Two newspapers A and B are published in a city. It is known that 25% of the city populations reads A

Q: Wait, why do we subtract the intersection? Isn't $P(A)$ just $25\%$?

Great question! $P(A)$ is indeed $25\%$, but that $25\%$ includes the people who read *both* newspapers. If we just used $25\%$ and $20\%$ directly, we would be counting the people who read both newspapers twice. By subtracting the intersection, we isolate the 'pure' $A$ readers, ensuring every person is counted exactly once.

Q: Can we use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$?

That formula is excellent for finding the total number of people who read *at least one* newspaper, but it doesn't help us here because the advertisement rates are different for each subgroup. We need to know the specific percentage of people in each disjoint region to apply the correct conditional probability.

Q: Why are we multiplying the percentages?

We are using the definition of conditional probability. We know $P(Ad \mid A \text{ only}) = 30\%$. By definition, $P(Ad \cap (A \text{ only})) = P(Ad \mid A \text{ only}) \times P(A \text{ only})$. We are essentially finding the fraction of the total population that falls into both categories.

Question

Two newspapers A and B are published in a city. It is known that 25% of the city populations reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is :-

Options:

(A) 12.8

(B) 13.5

(C) 13.9

(D) 13

Sigma Percentile · Accepted Answer

### The Architecture of Probability: Solving the Newspaper Puzzle

Welcome, future engineer. Today, we are not just solving a probability problem; we are learning the art of partitioning reality.

When you look at a problem involving overlapping sets, your first instinct should always be to visualize the architecture. Imagine a bustling city with two newspapers, $A$ and $B$. Some people read $A$, some read $B$, and a select few are well-informed enough to read both.

The trap that catches most students is treating these groups as simple, monolithic blocks. In set theory, as in life, the devil is in the details—specifically, in the overlaps.

### Phase 1: The Art of Partitioning

We are given $P(A) = 25\%$, $P(B) = 20\%$, and the intersection $P(A \cap B) = 8\%$. If you simply add $25\%$ and $20\%$, you are committing a cardinal sin of statistics: double counting.

The $8\%$ of people who read both are hidden inside the $25\%$ of $A$ readers AND inside the $20\%$ of $B$ readers. To solve this, we must create three mutually exclusive, disjoint regions.

First, we isolate the 'A only' readers. This is the set of people who read $A$ but not $B$:
$$P(A \setminus B) = P(A) - P(A \cap B) = 25\% - 8\% = 17\%$$

Next, we do the same for the 'B only' readers:
$$P(B \setminus A) = P(B) - P(A \cap B) = 20\% - 8\% = 12\%$$

Finally, we have the intersection, the $8\%$ who read both. We have partitioned the entire readership into three distinct, non-overlapping islands: $17\%$ (A only), $12\%$ (B only), and $8\%$ (Both). The sum of these is $37\%$, which represents the total percentage of the population that reads at least one newspaper.

### Phase 2: The Conditional Reality

Now that we have our disjoint regions, the problem becomes a simple weighted sum. We are told that the advertisement viewing rates differ based on reading habits: $30\%$ for 'A only', $40\%$ for 'B only', and $50\%$ for those who read both.

This is a classic application of the **Law of Total Probability**. We calculate the probability of the event $Ad$ (looking at advertisements) by summing the contributions from each partition:

$$P(Ad) = P(Ad \mid A 	ext{ only})P(A 	ext{ only}) + P(Ad \mid B 	ext{ only})P(B 	ext{ only}) + P(Ad \mid A \cap B)P(A \cap B)$$

Let's calculate the contribution from each group:

1. **From 'A only':** $0.30 	imes 17\% = 5.1\%$
2. **From 'B only':** $0.40 	imes 12\% = 4.8\%$
3. **From 'Both':** $0.50 	imes 8\% = 4.0\%$

### The Final Synthesis

When we sum these contributions, we aggregate the total percentage of the city's population that looks at advertisements:
$$5.1\% + 4.8\% + 4.0\% = 13.9\%$$

There it is. The complexity of the problem evaporates once you realize that the entire challenge was simply about correctly identifying the disjoint sets.

Never rush into calculations. Always pause, draw your Venn diagram, partition your sets, and then—and only then—apply the math. You have mastered the logic of the intersection. The final answer is **$13.9\%$**.

Select Answer:

Visualized Solution

Visualizing the Population

Mapping the Intersection

Setting up 'A Only' Population

Calculating 'A Only' Population

Setting up 'B Only' Population

Calculating 'B Only' Population

Advertisement Probabilities

Ads from 'A Only' Readers

Ads from 'B Only' Readers

Ads from 'Both' Readers

Total Percentage Calculation

The Sigma Insight: Total Probability Theorem

The Architecture of Probability

Phase 1

Phase 2

The Final Synthesis

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