Set A - Question 2

Question Details

• A. Explain Marginal and conditional probability with suitable examples. (5 Marks)
• B. A manufacturing company operates three production lines: Line A, Line B, and Line C. The proportion of total products manufactured by the three lines is 40%, 35%, and 25%, respectively. The probabilities that a product manufactured by each line is defective are 2%, 3%, and 5%, respectively. A product is randomly selected from the company's production and is found to be defective. Using Bayes' theorem, calculate the probability that the defective product was manufactured by Line A, Line B, and Line C. Based on your calculations, identify which production line is most likely to have produced the defective product. (10 Marks)
• C. A data center monitoring system records an average of 6 server alerts per hour. Find which type of distribution it will follow and also calculate the probability that the system will generate exactly 4 alerts in a particular hour. (5 Marks)

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Model Answer

A. Explain Marginal and Conditional Probability with Suitable Examples (5 Marks)

1. Marginal Probability (2 Marks)

• Definition: Marginal probability is the unconditional probability of an individual event occurring, without regard to the outcomes of any other events. In joint probability distributions, it is obtained by summing (marginalizing) the joint probabilities across all possible states of the second variable: $P(A) = \sum_{j=1}^m P(A \cap B_j)$
• Real-World Example: In a standard deck of 52 playing cards, the marginal probability of drawing an Ace across all four suits is: $P(\text{Ace}) = \frac{4}{52} = \frac{1}{13} \approx \mathbf{0.0769}$

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2. Conditional Probability (2 Marks)

• Definition: Conditional probability measures the probability of an event $A$ occurring given that another event $B$ has already occurred. This operation reduces the effective sample space from the universal set $\Omega$ to the subset $B$: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad \text{provided } P(B) > 0$
• Real-World Example: Suppose a card is drawn and announced to be a Red Card ($B$, containing 26 cards). The conditional probability that the card is an Ace given that it is red is: $P(\text{Ace} \mid \text{Red}) = \frac{P(\text{Ace} \cap \text{Red})}{P(\text{Red})} = \frac{2/52}{26/52} = \frac{2}{26} = \frac{1}{13} \approx \mathbf{0.0769}$ Conversely, the probability that the card is a Heart given it is Red is: $P(\text{Heart} \mid \text{Red}) = \frac{13/52}{26/52} = \frac{13}{26} = \mathbf{0.50}$

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3. Contingency Table Representation (1 Mark)

Consider a diagnostic dataset of 100 patient blood tests:

Test Result	Has Condition ($C$)	No Condition ($C^c$)	Marginal Probability
Positive ($T^+$)	$28$	$7$	$P(T^+) = 35/100 = \mathbf{0.35}$
Negative ($T^-$)	$2$	$63$	$P(T^-) = 65/100 = \mathbf{0.65}$
Marginal Probability	$P(C) = 30/100 = \mathbf{0.30}$	$P(C^c) = 70/100 = \mathbf{0.70}$	$\text{Total} = 1.00$

• Marginal Probability of Positive Test: $P(T^+) = 0.35$
• Conditional Probability of Having Condition given Positive Test: $P(C \mid T^+) = \frac{P(C \cap T^+)}{P(T^+)} = \frac{28/100}{35/100} = \frac{28}{35} = \mathbf{0.80}$

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B. Bayes' Theorem Defect Attribution Across Production Lines (10 Marks)

1. Prior Probabilities and Conditional Likelihoods (2 Marks)

Let $A$, $B$, and $C$ denote the events that a product was manufactured by Line A, Line B, and Line C, respectively:

• $P(A) = 40\% = 0.40$
• $P(B) = 35\% = 0.35$
• $P(C) = 25\% = 0.25$

Check for mutually exclusive and exhaustive partition of sample space: $P(A) + P(B) + P(C) = 0.40 + 0.35 + 0.25 = 1.00$

Let $D$ denote the event that a selected product is defective:

• $P(D \mid A) = 2\% = 0.02$
• $P(D \mid B) = 3\% = 0.03$
• $P(D \mid C) = 5\% = 0.05$

