Problem 44 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In a home theater system, the probability that the video components need repair within \(1 \mathrm{yr}\) is \(.01\), the probability that the electronic components need repair within 1 yr is \(.005\), and the probability that the audio components need repair within 1 yr is .001. Assuming that the events are independent, find the probability that a. At least one of these components will need repair within 1 yr. b. Exactly one of these components will need repair within 1 yr.

Step-by-Step Solution

Answer

The probability that at least one component will need repair within 1 year is approximately 0.01598, and the probability that exactly one component will need repair within 1 year is approximately 0.01594.

Step 1: Identify the probabilities of each component not needing repair

To solve this problem, it will be helpful to compute the probability of each component not needing repair within 1 year. We can find these values by subtracting the given probabilities from 1, since the sum of the probabilities that an event occurs and that an event doesn't occur equals 1. For video components: \(P(\text{not needing repair}) = 1 - P(\text{needing repair}) = 1 - 0.01 = 0.99\) For electronic components: \(P(\text{not needing repair}) = 1 - P(\text{needing repair}) = 1 - 0.005 = 0.995\) For audio components: \(P(\text{not needing repair}) = 1 - P(\text{needing repair}) = 1 - 0.001 = 0.999\)

Step 2: Find the probability of no components needing repair

Since the events are independent, we find the probability of no components needing repair by multiplying the probabilities of each component not needing repair within 1 year: \(P(\text{no components needing repair}) = P(\text{video not needing repair}) \times P(\text{electronic not needing repair}) \times P(\text{audio not needing repair})\) \(P(\text{no components needing repair}) = 0.99 \times 0.995 \times 0.999 = 0.984018015\)

Step 3a: Compute the probability of at least one component needing repair

To find the probability of at least one component needing repair, we subtract the probability of no components needing repair from 1: \(P(\text{at least one component needing repair}) = 1 - P(\text{no components needing repair}) = 1 - 0.984018015 = 0.015981985\)

Step 3b: Compute the probability of exactly one component needing repair

We can find the probability of exactly one component needing repair by considering the three possible cases: only video needs repair, only electronic needs repair, or only audio needs repair. For each case, we multiply the probabilities of the single component needing repair and the other two not needing repair. Then, we add the resulting probabilities: \(P(\text{exactly one component needing repair}) = P(\text{only video needing repair}) + P(\text{only electronic needing repair}) + P(\text{only audio needing repair})\) \(P(\text{exactly one component needing repair}) = (0.01 \times 0.995 \times 0.999) + (0.005 \times 0.99 \times 0.999) + (0.001 \times 0.99 \times 0.995) = 0.015939975\)

Step 5: Final Answer

The probability that at least one component will need repair within 1 year is approximately 0.01598, and the probability that exactly one component will need repair within 1 year is approximately 0.01594.

Key Concepts

Independent Events

Independent events are a fundamental concept in probability that can be intuitive yet sometimes tricky to grasp. When we say events are independent, it means the outcome of one event does not affect the outcome of any other event. In our problem, the probability that the video, electronic, and audio components need repair is seen as independent. This means that the likelihood of a video component needing repair has no effect on whether the electronic component or audio component needs repair.

For a more relatable example: think about flipping two different coins. The result of the first coin flip (heads or tails) does not change or impact the outcome of the second coin flip. Each event stands alone when events are independent. This concept allows us to calculate the joint probability of several events occurring by multiplying their individual probabilities. In this problem, by knowing the probability of each component not needing repair, we can easily calculate the combined likelihood of all components functioning without issues over the year.

Component Reliability

Component reliability refers to the likelihood that a component or system will perform its intended function without failure for a specified period. In our scenario, the provided probabilities tell us about the components’ reliability over one year. Specifically, a high probability indicates good reliability, while a low probability suggests a higher chance of needing repair.

• The video components have a 0.99 chance of functioning reliably.
• The electronic components have a 0.995 chance, slightly more reliable.
• The audio components boast the highest reliability with 0.999.

Reliability is crucial for ensuring smooth operations, especially in systems where multiple components work together. In computing overall system reliability, understanding individual component probabilities helps in forming a complete picture of system performance. Systems integrators often base maintenance schedules and preventive actions on these probabilities to prevent unexpected failures.

Probability of Repair

The probability of repair essentially tells us how likely it is for a component to fail and thus need service within the specified time, usually based on historical data or observed performance. In the given exercise, we calculate the probability of needing repair for each component, which involves understanding issues like wear and tear, environmental challenges, or design flaws.

When finding "at least one component needing repair," we are calculating the complement of all components working perfectly. This involves subtracting the probability of no component failures from one.

Additionally, the concept of having "exactly one component needing repair" introduces combinatorial considerations. We evaluate and sum the scenarios where each individual component alone needs repair, while the others don't. Being able to distinguish these probabilities aids in addressing repair logistics and resource allocation effectively. Properly applying these concepts can lead to more efficient budgeting and planning for anticipated maintenance operations.