CT1 Sample Questions — Batch 3: Probability Distributions & Bayes

This document provides rigorous, textbook-style solutions for the probability distribution and advanced Bayes' Theorem questions (Q10, Q11, Q12, Q13, and Q21) from the IFS UNIT - 1 & 2 sample questions sheet. All formulas are rendered in LaTeX, with step-by-step mathematical derivations and formal academic inferences.

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Question 10: Bayes' Theorem / Posterior Analysis of Electronic Devices

Given Data

Let $X, Y, Z$ represent the events that a produced device is of Type X, Type Y, or Type Z respectively. Let $D$ represent the event that a device is defective.

• Defective conditional probabilities: $P(D \mid X) = 0.01, \quad P(D \mid Y) = 0.04, \quad P(D \mid Z) = 0.02$
• Posteriors of device type given a defective is found: $P(X \mid D) = 0.5, \quad P(Y \mid D) = 0.3, \quad P(Z \mid D) = 0.2$

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Analysis & Derivations

The question asks: "If a defective device is picked up from the factory, what is the most likely type of device?"

Interpretation A: Direct Posterior Comparison (MAP Decision)

If we are given that a defective device has been picked up from the factory, we are seeking to find the maximum posterior probability: $\operatorname{argmax}_{k \in \{X,Y,Z\}} P(k \mid D)$ Since the posteriors are explicitly provided in the problem statement:

• $P(X \mid D) = 0.5$ (or $50\%$)
• $P(Y \mid D) = 0.3$ (or $30\%$)
• $P(Z \mid D) = 0.2$ (or $20\%$)

Conclusion A: Type X is the most likely type of device among the defectives, with a probability of $0.5$.

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Interpretation B: Deriving the Production Proportions (Priors)

To demonstrate advanced mastery, we can derive the overall production proportions (the priors $P(X), P(Y), P(Z)$) to see which device type the factory produces the most.

According to Bayes' Theorem: $P(k \mid D) = \frac{P(D \mid k)P(k)}{P(D)} \implies P(k) = \frac{P(k \mid D)P(D)}{P(D \mid k)}$ Let $P(D) = c$ represent the total probability of a defective device (constant). $P(X) = \frac{0.5 \cdot c}{0.01} = 50c$ $P(Y) = \frac{0.3 \cdot c}{0.04} = 7.5c$ $P(Z) = \frac{0.2 \cdot c}{0.02} = 10c$

Since the probabilities must sum to 1 ($P(X) + P(Y) + P(Z) = 1$): $50c + 7.5c + 10c = 1 \implies 67.5c = 1 \implies c = \frac{1}{67.5} = \frac{2}{135}$

Now, substitute $c$ back to find the exact priors:

• Prior of Type X: $P(X) = 50 \times \frac{2}{135} = \frac{100}{135} = \frac{20}{27} \approx 0.7407 \quad (74.07\%)$
• Prior of Type Y: $P(Y) = 7.5 \times \frac{2}{135} = \frac{15}{135} = \frac{3}{27} = \frac{1}{9} \approx 0.1111 \quad (11.11\%)$
• Prior of Type Z: $P(Z) = 10 \times \frac{2}{135} = \frac{20}{135} = \frac{4}{27} \approx 0.1481 \quad (14.81\%)$

Conclusion B: If the question refers to the overall production volume of the factory, Type X is the most likely type of device produced by the factory, accounting for approximately $74.07\%$ of total production.

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Question 11: Bayes' Theorem / Posterior Analysis of Food Products

Given Data

Let $D, E, F$ represent the events that an inspected food product is of Type D, Type E, or Type F respectively. Let $Def$ represent the event that a product is defective.

• Defective conditional probabilities: $P(Def \mid D) = 0.03, \quad P(Def \mid E) = 0.01, \quad P(Def \mid F) = 0.04$
• Posteriors of product type given a defective is found: $P(D \mid Def) = 0.2, \quad P(E \mid Def) = 0.5, \quad P(F \mid Def) = 0.3$

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Analysis & Derivations

The question asks: "If a defective product is selected, what is the most likely type of food product it belongs to?"

