GES 400/500 Engineering Statistics Homework #2 Notes

Problem 1

Event Definitions:

• Let ( A ) = Event where the first inspector found a defect. This event represents the likelihood of defects being detected during the initial quality control check.

• Let ( B ) = Event where the second inspector found a defect. This refers to the assessment by a second, independent inspector and may provide additional validation or indicate a pattern of defects.

Calculating Probabilities:

• Use formulas from probability theory to calculate:

  • ( P(A) ) = Probability of finding a defect during the first inspection. This represents the prevalence of defects in the initial production run.

  • ( P(B) ) = Probability of finding a defect during the second inspection, which could be influenced by both random chance and correlation with the first inspection.

  • ( P(A \cup B) ) = Probability that either the first or the second inspector finds a defect, showcasing the overall quality assurance effectiveness.

Problem 2

Venn Diagram Usage:

• Utilize Venn diagrams to visualize events A and B, providing an intuitive representation of the sample space and relationships between A and B when calculating probabilities like ( P(A \cap B) ). This tool helps to identify intersections and unions of events which are critical in understanding the overlap between inspections results.

Conditional Probability Calculations:

• Calculate utilizing Bayes' theorem:

  • ( P(A | B) = \frac{P(B | A) P(A)}{P(B)} )
    This calculation offers insights into the likelihood of the first inspector finding a defect given that the second inspector has already found one, thus indicating dependency between inspections.

Final Result:

• Example calculations provided with results, e.g., ( P(A \cap B) = 0.0435 ). This is a specific case showcasing the calculated probability of both inspectors finding a defect simultaneously.

Problem 3

Rolling Two Dice:

• Analyze the scenario of rolling two six-sided dice to calculate probabilities based on the sums. Specifically, conditions are set for sums greater than 3 and less than or equal to 6.

• Outcomes that meet the criteria include: (1,3), (2,2), (3,1), (1,2), (2,1).

Probability Calculation:

• Total outcomes ( = 36 ) represents the complete set of all possible combinations from rolling two dice.

• Successful outcomes ( M = 12 ) refers to the combinations that satisfy the given condition of sum constraints.

• Thus, ( P(3 < N1 + N2 \leq 6) = \frac{M}{36} = \frac{12}{36} = \frac{1}{3} ) illustrates the calculated probability of achieving a sum within the defined range.

Problem 4

System Functionality:

• Define tubes' operational parameters and failure probabilities:

  • Probability of a tube failing: ( P(S_i) = 1 - 0.04 = 0.96 ). This reflects the reliability of individual tubes in maintaining functionality within a system.

• Calculation of Apparatus's Overall Failure:

  • ( P(F) = 1 - P(S) = 1 - P(S1 \cap S2 \cap S_3) = 1 - (0.96)^3 = 0.1152 ). This formula is crucial for assessing the overall reliability of an assembly made up of multiple tubes, where failure of at least one tube results in system failure.

Problem 5

Event Definitions Regarding Computer Choices:

• Let ( A1 ) be choosing a laptop first, while ( B1 ) refers to choosing a desktop first. These events are representative of a decision-making process in a purchasing scenario.

Probabilities for Each Event:

• Calculate using conditional probability:

  • ( P(B1 | A1) = \frac{1}{2}, P(A2 | B1) = \frac{1}{4}, \, \ldots ). These represent the likelihoods associated with each choice, taking into account prior decisions and available options.

Total Probability Calculation:

• Apply the Law of Total Probability:

  • ( P(B1 \cup B2) = P(B1) + P(B2) - P(B1 \cap B2) ). This provides a comprehensive assessment of the likelihood of choosing a computer type across scenarios.

Problem 6

Model and Warranty Calculation:

• Let ( A1 ) be the basic model, while ( A2 ) denotes the deluxe model. The differentiation between these models could reflect varying features that affect customer purchase behavior.

• Use Bayes' theorem for conditional probabilities:

  • Example computations show respective probabilities of warranty purchase, yielding ( P(B | A_1) = \frac{0.12}{0.3} \rightarrow 0.29 ). This emphasizes the impact of model choice on warranty purchase likelihoods.

Problem 7

System with Components 1 to 4:

• Identify pathways: the connections are parallel for components 1 and 2, while components 3 and 4 are connected serially. This illustrates the importance of component arrangement in assessing system reliability.

• Use the formula for independent events:

  • ( P(A) = P(A1 \cup A2) P(A3 \cap A4) ). Here, probabilities of independent event arrangements provide insights into overall system functionality.

• Calculate effective probabilities from series against failures:

  • ( P(A) = 1 - P(B) = 1 - 0.1^3 \rightarrow \approx 0.9981 ). This calculation reflects the strong reliability of a system with redundant components in place.