exceeds 9 hours, is approximately \(0.360\).

Step 1: Exponential distribution formula

The probability density function (pdf) of an exponential distribution is given by: $$ f(x) = \lambda e^{-{ \lambda x}}, \quad x \geq 0 $$ And the cumulative distribution function (cdf) is given by: \( F(x) = P(X \leq x) = 1 - e^{ - \lambda x } \) where \(X\) is the random variable representing the repair time, and \(x\) is the specific repair time.

Step 2: Find the probability of repair taking more than 2 hours

We have to find \(P(X > 2)\), which can also be written as \(1 - P(X \leq 2)\). Using the cdf formula: \(P(X > 2) = 1 - F(2)\) Replace the values of \(\lambda\) and \(x\) in the cdf formula: \(P(X > 2) = 1 - (1 - e^{ - \frac{1}{2} \cdot 2 })\) Now, calculate the probability: \(P(X > 2) = 1 - (1 - e^{ -1 })\) \(P(X > 2) = e^{ -1 } \approx 0.368\) Therefore, the probability that a repair time exceeds 2 hours is approximately \(0.368\).

Step 3: Find the conditional probability

We are asked to find the conditional probability of a repair taking at least 10 hours, given that it exceeds 9 hours. Using the conditional probability formula: \(P(X \geq 10 \mid X > 9) = \frac{P(X \geq 10 \cap X > 9)}{P(X > 9)} = \frac{P(X \geq 10)}{P(X > 9)}\) Since \(X \geq 10\) implies \(X > 9\), we can write it as: \(P(X \geq 10 \mid X > 9) = \frac{P(X \geq 10)}{P(X > 9)}\) Now, we can find the probabilities using the same approach as in Step 2: \(P(X \geq 10) = 1 - F(10) = e^{-\frac{1}{2} \cdot 10} \approx 6.74 \times 10^{-5}\) \(P(X > 9) = 1 - F(9) = e^{-\frac{1}{2} \cdot 9} \approx 1.87 \times 10^{-4}\) Now we can find the conditional probability: \(P(X \geq 10 \mid X > 9) = \frac{6.74 \times 10^{-5}}{1.87 \times 10^{-4}} \approx 0.360\) Therefore, the conditional probability that a repair takes at least 10 hours, given its duration exceeds 9 hours, is approximately \(0.360\).

Key Concepts

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept in the world of probability and statistics, especially when dealing with continuous random variables. In an exponential distribution, which often models the time until an event occurs, the PDF helps us understand how the probabilities are distributed over time.
For an exponentially distributed variable with parameter \(\lambda\), the PDF is given by:

• \( f(x) = \lambda e^{-\lambda x} \), where \(x \geq 0\).

A key feature of this function is its memoryless property, meaning the past does not influence the future probabilities. In practical terms, this implies that the probability of adding another hour to the repair time does not depend on how long the repair already took. The function rapidly decreases as \(x\) increases, indicating decreasing probabilities for longer durations.
This characteristic makes the exponential distribution suitable for representing scenarios like machine repair times, where shorter durations are more expected than longer ones.

Cumulative Distribution Function

While the Probability Density Function provides insight into the likelihood of specific values, the Cumulative Distribution Function (CDF) gives the probability that a random variable is less than or equal to a specific value. In exponential distribution, the CDF is crucial for finding the probability of the variable falling within a particular range.
The CDF of an exponential distribution is expressed as:

• \( F(x) = 1 - e^{-\lambda x} \), where \(x \geq 0\).

Using the CDF, one can compute probabilities of ranges by considering the difference between \(F\) values at specific points. For instance, to find the probability that a repair occurs in less than 2 hours, one would use \(F(2)\). Conversely, to find the probability of the repair exceeding 2 hours, calculate \(1 - F(2)\).
This ability to quickly compute probabilities for various ranges makes the CDF a powerful tool in probability theory.

Conditional Probability

Conditional probability is a concept used when we need to find the probability of an event, given that another event has occurred. This is particularly useful for understanding dependent events and refining our belief based on new information.
In the context of exponential distributions, this can be computed using the formula:

• \(P(X \geq a \mid X > b) = \frac{P(X \geq a)}{P(X > b)}\), where \(a\) and \(b\) are specific times and \(a > b\).

This formula helps us answer questions like the probability of a repair lasting at least 10 hours if it has already lasted over 9 hours. It builds on the knowledge that the repair has already exceeded a specific time (9 hours, in this case), thus concentrating our probability calculations on the remaining time.
Utilizing the CDF, conditional probability simplifies problems by reducing the range of interest and providing more focused probabilities based on existing conditions.