In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the

Q: Wait, why do we divide by 400 in the probability formula if it just cancels out anyway?

That is a brilliant observation! You are absolutely right. In the expression $P(E_1|A) = \frac{(n(E_1)/N) \times P(A|E_1)}{\sum (n(E_i)/N) \times P(A|E_i)}$, the total population $N$ appears in every term of the numerator and the denominator. Mathematically, it cancels out perfectly. We include it initially to stay rigorous with the definition of probability, but in the exam hall, you can work directly with the counts $n(E_i)$ to save precious seconds. It is the same logic, just scaled up!

Q: How do I know when to use Bayes' Theorem versus simple conditional probability?

Think of it as a directionality problem. Simple conditional probability $P(A|B)$ asks: 'Given the cause, what is the effect?' Bayes' Theorem $P(B|A)$ asks: 'Given the effect, what was the cause?' If the problem tells you the outcome (the person is sick) and asks you to trace it back to a specific group, you are almost certainly looking at a Bayes' Theorem scenario. It is the bridge between observation and inference.

Q: Is it possible to solve this using a tree diagram instead of the formula?

Yes, and I highly encourage it! A tree diagram is just a visual representation of the law of total probability. Draw three branches from a starting point representing the three groups. From each branch, draw two sub-branches: 'Sick' and 'Healthy'. By filling in the probabilities on these branches, you can see the 'path' to the sick condition clearly. The probability you want is just the 'Sick' path for $E_1$ divided by the sum of all 'Sick' paths. It makes the math feel much more intuitive.

Question

In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35\%$, $20\%$ and $10\%$ respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is:

Options:

(A) $\frac{7}{45}$

(B) $\frac{8}{45}$

(C) $\frac{14}{45}$

(D) $\frac{28}{45}$

Sigma Percentile · Accepted Answer

### The Detective's Dilemma: Unraveling the Mystery of Probability

Imagine you are a detective in a bustling city of $400$ people. You have been tasked with a medical investigation regarding a specific chest disorder plaguing the population.

You have been handed a single, crucial piece of evidence: a person has been selected at random, and they are suffering from this disorder. Your mission is to trace this person back to their lifestyle group using the tools of probability.

### Phase 1: Partitioning the World

Before we can solve the mystery, we must understand the landscape. We have a total population of $N = 400$, divided into three distinct, mutually exclusive groups defined as events $E_1, E_2,$ and $E_3$.

$E_1$ represents the smokers who are non-vegetarians, with $n(E_1) = 160$.
$E_2$ represents the smokers who are vegetarians, with $n(E_2) = 100$.
$E_3$ represents the non-smokers who are vegetarians, with $n(E_3) = 140$.

These groups partition the entire population, as $160 + 100 + 140 = 400$. The probability of a randomly selected person belonging to any of these groups is:
$$P(E_1) = \frac{160}{400}, \quad P(E_2) = \frac{100}{400}, \quad P(E_3) = \frac{140}{400}$$

### Phase 2: The Conditional Landscape

We introduce the 'effect'—the chest disorder, denoted as event $A$. The disorder has a preference, represented by the following conditional probabilities:

$$P(A|E_1) = 35\% = \frac{35}{100}$$
$$P(A|E_2) = 20\% = \frac{20}{100}$$
$$P(A|E_3) = 10\% = \frac{10}{100}$$

These values are the 'likelihoods'. They bridge the gap between the lifestyle groups and the incidence of the disease.

### Phase 3: The Bayes' Insight

We know the person is sick (Event $A$ has occurred) and we want to find the probability that they belong to group $E_1$, denoted as $P(E_1|A)$. This is a classic inverse probability problem solved via Bayes' Theorem:

$$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)}$$

The total probability of being sick, $P(A)$, is the sum of all ways one could become sick:
$$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)$$

### Phase 4: The Beauty of Cancellation

Assembling our equation with the provided values, we obtain:

$$P(E_1|A) = \frac{\frac{160}{400} 	imes \frac{35}{100}}{\frac{160}{400} 	imes \frac{35}{100} + \frac{100}{400} 	imes \frac{20}{100} + \frac{140}{400} 	imes \frac{10}{100}}$$

By multiplying the numerator and denominator by $400 	imes 100$, the fractions vanish, leaving a cleaner expression based on raw counts:

$$P(E_1|A) = \frac{160 	imes 35}{(160 	imes 35) + (100 	imes 20) + (140 	imes 10)}$$

Calculating the components:
Numerator: $160 	imes 35 = 5600$.
Denominator: $5600 + 2000 + 1400 = 9000$.

### The Final Resolution

Our probability is $\frac{5600}{9000}$. Simplifying by canceling the zeros, we get $\frac{56}{90}$.

Dividing both the numerator and the denominator by $2$, we arrive at our final answer:

**$$P(E_1|A) = \frac{28}{45}$$**

We have successfully navigated through the conditional probabilities to reverse the logic. The probability of our sick person being a smoker and non-vegetarian is **$\frac{28}{45}$**.

Select Answer:

Visualized Solution

Defining the Sample Space

Identifying Events $E_{1}, E_{2}, E_{3}$

Calculating Base Probabilities $P (E_{i})$

The Condition: Chest Disorder $A$

Conditional Probabilities $P (A ∣ E_{i})$

The Goal: Bayes' Theorem

Setting up the Calculation

Simplifying the Expression

Calculating Numerator and Denominator

Final Summation

The Final Answer

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

The Final Resolution

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