If M and N are any two events, the probability that exactly one of them occurs is
(a) P(M) + P(N) -

Q: Wait, why do we subtract $2P(M \cap N)$ in Option (a)? Shouldn't it just be one intersection?

That is a brilliant observation! When we add $P(M)$ and $P(N)$, we count the intersection region twice—once inside $M$ and once inside $N$. To isolate the 'exactly one' region, we must remove that intersection twice: once to remove the 'both' case, and once to correct the double counting. It is all about precision!

Q: Option (c) looks so complicated with all those complements. Is there a shortcut?

The shortcut is De Morgan's Law! Whenever you see complements in probability, think of the total sample space. By converting $P(M^c \cap N^c)$ into $P((M \cup N)^c)$, you simplify the expression significantly. Don't let the notation scare you; it is just a different perspective on the same truth.

Q: Why are there multiple correct answers? I thought JEE Advanced questions only had one.

In JEE Advanced, 'Multiple Correct' questions are a staple! They test your ability to see the same physical reality through different mathematical lenses. Here, options (a), (c), and (d) are all mathematically equivalent descriptions of the same event. Embrace the versatility!

Question

If $M$ and $N$ are any two events, the probability that exactly one of them occurs is

Options:

(A) $P(M) + P(N) - 2P(M \cap N)$

(B) $P(M) + P(N) - P(M \cup N)$

(C) $P(M^c) + P(N^c) - 2P(M^c \cap N^c)$

(D) $P(M \cap N^c) + P(M^c \cap N)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, warriors of JEE Advanced. Today, we are not just solving a probability problem; we are dissecting the very anatomy of events. When you look at a problem like this, do not just see symbols. See the geometry. See the logic.

Probability is not just about numbers; it is about the space where possibilities collide. Imagine you are standing in front of a whiteboard. You draw a large rectangle representing the sample space $S$. Inside, you draw two circles, $M$ and $N$. This is the classic Venn diagram.

The question asks for the probability that 'exactly one' of these events occurs. Physically, this means we want the regions where $M$ happens but $N$ does not, OR where $N$ happens but $M$ does not.

If you look at the diagram, these are the two 'crescent' shapes on the sides. The central overlapping region—the intersection $M \cap N$—is strictly forbidden territory because that represents 'both' events occurring.

### The Geometry of the Crescent

In set notation, this is elegantly written as $(M \cap N^c) \cup (M^c \cap N)$. Since these two regions are disjoint (they do not touch), the probability of their union is simply the sum of their individual probabilities:

$$P(	ext{Exactly one}) = P(M \cap N^c) + P(M^c \cap N)$$

This immediately validates **Option (d)**. We have our first victory!

### The Algebraic Transformation

A true JEE aspirant does not stop at the first answer. We must explore the landscape. Let us look at the union $P(M \cup N)$, which covers the left crescent, the right crescent, and the intersection.

If we want only the crescents, we must take the entire union and subtract the intersection. Mathematically, this gives us:

$$P(	ext{Exactly one}) = P(M \cup N) - P(M \cap N)$$

Now, recall the fundamental addition theorem: $P(M \cup N) = P(M) + P(N) - P(M \cap N)$. If we substitute this into our expression, we get:

$$P(	ext{Exactly one}) = [P(M) + P(N) - P(M \cap N)] - P(M \cap N)$$

Look closely at the terms. We have two negative $P(M \cap N)$ terms. Combining them gives us:

$$P(	ext{Exactly one}) = P(M) + P(N) - 2P(M \cap N)$$

This is **Option (a)**. The algebra flows like water, and we have successfully transformed a geometric concept into a powerful algebraic tool.

### The De Morgan's Challenge

Now, let us tackle the final boss: **Option (c)**, which is $P(M^c) + P(N^c) - 2P(M^c \cap N^c)$. When you see complements, think of De Morgan's Law. It is your best friend in probability.

We know that $M^c \cap N^c$ is the same as $(M \cup N)^c$. Furthermore, the probability of any complement is $1$ minus the probability of the event. So, let us rewrite the expression:

$$P(M^c) + P(N^c) - 2P(M^c \cap N^c) = (1 - P(M)) + (1 - P(N)) - 2(1 - P(M \cup N))$$

Let us expand this carefully. We get $1 + 1 - P(M) - P(N) - 2 + 2P(M \cup N)$. The constants $1 + 1 - 2$ cancel out perfectly to zero! We are left with:

$$2P(M \cup N) - (P(M) + P(N))$$

Now, use the addition theorem one last time: $P(M) + P(N) = P(M \cup N) + P(M \cap N)$. Substitute this back into our expression:

$$2P(M \cup N) - [P(M \cup N) + P(M \cap N)] = P(M \cup N) - P(M \cap N)$$

And there it is! We have arrived back at our original formula. **Option (c)** is also correct.

### Conclusion: The Beauty of Equivalence

We have proven that the probability of 'exactly one' event can be expressed in three distinct, beautiful ways. This is the essence of JEE Advanced.

It is not about memorizing one formula; it is about understanding the connections between them. Whether you use the disjoint sets, the union-minus-intersection, or the complement-based approach, the truth remains the same. Keep practicing, keep visualizing, and keep falling in love with the logic behind the math.

Animated Solution for Mathematics - Probability: If $M$ and $N$ are any two events, the probability that exactly one of them occurs is

Select Answer:

Visualized Solution

Visualizing the Sample Space

Defining 'Exactly One' Event

Evaluating Option (d)

Using the Union $P (M \cup N)$

Subtracting the Intersection

The Addition Theorem

Evaluating Option (a)

Evaluating Option (c)

Applying De Morgan's Law

Substituting Complements

Expanding the Expression

Final Simplification for (c)

Summary of Correct Options

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Geometry of the Crescent

The Algebraic Transformation

The De Morgan's Challenge

Conclusion

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Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If M and N are any two events, the probability that exactly one of them occurs is