Interpretation A: Direct Posterior Comparison (MAP Decision)

Since we are given that a defective product has been selected, we compare the given posterior probabilities:

• $P(D \mid Def) = 0.2$ (or $20\%$)
• $P(E \mid Def) = 0.5$ (or $50\%$)
• $P(F \mid Def) = 0.3$ (or $30\%$)

Conclusion A: Type E is the most likely type of food product to have been selected among defectives, with a probability of $0.5$.

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Interpretation B: Deriving the Production Proportions (Priors)

Let us compute the total production distribution (the priors $P(D), P(E), P(F)$) across the factory.

According to Bayes' Theorem: $P(k) = \frac{P(k \mid Def)P(Def)}{P(Def \mid k)}$ Let $P(Def) = c$ represent the total probability of a defective product. $P(D) = \frac{0.2 \cdot c}{0.03} = \frac{20}{3}c = \frac{80}{12}c$ $P(E) = \frac{0.5 \cdot c}{0.01} = 50c = \frac{600}{12}c$ $P(F) = \frac{0.3 \cdot c}{0.04} = \frac{30}{4}c = \frac{90}{12}c$

Summing the probabilities to 1: $\frac{80 + 600 + 90}{12} c = 1 \implies \frac{770}{12}c = 1 \implies c = \frac{12}{770} = \frac{6}{385}$

Now, substitute $c$ back to find the priors:

• Prior of Type D: $P(D) = \frac{80}{12} \times \frac{12}{770} = \frac{80}{770} = \frac{8}{77} \approx 0.1039 \quad (10.39\%)$
• Prior of Type E: $P(E) = \frac{600}{12} \times \frac{12}{770} = \frac{600}{770} = \frac{60}{77} \approx 0.7792 \quad (77.92\%)$
• Prior of Type F: $P(F) = \frac{90}{12} \times \frac{12}{770} = \frac{90}{770} = \frac{9}{77} \approx 0.1169 \quad (11.69\%)$

Conclusion B: If the question refers to the overall production volume, Type E is the most likely type of food product produced by the facility, accounting for approximately $77.92\%$ of total production.

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Question 12: Poisson Distribution - Fire Alarm Activations

Given Data

Let $X$ denote the number of fire alarm activations in a week.

• Average rate of activation ($\lambda$) = $2$ activations/week
• Distribution model: $X \sim \operatorname{Pois}(\lambda = 2)$

We seek the probability of exactly 3 alarm activations in a week: $P(X = 3)$.

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Step-by-Step Derivation

The Probability Mass Function (PMF) of a Poisson random variable is: $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Substitute $\lambda = 2$ and $x = 3$: $P(X = 3) = \frac{e^{-2} 2^3}{3!}$ $P(X = 3) = \frac{e^{-2} \times 8}{6} = \frac{4}{3} e^{-2}$

Using the approximation $e^{-2} \approx 0.135335$: $P(X = 3) \approx \frac{4}{3} \times 0.135335 \approx 0.180447$

Inference: The probability that exactly 3 fire alarms will activate in a given week is approximately $18.04\%$.

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Question 13: Poisson Distribution - Coffee Shop Customer Arrivals

Given Data

Let $X$ denote the number of customers arriving at a coffee shop in a given hour.

• Average rate of arrival ($\lambda$) = $10$ customers/hour
• Distribution model: $X \sim \operatorname{Pois}(\lambda = 10)$

We seek the probability of exactly 15 customer arrivals in an hour: $P(X = 15)$.

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Step-by-Step Derivation

Using the Poisson PMF: $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$

Substitute $\lambda = 10$ and $x = 15$: $P(X = 15) = \frac{e^{-10} 10^{15}}{15!}$

Given:

• $10^{15} = 10^{15}$
• $15! = 1,307,674,368,000 = 1.307674368 \times 10^{12}$
• $e^{-10} \approx 0.0000453999$

Substitute these numerical values: $P(X = 15) \approx \frac{0.0000453999 \times 10^{15}}{1.307674368 \times 10^{12}} = \frac{4.53999 \times 10^{10}}{1.307674368 \times 10^{12}}$ $P(X = 15) \approx \frac{4.53999}{130.7674368} \approx 0.034718$

Inference: The probability that exactly 15 customers arrive in a given hour is approximately $3.47\%$.

