Let E1 and E2 be two events such that the conditional probabilities P(E1 | E2) = 1/2, P(E2 | E1) = 3

Q: Wait, why can't we just assume the events are independent?

Independence is a special condition where $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$. In this problem, we are given the intersection and conditional probabilities, which define the relationship. We must calculate the individual probabilities first to test for independence, not assume it.

Q: How do I know that $P(E_1 \cap E_2') = P(E_1) - P(E_1 \cap E_2)$?

Think of the Venn diagram. The circle $E_1$ is composed of two disjoint parts: the part inside $E_2$ (which is $E_1 \cap E_2$) and the part outside $E_2$ (which is $E_1 \cap E_2'$). Therefore, the total probability $P(E_1)$ is the sum of these two parts. Subtracting the intersection gives you exactly what you need.

Q: Why did we get $1/24$ for both $P(E_1 \cap E_2')$ and $P(E_1) \cdot P(E_2)$? Does this mean they are independent?

No, that is a common trap! Independence requires $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$. Here, $P(E_1 \cap E_2) = 1/8$, while $P(E_1) \cdot P(E_2) = 1/24$. They are not equal, so the events are dependent. The equality $P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$ is just a specific numerical property of these events, not a definition of independence.

Question

Let $E_1$ and $E_2$ be two events such that the conditional probabilities $P(E_1 | E_2) = \frac{1}{2}$, $P(E_2 | E_1) = \frac{3}{4}$ and $P(E_1 \cap E_2) = \frac{1}{8}$. Then:

Options:

(A) $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$

(B) $P(E_1' \cap E_2') = P(E_1') \cdot P(E_2)$

(C) $P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$

(D) $P(E_1' \cap E_2) = P(E_1) \cdot P(E_2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a probability space to understand how events interact.

When you look at a problem involving $P(E_1 | E_2)$ and $P(E_1 \cap E_2)$, do not see them as abstract symbols. See them as a map. Imagine a vast sample space $S$, a universe of possibilities.

Within this universe, we have two events, $E_1$ and $E_2$, represented as two circles. The overlap, the intersection $E_1 \cap E_2$, is the region where both events occur simultaneously. We are given that this overlap has a probability of:

$$P(E_1 \cap E_2) = \frac{1}{8}$$

This is our anchor point.

### The Master Key: Conditional Probability

Now, let us talk about the conditional probability formula:

$$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

This is not just a formula; it is a lens. It tells us how the probability of an event changes when we restrict our universe to only those outcomes where $B$ has already occurred.

We are given $P(E_1 | E_2) = \frac{1}{2}$. By rearranging our master key, we get:

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

Substituting our known values, we find:

$$P(E_2) = \frac{1/8}{1/2} = \frac{1}{4}$$

Similarly, using $P(E_2 | E_1) = \frac{3}{4}$, we find:

$$P(E_1) = \frac{P(E_1 \cap E_2)}{P(E_2 | E_1)} = \frac{1/8}{3/4} = \frac{1}{6}$$

We have now successfully mapped the individual probabilities of our events.

### The Geometry of the Crescent

Here is where the intuition of a true physicist comes into play. We are asked to evaluate expressions involving $E_1 \cap E_2'$.

What is $E_2'$? It is the complement of $E_2$, everything outside the circle of $E_2$. So, $E_1 \cap E_2'$ is the region inside $E_1$ but outside $E_2$. Visually, this is a crescent shape.

How do we calculate its probability? We take the entire circle $E_1$ and subtract the part that overlaps with $E_2$. Mathematically, this is:

$$P(E_1 \cap E_2') = P(E_1) - P(E_1 \cap E_2)$$

Plugging in our values, we get:

$$P(E_1 \cap E_2') = \frac{1}{6} - \frac{1}{8}$$

Finding a common denominator of $24$, we get:

$$\frac{4}{24} - \frac{3}{24} = \frac{1}{24}$$

### The Final Revelation

Finally, let us look at the product $P(E_1) \cdot P(E_2)$. We have $P(E_1) = \frac{1}{6}$ and $P(E_2) = \frac{1}{4}$.

Their product is:

$$P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$$

Look at that! The probability of our crescent, $P(E_1 \cap E_2')$, is exactly equal to the product $P(E_1) \cdot P(E_2)$.

This is a beautiful result. It shows us that while the events themselves are not independent—because $P(E_1 \cap E_2) 
eq P(E_1) \cdot P(E_2)$—they satisfy a specific algebraic relationship that leads us directly to the correct option.

You have navigated the logic, visualized the geometry, and executed the algebra. This is how you master JEE Advanced: one step, one insight, and one logical leap at a time.

Animated Solution for Mathematics - Probability: Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then:

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Visualized Solution

Visualizing the Events

The Conditional Probability Rule

Setting up for $P (E_{2})$

Calculating $P (E_{2})$

Setting up for $P (E_{1})$

Calculating $P (E_{1})$

Computing the Product $P (E_{1}) \cdot P (E_{2})$

Understanding $E_{1} \cap E_{2}^{'}$

Formula for $P (E_{1} \cap E_{2}^{'})$

Calculating $P (E_{1} \cap E_{2}^{'})$

Final Comparison

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Master Key

The Geometry of the Crescent

The Final Revelation

Similar Questions

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If two events $A$ and $B$ are such that $P (A^{c}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P (B ∣ A \cup B^{c}) = .........$

Let A, B and C be three events, which are pair-wise independent and $\overset{ˉ}{E}$ denotes the complement of an event E. If $P (A \cap B \cap C) = 0$ and $P (C) > 0$ , then $P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]$ is equal to :

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If $A$ and $B$ are two events such that $P (A) = 0.7$ , $P (B) = 0.4$ and $P (A \cap \overset{ˉ}{B}) = 0.5$ , where $\overset{ˉ}{B}$ denotes the complement of $B$ , then $P (B ∣ (A \cup \overset{ˉ}{B}))$ is equal:-

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Visualized Solution

Visualizing the Events

The Conditional Probability Rule

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E2​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E2​)

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E1​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E1​)

Computing the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E1​)⋅P(E2​)

Understanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​∩E2′​

Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E1​∩E2′​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E1​∩E2′​)