If E̅ and F̅ are the complementary events of events E and F respectively and if 0

Q: Wait, why can't we just add $P(E|F) + P(E|\bar{F})$? Shouldn't that cover the whole space?

This is a classic trap! You are trying to add probabilities from two completely different 'universes.' When you calculate $P(E|F)$, you have zoomed into the world of $F$. When you calculate $P(E|\bar{F})$, you have zoomed into the world of $\bar{F}$. These are disjoint sets. Adding them is like trying to add the probability of rain in Mumbai to the probability of rain in London—they are independent contexts. The complement rule $P(A) + P(\bar{A}) = 1$ only works when you are operating within the *same* sample space.

Q: Why does the problem explicitly state $0 < P(F) < 1$?

That is a brilliant observation. In the definition of conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$, we are dividing by $P(B)$. If $P(B)$ were zero, the expression would be undefined. By stating $0 < P(F) < 1$, the problem guarantees that both $P(F)$ and $P(\bar{F})$ are non-zero, ensuring that all conditional probabilities mentioned in the options are mathematically valid and well-defined.

Q: Is $P(\bar{E}|F)$ the same as $1 - P(E|F)$?

Absolutely! This is the 'Complement Rule' applied to conditional probability. Since $E$ and $\bar{E}$ are complementary events, they partition the entire sample space. When you restrict your view to the subset $F$, they still partition that subset. Therefore, the sum of their conditional probabilities must be $1$. So, $P(\bar{E}|F) = 1 - P(E|F)$ is a perfectly valid and powerful identity.

Question

If $\bar{E}$ and $\bar{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P(F) < 1$, then

Options:

(A) $P(E/F) + P(\bar{E}/F) = 1$

(B) $P(E/F) + P(E/\bar{F}) = 1$

(C) $P(\bar{E}/F) + P(E/\bar{F}) = 1$

(D) $P(E/\bar{F}) + P(\bar{E}/\bar{F}) = 1$

Sigma Percentile · Accepted Answer

### The Geometry of Uncertainty: Mastering Conditional Probability

Welcome, future engineers. Today, we are going to peel back the layers of a problem that seems simple on the surface but tests the very foundation of how you perceive probability.

In the JEE Advanced arena, the difference between a top ranker and a struggler often lies in the ability to visualize the 'Sample Space.' Let us embark on this journey to understand why conditional probability is not just a formula, but a change of perspective.

### Phase 1: The Art of the Zoom

Imagine you are looking at a map of the entire universe, which we call the sample space $S$. Inside this map, there are two events, $E$ and $F$.

Now, suppose I tell you, 'Event $F$ has occurred.' What happens to your map? You don't care about the rest of the universe anymore; you zoom in. Your new universe is no longer $S$; it is now $F$.

This is the essence of conditional probability. When we write $P(E|F)$, we are asking: 'Given that we are strictly inside the region $F$, what is the likelihood that we are also inside $E$?'

Mathematically, we define this as:

$$P(E|F) = \frac{P(E \cap F)}{P(F)}$$

This formula is a ratio of areas. It tells us how much of the 'F-world' is occupied by $E$.

The problem gives us the constraint $0 < P(F) < 1$. If $P(F)$ were zero, we would be dividing by zero—a mathematical impossibility. If $P(F)$ were one, $F$ would be the entire sample space, and the condition would be trivial.

### Phase 2: The Complementary Law in a Restricted World

Now, let us consider the complement of $E$, denoted as $\bar{E}$. In the original sample space $S$, we know that $P(E) + P(\bar{E}) = 1$. This is the law of total probability; you are either in $E$ or you are not.

But what happens when we restrict our vision to $F$? If you are standing inside the region $F$, you are either in the part of $F$ that overlaps with $E$ (which is $E \cap F$), or you are in the part of $F$ that overlaps with $\bar{E}$ (which is $\bar{E} \cap F$).

Because these two regions, $(E \cap F)$ and $(\bar{E} \cap F)$, partition the entire set $F$, their probabilities must sum to the probability of $F$. If we divide this entire equation by $P(F)$, we get:

$$\frac{P(E \cap F)}{P(F)} + \frac{P(\bar{E} \cap F)}{P(F)} = \frac{P(F)}{P(F)} = 1$$

This simplifies beautifully to:

$$P(E|F) + P(\bar{E}|F) = 1$$

This is the golden rule of conditional complements: **The conditioning event must remain identical.** As long as you are conditioning on $F$, the sum of the probabilities of complementary events $E$ and $\bar{E}$ will always be $1$.

### Phase 3: Navigating the Traps

Now, let us look at the options provided in the problem. We have identified that $P(E|F) + P(\bar{E}|F) = 1$ is a truth. Similarly, if we replace $F$ with $\bar{F}$, the logic holds perfectly.

Thus, $P(E|\bar{F}) + P(\bar{E}|\bar{F}) = 1$ is also true.

But what about Options B and C? Option B suggests $P(E|F) + P(E|\bar{F}) = 1$.

Let us test this. Imagine $E$ and $F$ are independent. Then $P(E|F) = P(E)$ and $P(E|\bar{F}) = P(E)$. The sum would be $2P(E)$.

Unless $P(E) = 0.5$, this sum is not $1$. This proves that mixing conditioning events is a dangerous path. You cannot add probabilities from different universes and expect them to sum to unity; it is a category error in probability theory.

### Conclusion: The Elegance of Consistency

In your JEE Advanced journey, you will encounter many problems that test your ability to distinguish between valid identities and common misconceptions. The lesson here is simple yet profound: **Consistency is key.**

When you see conditional probabilities, always check the conditioning event (the part after the vertical bar). If the conditioning events are the same, you can apply the complement rule. If they are different, you are dealing with two separate scenarios, and you cannot simply add them.

By understanding the geometry of the sample space—by seeing how we 'zoom' into $F$ or $\bar{F}$—you move beyond rote memorization. You start to see the math as a living, breathing structure. Keep this intuition sharp, stay curious, and remember: every complex problem is just a collection of simple, elegant truths waiting to be connected.

Animated Solution for Mathematics - Probability: If $\overset{ˉ}{E}$ and $\overset{ˉ}{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P (F) < 1$ , then

Select Answer:

Visualized Solution

Visualizing the Sample Space $S$

What are Complementary Events?

The Concept of Conditional Probability

Probability of Complements under Conditioning

Testing Option A: Conditioning on $F$

Verification of Option A

Testing Option D: Conditioning on $\overset{ˉ}{F}$

Verification of Option D

Why Option B is Incorrect

Why Option C is Incorrect

Summary of Correct Options

The Sigma Insight: Conditional Probability

The Geometry of Uncertainty

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

Let $E^{C}$ denote the complement of an event $E$ . Let $E_{1}, E_{2}$ and $E_{3}$ be any pairwise independent events with $P (E_{1}) > 0$ and $P (E_{1} \cap E_{2} \cap E_{3}) = 0$ . Then $P (E_{2}^{C} \cap E_{3}^{C} / E_{1})$ is equal to :

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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 Eˉ and Fˉ are the complementary events of events E and F respectively and if 0<P(F)<1, then