In a certain city only two newspapers A and B are published, it is known that 25% of the city popula

Q: Why can't I just calculate $P(E) = 0.30 \times 0.25 + 0.40 \times 0.20 + 0.50 \times 0.08$?

This is a classic trap! If you do that, you are double-counting the people who read both newspapers. The groups must be mutually exclusive (disjoint) before you apply the conditional probabilities. You must subtract the intersection from the individual probabilities first to isolate the 'Only A' and 'Only B' groups.

Q: What if the problem didn't give me the intersection? Could I still solve it?

If the intersection wasn't given, you would need more information to define the disjoint sets. The beauty of this problem is that it provides the exact pieces needed to partition the sample space. Always look for the 'overlap' first in any set-based probability problem.

Q: Is there a way to visualize this without a Venn diagram?

While you could use algebraic set notation, the Venn diagram is the most intuitive way to 'see' the partitioning. It prevents the common mistake of double-counting. I highly recommend drawing it every time you face a set-based probability problem.

Question

In a certain city only two newspapers $A$ and $B$ are published, it is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$ while $8\%$ reads both $A$ and $B$. It is also known that $30\%$ of those who read $A$ but not $B$ look into advertisements and $40\%$ of those who read $B$ but not $A$ look into advertisements while $50\%$ of those who read both $A$ and $B$ look into advertisements. What is the percentage of the population that reads an advertisement?

Sigma Percentile · Accepted Answer

### The Art of Partitioning: A Journey Through Probability

Imagine you are standing in the center of a bustling city, observing the reading habits of its citizens. You have two newspapers, $A$ and $B$, and you want to know how many people are reading advertisements.

This isn't just a math problem; it's a study of human behavior, partitioned into neat, logical segments. To solve this, we must first master the art of partitioning.

We cannot simply look at the total percentage of readers for $A$ and $B$ and jump to conclusions. Because some people are overachievers—they read both—we must account for this overlap to avoid double-counting, which would skew our final answer.

### Visualizing the City: The Venn Diagram

Let's draw a map of our city using a Venn diagram. We have two circles, $A$ and $B$, overlapping in the middle.

The problem provides the raw data: $P(A) = 0.25$, $P(B) = 0.20$, and the intersection, $P(A \cap B) = 0.08$. This intersection represents the 'Both' group.

Now, we isolate the segments. The people who read only $A$ are those in circle $A$ but outside the intersection:
$$P(A \cap B^c) = P(A) - P(A \cap B) = 0.25 - 0.08 = 0.17$$

Similarly, for those who read only $B$:
$$P(A^c \cap B) = P(B) - P(A \cap B) = 0.20 - 0.08 = 0.12$$

We now have three distinct, mutually exclusive groups: the 'Only A' readers ($0.17$), the 'Only B' readers ($0.12$), and the 'Both' readers ($0.08$). These three groups perfectly partition our reading population.

### The Total Probability Theorem: The Engine of Our Solution

Now that we have our segments, we introduce the event $E$: reading an advertisement. We are given the conditional probabilities: $P(E | A \cap B^c) = 0.30$, $P(E | A^c \cap B) = 0.40$, and $P(E | A \cap B) = 0.50$.

To find the total probability $P(E)$, we use the **Total Probability Theorem**. Think of this as a weighted average where the total percentage of ad-readers is the sum of the contributions from each group.

The master equation is:
$$P(E) = P(E|A \cap B^c)P(A \cap B^c) + P(E|A^c \cap B)P(A^c \cap B) + P(E|A \cap B)P(A \cap B)$$

This equation acts as the engine of our solution. It takes the probability of being in a specific group and multiplies it by the likelihood of that group reading an ad.

### The Final Calculation: Precision and Patience

Let's plug in our numerical values:
$$P(E) = (0.30)(0.17) + (0.40)(0.12) + (0.50)(0.08)$$

We calculate each term with care:
$0.30 	imes 0.17 = 0.051$
$0.40 	imes 0.12 = 0.048$
$0.50 	imes 0.08 = 0.040$

Adding these together:
$$0.051 + 0.048 + 0.040 = 0.139$$

To convert this to a percentage, we multiply by $100$, giving us a final result of **$13.9\%$**.

### Conclusion: The Elegance of Systematic Thinking

We have successfully navigated the city of readers. By breaking the problem down into mutually exclusive segments, we avoided the common traps that catch many students.

We didn't just calculate; we visualized, we partitioned, and we applied the theorem with precision. Remember, in JEE Advanced, the complexity isn't in the arithmetic—it's in the setup.

Once you have the right structure, the math flows naturally. Keep this systematic approach in your toolkit, and you will conquer any probability problem that comes your way.

Enter Numerical Value:

Visualized Solution

Visualizing the Population Segments

Finding 'Only $A$ ' Readers

Finding 'Only $B$ ' Readers

Highlighting 'Both' Readers

Defining Advertisement Probabilities

Applying Total Probability Theorem

Substituting the Values

Calculating the Terms

Final Addition

Final Conclusion

The Sigma Insight: Total Probability Theorem

The Art of Partitioning

Visualizing the City

The Total Probability Theorem

The Final Calculation

Conclusion

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