91Ó°ÊÓ

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

Step by step solution

01

Find Probability of No Type 1 Defect

The probability that the system does not have a type 1 defect is given by the complement of the probability of having a type 1 defect. This is calculated as follows:\[ P(A_1') = 1 - P(A_1) = 1 - 0.12 = 0.88 \]

02

Find Probability of Both Type 1 and Type 2 Defects

To find the probability of having both type 1 and type 2 defects, use the inclusion-exclusion principle:\[ P(A_1 \cap A_2) = P(A_1) + P(A_2) - P(A_1 \cup A_2) = 0.12 + 0.07 - 0.13 = 0.06 \]

03

Find Probability of Both Type 1 and Type 2 Defects but Not Type 3

To find this probability, we require both defects type 1 and type 2, but not defect type 3. Use the following:\[ P(A_1 \cap A_2 \cap A_3') = P(A_1 \cap A_2) - P(A_1 \cap A_2 \cap A_3) \]Substituting the values:\[ P(A_1 \cap A_2 \cap A_3') = 0.06 - 0.01 = 0.05 \]

04

Find Probability of At Most Two Defects

The probability that the system has at most two types of defects includes the events where there is no defect, one defect, or any two defects but not all three defects. Use the inclusion-exclusion principle for three events, adjusting for no defect:\[ P(\text{At most 2 defects}) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_1 \cap A_3) - P(A_2 \cap A_3) + P(A_1 \cap A_2 \cap A_3) \]Substitute the given probabilities:\[ P(\text{At most 2 defects}) = 0.12 + 0.07 + 0.05 - 0.06 - 0.04 - 0.02 + 0.01 = 0.13 \]

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Key Concepts

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

Inclusion-Exclusion Principle

Sometimes, events overlap, meaning they can occur simultaneously. The inclusion-exclusion principle helps determine the probability of the union of multiple events while accounting for these overlaps. When you want the probability that at least one of several events occurs, merely adding their probabilities would count the intersection of events multiple times.
Using the inclusion-exclusion principle allows you to adjust for such intersections:

• Subtract the probability of any two events occurring together.
• Add back the probability if all events occur.

For example, consider calculating the probability of having either a defect of type 1 or type 2, you use the formula:\[ P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) \]
This ensures each event's contribution is counted just once.

Complementary Events

The concept of complementary events is central to probability. If you have an event, its complement is the possibility of the event not occurring. The probabilities of an event and its complement always add up to 1 because one of them must happen.
A common notation for the complement of event \( A \) is \( A' \). For instance, given \( P(A_1) = 0.12 \), the probability that event \( A_1 \) does not happen, or \( A_1' \), would be:

• \( P(A_1') = 1 - P(A_1) \)
• This results in \( P(A_1') = 0.88 \), meaning there's an 88% chance the system does not have a type 1 defect.

Probability of Intersections

The probability of intersections relates to the chance that multiple events will occur at the same time. When dealing with events \( A \) and \( B \), \( P(A \cap B) \) stands for the probability that both \( A \) and \( B \) happen.
To find this probability, especially when events overlap but aren't entirely independent, you often use the inclusion-exclusion:

• For example, \( P(A_1 \cap A_2) = P(A_1) + P(A_2) - P(A_1 \cup A_2) \).
• Insert the given values: \( P(A_1 \cap A_2) = 0.12 + 0.07 - 0.13 = 0.06 \).

Thus, there's a 6% chance the system has both type 1 and type 2 defects.

Independent and Dependent Events

Events can be classified as independent or dependent. Independent events do not affect each other's occurrence. If knowing whether one event occurs doesn't change the probability of another, they're independent. However, if they do impact each other, we're dealing with dependent events.
When events are dependent, you need special methods, like the inclusion-exclusion principle, to calculate combined probabilities effectively. In this example, defects may coexist and influence each other's probabilities, making them dependent.
This means when you find the probability for occurrences like "both type 1 and 2" defects without type 3, you adjust the calculations to account for these dependent relationships.