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

Probability of No Type 1 Defect

To find the probability that the system does not have a type 1 defect, calculate the complement of the probability of having a type 1 defect: \( P(A_1^c) = 1 - P(A_1) = 1 - 0.12 = 0.88 \). Thus, the probability that the system does not have a type 1 defect is 0.88.

02

Probability of Both Type 1 and Type 2 Defects

Use the inclusion-exclusion principle to find the probability of having both type 1 and type 2 defects: \( P(A_1 \cap A_2) = P(A_1) + P(A_2) - P(A_1 \cup A_2) = 0.12 + 0.07 - 0.13 = 0.06 \). So, the probability that the system has both type 1 and type 2 defects is 0.06.

03

Probability of Both Type 1 and Type 2 but Not Type 3 Defect

Use the relation between probabilities to find \( P(A_1 \cap A_2 \cap A_3^c) \): From \( P(A_1 \cap A_2) = P(A_1 \cap A_2 \cap A_3) + P(A_1 \cap A_2 \cap A_3^c) \), we have \( 0.06 = 0.01 + P(A_1 \cap A_2 \cap A_3^c) \). Solving this gives \( P(A_1 \cap A_2 \cap A_3^c) = 0.05 \). Hence, the probability is 0.05.

04

Probability of At Most Two Defects

The probability of having at most two defects is found by excluding the probability of having all three defects from 1: \( P( ext{At most two defects}) = 1 - P(A_1 \cap A_2 \cap A_3) = 1 - 0.01 = 0.99 \). Therefore, the probability is 0.99.

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Key Concepts

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

Complementary Probability

Understanding complementary probability is an important part of probability theory. It helps us determine the chance of an event not happening, given its probability of occurrence. In our exercise, we explored how to find the probability that the system does not have a type 1 defect.
The formula for complementary probability is simple: if you know the probability of an event, the probability of its complement (the event not happening) is:

• \( P(A^c) = 1 - P(A) \)

For the system, where the probability of having a type 1 defect \( P(A_1) \) is 0.12, the probability of not having a type 1 defect is:

• \( P(A_1^c) = 1 - 0.12 = 0.88 \)

This means there is a 0.88 probability that the system is free from this particular defect type.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is useful when calculating the probability of either or both of two events occurring. It allows us to account for the overlap between events, preventing double counting. In the exercise, we calculated the probability of both type 1 and type 2 defects occurring in the system.
To find the probability of the intersection of two events \( A_1 \) and \( A_2 \), we apply this principle:

• \( P(A_1 \cap A_2) = P(A_1) + P(A_2) - P(A_1 \cup A_2) \)

By using the given probabilities:

• \( 0.06 = 0.12 + 0.07 - 0.13 \)

This calculation tells us the probability of both defects being present is 0.06, accurately accounting for their intersection.

Intersection of Events

The intersection of events represents scenarios where multiple events happen simultaneously. It's symbolized by \( \cap \) and is crucial for finding probabilities of events occurring together.
In our exercise, we wanted to find the probability of the system having both type 1 and type 2 defects, but not type 3. To do this, we had to calculate the probability for \( A_1 \cap A_2 \cap A_3^c \). Given that:

• \( P(A_1 \cap A_2) = P(A_1 \cap A_2 \cap A_3) + P(A_1 \cap A_2 \cap A_3^c) \)

Substituting the known values:

• \( 0.06 = 0.01 + P(A_1 \cap A_2 \cap A_3^c) \)
• \( P(A_1 \cap A_2 \cap A_3^c) = 0.05 \)

Thus, there is a 0.05 probability that only these two defects occur, excluding the type 3 defect.

Union of Events

The union of events, denoted by \( \cup \), signifies at least one of the included events occurs. Calculating the probability of a union involves considering all involved probabilities. If events overlap, the inclusion-exclusion principle might be used.
In this exercise, we were interested in the probability of the system having at most two of the three possible defects. This means finding the complement of all three defects occurring together. Calculating for the union:

• \( P(\text{At most two defects}) = 1 - P(A_1 \cap A_2 \cap A_3) \)

From the given:

• \( P(A_1 \cap A_2 \cap A_3) = 0.01 \)
• \( P(\text{At most two defects}) = 1 - 0.01 = 0.99 \)

Therefore, almost certainly (0.99 probability), the system will have no more than two defects.