Let A and B be two events such that the probability that exactly one of them occurs is 2/5 and the p

Q: Wait, why do we subtract the intersection from the union to get 'exactly one'?

Think of the union $P(A \cup B)$ as the total area covered by both circles. This area consists of three distinct parts: the part of $A$ only, the part of $B$ only, and the overlap $P(A \cap B)$. When we say 'exactly one', we mean the 'A only' and 'B only' parts. If you take the total union and remove the overlap, you are left exactly with those two parts!

Q: Is there a formula involving $P(A)$ and $P(B)$ that I should use instead?

You could use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, but you don't know $P(A)$ or $P(B)$ individually! This is a classic trap. The beauty of this problem is that you don't need the individual probabilities; you only need the relationship between the regions of the Venn diagram.

Q: What if the events were mutually exclusive? Would the answer change?

If they were mutually exclusive, $P(A \cap B)$ would be $0$. In that case, the probability of 'exactly one' would be equal to the probability of the union. Since our given values don't satisfy that ($2/5 \neq 1/2$), we know they must overlap!

Q: Why is the probability of 'exactly one' defined as $P(A \cup B) - P(A \cap B)$?

It is a direct consequence of set theory. The set of elements in the union but not in the intersection is exactly the symmetric difference of the two sets. In probability terms, this is the sum of the probabilities of the regions that do not overlap.

Q: I keep getting confused with the fractions. Is there a trick?

Always convert to a common denominator immediately! Here, the denominators are $2$ and $5$, so $10$ is your best friend. Once you have $\frac{5}{10}$ and $\frac{4}{10}$, the subtraction becomes trivial. Never fear the arithmetic; focus on the geometry!

Question

Let $A$ and $B$ be two events such that the probability that exactly one of them occurs is $2/5$ and the probability that $A$ or $B$ occurs is $1/2$, then the probability of both of them occur together is

Options:

(A) 1/10

(B) 2/9

(C) 1/8

(D) 1/12

Sigma Percentile · Accepted Answer

### The Geometry of Chance: Unlocking the Venn Diagram

Welcome, future engineer! Today, we are going to peel back the layers of a classic probability problem.

Often, when students see probability, they immediately reach for complex formulas like the Inclusion-Exclusion Principle. While those are powerful, the true secret to mastering JEE-level probability lies in your ability to visualize the sample space.

Let us embark on this journey together.

### Phase 1: Visualizing the Landscape

Imagine you have two events, $A$ and $B$, living inside a sample space. We represent these as two circles in a Venn diagram.

The entire area covered by both circles is the union, denoted as $P(A \cup B)$. We are told that this total area is $\frac{1}{2}$.

Now, consider the 'Exactly One' region. This is the part of the diagram where you are inside $A$ but not $B$, or inside $B$ but not $A$.

Visually, this is the entire union minus that central, shared 'almond' shape where the circles overlap. That overlap is our intersection, $P(A \cap B)$.

### Phase 2: The Logic of Subtraction

We are given that the probability of exactly one event occurring is $\frac{2}{5}$. Think about the relationship between these regions.

If you take the union $P(A \cup B)$ and you subtract the intersection $P(A \cap B)$, what are you left with? You are left with exactly the parts of $A$ and $B$ that do not overlap.

This is the core realization:

$$P(	ext{Exactly One}) = P(A \cup B) - P(A \cap B)$$

This equation is not just a formula; it is a geometric truth. It tells us that the 'Exactly One' region is simply the union stripped of its shared core.

### Phase 3: The Algebraic Execution

Now that we have our conceptual map, the math becomes a simple matter of substitution. We know $P(	ext{Exactly One}) = \frac{2}{5}$ and $P(A \cup B) = \frac{1}{2}$.

Plugging these into our equation, we get:

$$\frac{2}{5} = \frac{1}{2} - P(A \cap B)$$

To find the intersection, we rearrange the terms. We move $P(A \cap B)$ to the left and $\frac{2}{5}$ to the right:

$$P(A \cap B) = \frac{1}{2} - \frac{2}{5}$$

### Phase 4: The Final Calculation

Here is where many students rush and make silly errors. Do not subtract blindly! We need a common denominator.

The least common multiple of $2$ and $5$ is $10$. We convert our fractions:

$$\frac{1}{2} = \frac{5}{10}$$

$$\frac{2}{5} = \frac{4}{10}$$

Now, the subtraction is elegant and clear:

$$P(A \cap B) = \frac{5}{10} - \frac{4}{10} = \frac{1}{10}$$

### Conclusion: The Beauty of Simplicity

And there we have it! The probability that both events occur together is **$\frac{1}{10}$**.

Notice how we didn't need to know the individual probabilities of $A$ or $B$. By focusing on the relationship between the regions of the Venn diagram, we bypassed the need for extra variables.

This is the hallmark of a great problem-solver: finding the most direct path through the logic. Keep this visualization in your toolkit, and no probability problem will ever intimidate you again.

Animated Solution for Mathematics - Probability: Let $A$ and $B$ be two events such that the probability that exactly one of them occurs is $2/5$ and the probability that $A$ or $B$ occurs is $1/2$ , then the probability of both of them occur together is

Select Answer:

Visualized Solution

Visualizing Events $A$ and $B$

Probability of $A$ or $B$

Probability of Exactly One Event

Relating the Regions

The Core Equation

Substituting the Values

Isolating $P (A \cap B)$

Finding the Common Denominator

Final Subtraction

Conclusion

The Sigma Insight: Addition and Multiplication Theorems

The Geometry of Chance

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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