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

Probability Definitions

First, let's identify the given probabilities: - Probability a product is used for backup, \(P(B) = 0.25\).- Probability a product is used for main storage, \(P(B') = 0.75\).- Probability the useful life exceeds five years for backup, \(P(A \mid B) = 0.95\).- Probability the useful life exceeds five years for main storage, \(P(A \mid B') = 0.995\).

02

Calculate Joint Probabilities

The joint probability for products used for backup and surpassing five years is:\[P(A \cap B) = P(A \mid B) \times P(B) = 0.95 \times 0.25 = 0.2375\]Similarly, for main storage it is:\[P(A \cap B') = P(A \mid B') \times P(B') = 0.995 \times 0.75 = 0.74625\]

03

Total Probability of Surpassing Five Years

The probability that a laser's useful life exceeds five years, accounting for all scenarios, is:\[P(A) = P(A \cap B) + P(A \cap B') = 0.2375 + 0.74625 = 0.98375\]

04

Conditional Probability of Backup Failure

First calculate the probability that a laser fails before five years, \(P(A')\):\[P(A') = 1 - P(A) = 1 - 0.98375 = 0.01625\]Then, find the probability a laser from backup fails before five years:\[P(A' \cap B) = P(B) - P(A \cap B) = 0.25 - 0.2375 = 0.0125\]Finally, calculate conditional probability:\[P(B \mid A') = \frac{P(A' \cap B)}{P(A')} = \frac{0.0125}{0.01625} = 0.7692\]

05

Summary

With the calculations done, we can answer the questions:- Probability a laser is for backup, \(P(B) = 0.25\).- Probability a laser exceeds 5 years in backup, \(P(A \mid B) = 0.95\).- Probability a laser exceeds 5 years in main storage, \(P(A \mid B') = 0.995\).- Joint probability exceeding 5 years in backup, \(P(A \cap B) = 0.2375\).- Joint probability exceeding 5 years in main storage, \(P(A \cap B') = 0.74625\).- Overall probability exceeding 5 years, \(P(A) = 0.98375\).- Probability a failed laser is from backup, \(P(B \mid A') = 0.7692\).

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Key Concepts

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

Conditional Probability

Conditional probability helps determine the likelihood of an event occurring given that another event has already occurred. In our exercise, we are particularly interested in the useful life of semiconductor lasers exceeding five years, given different usage situations.
For the lasers used in backup, the conditional probability is represented as \( P(A \mid B) \). This reads as the probability of event \( A \), the useful life exceeding five years, given event \( B \), which is being used for backup. The exercise gives us \( P(A \mid B) = 0.95 \), meaning there is a 95% chance the laser will last over five years when used for backup.
Similarly, for lasers in main storage, \( P(A \mid B') = 0.995 \), indicating a 99.5% probability of exceeding five years. Conditional probability is useful in determining how different conditions influence the outcome.

Tree Diagram

A tree diagram is a handy visual aid that assists in mapping out complex probability scenarios in a structured and hierarchical manner. Think of it as a way to visually break down a series of dependent or independent events.
In the context of our laser example, a tree diagram can help show all possible outcomes of a laser's use and useful life. The first branch divides into two: backup and main storage (with probabilities \( P(B) = 0.25 \) and \( P(B') = 0.75 \)).
Each of these branches further splits depending on whether the useful life exceeds five years or not, showing both the conditional probabilities and their complements. This diagram is helpful for visual learners, illustrating how each product type affects probability outcomes.

Joint Probability

Joint probability refers to the probability of two events happening simultaneously. It is represented as \( P(A \cap B) \), for instance, which defines the probability of both event \( A \) occurring (laser lasting over five years) and event \( B \) occurring (laser used for backup) simultaneously.
In the exercise, the joint probability \( P(A \cap B) \) is calculated as \( P(A \mid B) \times P(B) \). For the backup lasers, this is \( 0.95 \times 0.25 = 0.2375 \). It indicates a 23.75% probability that a laser used for backup will surpass five years.
Similarly, for lasers in main storage, the joint probability \( P(A \cap B') \) is \( 0.995 \times 0.75 = 0.74625 \), indicating 74.625% likelihood of exceeding five years in this scenario. Joint probability is crucial in understanding the overlap of conditions.

Total Probability

The Total Probability Rule gives us the overall probability of a single event occurring by considering all possible scenarios. This formula involves summing the joint probabilities of the event occurring under different conditions.
For our exercise, the rule gives us \( P(A) \), the probability of a laser lasting over five years regardless of its use type. We compute it by adding the joint probabilities: \( P(A \cap B) + P(A \cap B') = 0.2375 + 0.74625 = 0.98375 \).
This result indicates that there's a high overall probability, 98.375%, that any laser will last beyond five years. The Total Probability Rule is valuable for accounting for all probabilities when there are multiple methods or paths an event can occur through.