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2. Law of Total Probability for Defective Item (3 Marks)

The overall marginal probability of selecting a defective item from production is: $P(D) = P(A) \cdot P(D \mid A) + P(B) \cdot P(D \mid B) + P(C) \cdot P(D \mid C)$ $P(D) = (0.40 \times 0.02) + (0.35 \times 0.03) + (0.25 \times 0.05)$ $P(D) = 0.0080 + 0.0105 + 0.0125 = \mathbf{0.0310} \quad (3.10\%)$

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3. Posterior Probability Calculations via Bayes' Theorem (4 Marks)

Bayes' Theorem updates the prior probability given the observed evidence of a defect: $P(E_i \mid D) = \frac{P(E_i) \cdot P(D \mid E_i)}{P(D)}$

1. Probability that defective item originated from Line A:

$P(A \mid D) = \frac{P(A) \cdot P(D \mid A)}{P(D)} = \frac{0.0080}{0.0310} = \frac{80}{310} = \frac{8}{31} \approx \mathbf{0.2581} \quad (25.81\%)$

2. Probability that defective item originated from Line B:

$P(B \mid D) = \frac{P(B) \cdot P(D \mid B)}{P(D)} = \frac{0.0105}{0.0310} = \frac{105}{310} = \frac{21}{62} \approx \mathbf{0.3387} \quad (33.87\%)$

3. Probability that defective item originated from Line C:

$P(C \mid D) = \frac{P(C) \cdot P(D \mid C)}{P(D)} = \frac{0.0125}{0.0310} = \frac{125}{310} = \frac{25}{62} \approx \mathbf{0.4032} \quad (40.32\%)$

Verification Check:

$\sum_{i} P(E_i \mid D) = \frac{16}{62} + \frac{21}{62} + \frac{25}{62} = \frac{62}{62} = 1.0000 \quad (100.00\%)$

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4. Identification of Most Likely Production Line (1 Mark)

Comparing the posterior probabilities: $P(C \mid D) = 40.32\% > P(B \mid D) = 33.87\% > P(A \mid D) = 25.81\%$

• Conclusion: Line C is most likely to have produced the defective product.
• Managerial Insight: While Line A manufactures the largest product volume ($40\%$), its strict quality control ($2\%$ defect rate) keeps its share of defects low. In contrast, Line C produces only $25\%$ of total units but contributes $40.32\%$ of all defective inventory due to its $5\%$ defect rate.

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C. Poisson Distribution: Server Alert Modeling (5 Marks)

1. Identification and Justification of Distribution (2 Marks)

• Identified Distribution: Poisson Distribution, parameterized by arrival rate $\lambda = 6$ alerts/hour: $X \sim \text{Poisson}(\lambda = 6)$
• Justification:
  1. Discrete Count Process: Alerts are non-negative integer events ($k \in \{0, 1, 2, \dots\}$).
  2. Continuous Fixed Interval: The events occur over a fixed observation window (one hour).
  3. Independence: The occurrence of an alert in one interval is statistically independent of occurrences in disjoint intervals.
  4. Stationary Arrival Rate: The underlying mean rate of $\lambda = 6$ alerts per hour remains constant throughout the observation period.

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2. Calculation of Probability of Exactly 4 Alerts (3 Marks)

The Poisson probability mass function (PMF) is: $P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}$

Substituting $\lambda = 6$ and $k = 4$: $P(X = 4) = \frac{e^{-6} \cdot 6^4}{4!}$

Evaluating the numerical terms:

• $6^4 = 1296$
• $4! = 4 \times 3 \times 2 \times 1 = 24$
• $\frac{1296}{24} = 54$

Using $e \approx 2.7182818 \implies e^{-6} \approx 0.00247875$: $P(X = 4) = 54 \times 0.00247875 = 0.1338526 \approx \mathbf{0.1339}$

• Final Result: The probability that the monitoring system generates exactly 4 alerts in a specific hour is $0.1339$ (or $13.39\%$).