Let E^C denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events wi

Q: Wait, why can't we just assume $P(E_2^C \cap E_3^C | E_1) = P(E_2^C)P(E_3^C)$? Isn't that what independence means?

That is a common trap! Pairwise independence means $P(E_i \cap E_j) = P(E_i)P(E_j)$, but it does not automatically mean that the complements are independent given $E_1$. You must derive the expression using the definition of conditional probability and set theory to be certain.

Q: How do we know $P(E_1 \cap E_2 \cap E_3) = 0$ is useful? It seems like a random piece of information.

In probability problems, every piece of information is a key to a door. This condition is crucial because it simplifies the Inclusion-Exclusion Principle. Without it, the term $P(E_1 \cap E_2 \cap E_3)$ would remain in your equation, making it impossible to simplify to the final answer.

Q: Is there a simpler way to write $P(E_1 \cap (E_2 \cup E_3))$?

Yes! You can use the Distributive Law of sets: $E_1 \cap (E_2 \cup E_3) = (E_1 \cap E_2) \cup (E_1 \cap E_3)$. This allows you to break the complex intersection into two simpler parts, which you can then evaluate using the pairwise independence property.

Question

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 :

Options:

(A) P(E_3^C) - P(E_2^C)

(B) P(E_2^C) + P(E_3)

(C) P(E_3) - P(E_2^C)

(D) P(E_3^C) - P(E_2)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the road to JEE mastery! Today, we are going to dissect a problem that, at first glance, might seem like a tangled web of set theory and probability.

We have three events: $E_1$, $E_2$, and $E_3$. We are told they are pairwise independent, and that their triple intersection is a ghost—it has a probability of zero, i.e., $P(E_1 \cap E_2 \cap E_3) = 0$.

Our mission is to find the conditional probability $P(E_2^C \cap E_3^C | E_1)$.

### The Conditional Lens

First, let us ground ourselves in the definition of conditional probability. When we ask for the probability of an event given another, we are essentially zooming in on a specific part of our sample space.

The formula is our compass:

$$P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1 \cap E_2^C \cap E_3^C)}{P(E_1)}$$

We are looking for the probability of $E_1$ occurring alongside the non-occurrence of both $E_2$ and $E_3$, all relative to the occurrence of $E_1$.

### The De Morgan Transformation

Now, look at that numerator: $P(E_1 \cap E_2^C \cap E_3^C)$. This is where the elegance of De Morgan's Law comes to our rescue.

We know that $E_2^C \cap E_3^C$ is equivalent to $(E_2 \cup E_3)^C$. So, our numerator transforms into $P(E_1 \cap (E_2 \cup E_3)^C)$.

Mathematically, we can express this as:

$$P(E_1 \cap (E_2 \cup E_3)^C) = P(E_1) - P(E_1 \cap (E_2 \cup E_3))$$

It is like taking the entire area of $E_1$ and subtracting the part that overlaps with either $E_2$ or $E_3$.

### The Inclusion-Exclusion Dance

We need to evaluate $P(E_1 \cap (E_2 \cup E_3))$. Using the Distributive Law, this becomes $P((E_1 \cap E_2) \cup (E_1 \cap E_3))$.

Now, we apply the Inclusion-Exclusion Principle:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Applying this to our expression, we get:

$$P(E_1 \cap (E_2 \cup E_3)) = P(E_1 \cap E_2) + P(E_1 \cap E_3) - P(E_1 \cap E_2 \cap E_3)$$

### The Magic of Independence

Remember that crucial detail from the problem statement? $P(E_1 \cap E_2 \cap E_3) = 0$. This is our golden ticket!

The last term in our expression vanishes into thin air. We are left with $P(E_1 \cap E_2) + P(E_1 \cap E_3)$.

Because the events are pairwise independent, we can break these intersections down into simple products:

$$P(E_1 \cap E_2) = P(E_1)P(E_2)$$
$$P(E_1 \cap E_3) = P(E_1)P(E_3)$$

### The Final Synthesis

Let us put it all back together. Our numerator becomes:

$$P(E_1) - [P(E_1)P(E_2) + P(E_1)P(E_3)]$$

When we divide this by $P(E_1)$, the $P(E_1)$ terms cancel out beautifully:

$$\frac{P(E_1) - P(E_1)P(E_2) - P(E_1)P(E_3)}{P(E_1)} = 1 - P(E_2) - P(E_3)$$

The final result is **$1 - P(E_2) - P(E_3)$**. You have navigated the logic, applied the theorems, and arrived at the truth. Keep this confidence, and you will conquer any problem the JEE throws your way!

Animated Solution for Mathematics - Probability: 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 :

Select Answer:

Visualized Solution

Identify the Given Conditions

Apply Conditional Probability Definition

Simplify the Numerator using Set Theory

Express as Set Difference

Distribute the Intersection

Apply Inclusion-Exclusion Principle

Use the Triple Intersection Condition

Apply Pairwise Independence

Substitute Back into Numerator

Solve for the Conditional Probability

Simplify the Expression

Match with Options

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Conditional Lens

The De Morgan Transformation

The Inclusion-Exclusion Dance

The Magic of Independence

The Final Synthesis

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