91Ó°ÊÓ

Semiconductor lasers used in optical storage products require higher power levels for write operations than for read operations. High-power-level operations lower the useful life of the laser. Lasers in products used for backup of higher-speed magnetic disks primarily write, and the probability that the useful life exceeds five years is \(0.95 .\) Lasers that are in products that are used for main storage spend approximately an equal amount of time reading and writing, and the probability that the useful life exceeds five years is \(0.995 .\) Now, \(25 \%\) of the products from a manufacturer are used for backup and \(75 \%\) of the products are used for main storage. Let \(A\) denote the event that a laser's useful life exceeds five years, and let \(B\) denote the event that a laser is in a product that is used for backup. Use a tree diagram to determine the following: (a) \(P(B)\) (b) \(P(A \mid B)\) (c) \(P\left(A \mid B^{\prime}\right)\) (d) \(P(A \cap B)\) (e) \(P\left(A \cap B^{\prime}\right)\) (f) \(P(A)\) (g) What is the probability that the useful life of a laser exceeds five years? (h) What is the probability that a laser that failed before five years came from a product used for backup?

Step by step solution

01

Understanding the Problem

We have two types of products: backup (event B) and main storage \(B'\). For backup products, the probability of a laser exceeding 5 years (event A) is \(P(A | B) = 0.95\). For main storage, \(P(A | B') = 0.995\). The probability a product is for backup is \(P(B) = 0.25\) and for main storage is \(P(B') = 0.75\). We will use a tree diagram and basic probability rules to find the required probabilities.

02

Find Probability of Event B

The probability that a product is used for backing up, denoted as \(P(B)\), is already given as \(0.25\). Thus, \(P(B) = 0.25\).

03

Find \(P(A | B)\)

The probability that a laser used for backup exceeds five years is given by \(P(A | B) = 0.95\). Thus, \(P(A | B) = 0.95\).

04

Find \(P(A | B')\)

The probability that a laser used for main storage exceeds five years is given by \(P(A | B') = 0.995\). Thus, \(P(A | B') = 0.995\).

05

Calculate \(P(A \cap B)\)

Using the formula: \(P(A \cap B) = P(A|B) \times P(B)\), we substitute the given values:\[P(A \cap B) = 0.95 \times 0.25 = 0.2375.\]

06

Calculate \(P(A \cap B')\)

Similarly, we use the formula: \(P(A \cap B') = P(A|B') \times P(B')\):\[P(A \cap B') = 0.995 \times 0.75 = 0.74625.\]

07

Calculate \(P(A)\)

\(P(A)\) can be found using the law of total probability:\[P(A) = P(A \cap B) + P(A \cap B') = 0.2375 + 0.74625 = 0.98375.\]

08

Probability exceeding five years

The probability that the useful life of a laser exceeds five years is \(P(A) = 0.98375\). This means that about 98.375% of all lasers exceed five years.

09

Probability failed before five years from backup

To find \(P(B|A')\), we use Bayes' theorem.First, find \(P(A') = 1 - P(A) = 1 - 0.98375 = 0.01625\).Use Bayes’ Theorem: \[P(B|A') = \frac{P(A'|B) \times P(B)}{P(A')} = \frac{(1-0.95) \times 0.25}{0.01625} = \frac{0.0125}{0.01625} \approx 0.76923.\] The probability a laser that failed before five years came from a backup product is approximately 76.923%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability

Conditional probability is a fundamental concept in probability theory that focuses on determining the likelihood of an event happening, given that another event has already occurred. For example, in the context of semiconductor lasers, we are interested in finding the probability that a laser exceeds five years (

• for backup: \(P(A | B) = 0.95\)
• for main storage: \(P(A | B') = 0.995\)

It specifies the probability of event A occurring under the condition that event B has occurred. This relationship is essential in various engineering fields where the conditions significantly affect outcomes. In practice, you can calculate the conditional probability using the formula:\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]where \(P(A \cap B)\) is the probability that both events A and B occur, and \(P(B)\) is the probability of event B.

Tree Diagram

A tree diagram is a valuable tool for visualizing probability problems, particularly those involving sequences of events or conditional probabilities. It allows a structured way to map out different scenarios and calculate probabilities step-by-step. In the laser example, the first branching could represent the choice between backup (B) and main storage (

• Branch for Backup: \(P(B) = 0.25\)
• Branch for Main Storage: \(P(B') = 0.75\)

Each of these branches further splits according to whether the laser exceeds five years or not, depending on their respective conditional probabilities, helping us calculate results like \(P(A \cap B)\) and \(P(A \cap B')\).

• For Backup, Exceed: \(P(A \cap B) = 0.2375\)
• For Storage, Exceed: \(P(A \cap B') = 0.74625\)

These structured divisions lead to a comprehensive grid of outcomes where probabilities can be clearly calculated as depicted in the step-by-step solution.

Bayes' Theorem

Bayes' Theorem is a powerful mathematical tool that relates current probabilities to prior knowledge. It is especially useful for predicting the likelihood of an event, given new evidence. In our scenario, we calculate the probability that a laser that failed before five years is from a backup product. This is accomplished using Bayes' Theorem:\[ P(B|A') = \frac{P(A'|B) \times P(B)}{P(A')} \]where

• \(P(A')\) represents the probability that it doesn't last five years.
• \(P(A'|B)\) indicates the probability of failing before five years given it's a backup laser.
• \(P(B|A')\) provides the updated probability, around 76.923% in our problem.

This theorem enables a nuanced analysis as it takes into account both the current condition and prior probabilities, yielding a comprehensive understanding of how likely certain events are, given the present evidence.

Probability Distribution

Probability distribution is a concept that describes the likelihood of various outcomes in a random experiment. This exercise deals indirectly with a discrete probability distribution through understanding the different classifications of lasers based on their usage and their respective probabilities of exceeding five years. In a simplified form, we determine:

• The fraction of products used for backup
• The fraction used for main storage
• The probabilities associated with each category exceeding five years of use

A probability distribution must sum to 1, indicating that one of the possible outcomes must occur. The combined probabilities \(P(A) = 0.98375\) and \(P(A') = 0.01625\) reveal almost certainty for lasers exceeding five years, providing a profound insight into their life expectancy under given conditions. Understanding distributions like this is crucial for assessing and predicting failure rates in engineering applications.