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Question 21: Defective Bulbs - Binomial vs. Hypergeometric Models

Given Data

• Total production batch size ($N$) = $1000$
• Probability of a bulb being defective ($p$) = $0.01$
• Number of defective bulbs in the run ($D$) = $1000 \times 0.01 = 10$
• Number of non-defective bulbs ($N - D$) = $990$
• Sample size selected ($n$) = $2$

Let $X$ denote the number of defective bulbs in our sample of size $2$. We solve this using both standard models to ensure complete accuracy.

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Model 1: Binomial Distribution (With Replacement Approximation)

Since $N$ is large relative to $n$, we can model this as $X \sim \operatorname{Bin}(n=2, p=0.01)$.

1. Probability that both are defective ($P(X = 2)$)

$P(X = 2) = \binom{2}{2} p^2 (1-p)^0 = (0.01)^2 = 0.0001$

2. Probability that both are non-defective ($P(X = 0)$)

$P(X = 0) = \binom{2}{0} p^0 (1-p)^2 = (0.99)^2 = 0.9801$

3. Probability that one is defective and the other is non-defective ($P(X = 1)$)

$P(X = 1) = \binom{2}{1} p^1 (1-p)^1 = 2 \times 0.01 \times 0.99 = 0.0198$

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Model 2: Hypergeometric Distribution (Exact Without Replacement)

Since the bulbs are drawn without replacement from a finite batch of $1000$, the exact model is $X \sim \operatorname{Hypergeom}(N=1000, D=10, n=2)$. $P(X = x) = \frac{\binom{D}{x}\binom{N-D}{n-x}}{\binom{N}{n}}$

Total combinations of choosing 2 bulbs out of 1000: $\binom{1000}{2} = \frac{1000 \times 999}{2} = 499500$

1. Probability that both are defective ($P(X = 2)$)

$P(X = 2) = \frac{\binom{10}{2}\binom{990}{0}}{499500} = \frac{45 \times 1}{499500} = \frac{1}{11100} \approx 0.00009009$

2. Probability that both are non-defective ($P(X = 0)$)

$P(X = 0) = \frac{\binom{10}{0}\binom{990}{2}}{499500} = \frac{1 \times \frac{990 \times 989}{2}}{499500} = \frac{489555}{499500} = \frac{10879}{11100} \approx 0.980090$

3. Probability that one is defective and the other is non-defective ($P(X = 1)$)

$P(X = 1) = \frac{\binom{10}{1}\binom{990}{1}}{499500} = \frac{10 \times 990}{499500} = \frac{9900}{499500} = \frac{22}{1110} \approx 0.0198198$

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Academic Summary of Results

Event	Binomial Model (Approx)	Hypergeometric Model (Exact)
Both Defective	$0.000100$ ($0.01\%$)	$0.000090$ ($0.009\%$)
Both Non-Defective	$0.980100$ ($98.01\%$)	$0.980090$ ($98.01\%$)
One Defective, One Non-Defective	$0.019800$ ($1.98\%$)	$0.019820$ ($1.982\%$)

Exam tip: Mentioning both models in your answer shows exceptional depth. Explain that because the sample size ($n=2$) is extremely small compared to the population ($N=1000$), the Binomial model is an incredibly accurate approximation of the exact Hypergeometric process.

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Quick-Facts & Formula Sheet (Probability Distributions)

Distribution Model	Probability Mass Function (PMF)	Mean ($\mu$)	Variance ($\sigma^2$)	Characteristics / Conditions
Binomial	$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$	$np$	$np(1-p)$	$n$ trials; binary outcomes; independent; constant success probability $p$.
Poisson	$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$	$\lambda$	$\lambda$	Rare event count; continuous interval; rate $\lambda$ is constant; independence.
Hypergeometric	$P(X=x) = \frac{\binom{D}{x}\binom{N-D}{n-x}}{\binom{N}{n}}$	$n \left(\frac{D}{N}\right)$	$n\left(\frac{D}{N}\right)\left(\frac{N-D}{N}\right)\left(\frac{N-n}{N-1}\right)$	Finite population $N$; $D$ items of interest; size $n$ sampled without replacement.