JEE Advanced 2021

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $E, F$ and $G$ be three events having probabilities $P (E) = \frac{1}{8}, P (F) = \frac{1}{6}$ and $P (G) = \frac{1}{4}$ , and let $P (E \cap F \cap G) = \frac{1}{10}$ . For any event $H$ , if $H^{c}$ denotes its complement, then which of the following statements is(are) TRUE ?

Q: Why does the inequality sign flip when we evaluate Option D?

Think of the function $f(x) = 1 - x$. This is a strictly decreasing function. When you have an inequality $x \le A$, applying a decreasing function reverses the inequality. Since $P(E \cup F \cup G) \le \frac{13}{24}$, subtracting this from $1$ gives $1 - P(E \cup F \cup G) \ge 1 - \frac{13}{24}$. It's a classic trap, so stay alert!

Select Answer:

* Multiple Correct

Visualized Solution

Visual Anchor: The Venn Diagram Setup

Given probabilities:
$P (E) = \frac{1}{8}$
$P (F) = \frac{1}{6}$
$P (G) = \frac{1}{4}$
Intersection of all three events:
$P (E \cap F \cap G) = \frac{1}{10}$

Logic Bridge: Analyzing Option A

Let $P (E \cap F \cap G^{c}) = x$
From the Venn diagram, Circle $E$ is composed of 4 disjoint regions.
$P (E) = a + x + y + \frac{1}{10}$

Raw Setup: Non-Negativity Constraint

Since probabilities cannot be negative, $a \geq 0$ and $y \geq 0$ .
Therefore, the sum of the remaining parts must be at most $P (E)$ :
$x + \frac{1}{10} \leq P (E)$
$x + \frac{1}{10} \leq \frac{1}{8}$

Atomic Compute: Solving for Option A

$x \leq \frac{1}{8} - \frac{1}{10}$
$x \leq \frac{5}{40} - \frac{4}{40}$
$P (E \cap F \cap G^{c}) \leq \frac{1}{40}$
Option A is TRUE.

Logic Bridge: Analyzing Option B

Let $P (E^{c} \cap F \cap G) = z$
Focus on Circle $F$ . It contains 4 disjoint regions.
$P (F) = b + x + z + \frac{1}{10}$

Raw Setup: Constraint for Circle F

Since $b \geq 0$ and $x \geq 0$ , we have:
$z + \frac{1}{10} \leq P (F)$
$z + \frac{1}{10} \leq \frac{1}{6}$

Atomic Compute: Solving for Option B

$z \leq \frac{1}{6} - \frac{1}{10}$
$z \leq \frac{5}{30} - \frac{3}{30} = \frac{2}{30}$
$P (E^{c} \cap F \cap G) \leq \frac{1}{15}$
Option B is TRUE.

Logic Bridge: Boole's Inequality for Option C

We need to evaluate $P (E \cup F \cup G)$ .
Using Boole's Inequality (Union Bound):
The probability of a union is at most the sum of individual probabilities.
$P (E \cup F \cup G) \leq P (E) + P (F) + P (G)$

Raw Setup: Substituting Probabilities

Substitute the given values:
$P (E \cup F \cup G) \leq \frac{1}{8} + \frac{1}{6} + \frac{1}{4}$

Atomic Compute: Solving for Option C

Common denominator is $24$ :
$P (E \cup F \cup G) \leq \frac{3}{24} + \frac{4}{24} + \frac{6}{24}$
$P (E \cup F \cup G) \leq \frac{13}{24}$
Option C is TRUE.

Logic Bridge: De Morgan's Laws for Option D

Evaluate $P (E^{c} \cap F^{c} \cap G^{c})$ .
By De Morgan's Laws, this is the complement of the union:
$P (E^{c} \cap F^{c} \cap G^{c}) = 1 - P (E \cup F \cup G)$

Raw Setup: Bounding the Outside Region

Since $P (E \cup F \cup G) \leq \frac{13}{24}$ :
$1 - P (E \cup F \cup G) \geq 1 - \frac{13}{24}$

Atomic Compute: Disproving Option D

$P (E^{c} \cap F^{c} \cap G^{c}) \geq \frac{11}{24}$
Option D claims it is $\leq \frac{5}{12}$ (which is $\frac{10}{24}$ ).
Since $\frac{11}{24} > \frac{10}{24}$ , the claim is impossible.
Option D is FALSE.

The Way Forward

Final Answer: Statements A, B, and C are TRUE.
Key Takeaway: Venn diagrams visually enforce non-negativity constraints ( $P \geq 0$ ) on individual regions.

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

We are given three events

E

F

, and

G

with the following probabilities:

P (E) = \frac{1}{8}

P (F) = \frac{1}{6}

, and

P (G) = \frac{1}{4}

The intersection of all three events is given as:

P (E \cap F \cap G) = \frac{1}{10}

The Non-Negativity Constraint

The fundamental rule of probability dictates that no disjoint region within a set can have a negative probability. For event

E

, the sum of its disjoint parts must satisfy:

P (E \cap F \cap G^{c}) + P (E \cap F \cap G) \leq P (E)

Substituting the known values:

P (E \cap F \cap G^{c}) + \frac{1}{10} \leq \frac{1}{8}

Solving for the intersection, we find:

P (E \cap F \cap G^{c}) \leq \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40}

Thus, Option A is verified.

Applying the same logic to event

F

P (E^{c} \cap F \cap G) + P (E \cap F \cap G) \leq P (F)

P (E^{c} \cap F \cap G) + \frac{1}{10} \leq \frac{1}{6}

Solving this yields:

P (E^{c} \cap F \cap G) \leq \frac{1}{6} - \frac{1}{10} = \frac{5 - 3}{30} = \frac{2}{30} = \frac{1}{15}

Thus, Option B is verified.

The Power of Boole's Inequality

To evaluate the union

P (E \cup F \cup G)

, we utilize Boole's Inequality, which states that the probability of a union is at most the sum of the individual probabilities:

P (E \cup F \cup G) \leq P (E) + P (F) + P (G)

Substituting the given values:

P (E \cup F \cup G) \leq \frac{1}{8} + \frac{1}{6} + \frac{1}{4} = \frac{3 + 4 + 6}{24} = \frac{13}{24}

Thus, Option C is verified.

The De Morgan Trap

Finally, we examine the region outside all events, denoted as

P (E^{c} \cap F^{c} \cap G^{c})

. By De Morgan's Laws, this is equivalent to

1 - P (E \cup F \cup G)

Since

P (E \cup F \cup G) \leq \frac{13}{24}

, it follows that:

P (E^{c} \cap F^{c} \cap G^{c}) \geq 1 - \frac{13}{24} = \frac{11}{24}

Option D claims this probability is

\leq \frac{5}{12}

(which is

\frac{10}{24}

). Since

\frac{11}{24} > \frac{10}{24}

, this statement is mathematically impossible.

Option D is false.

Common Traps & FAQs

Wait, why do we assume the other regions are zero? Isn't that an assumption?

Great question! We aren't assuming they are zero. We are using the fact that they are non-negative (

\geq 0

). If

A + B = C

and

B \geq 0

, then

A \leq C

. We are finding the *maximum* possible value for a region by setting the other unknown regions to their minimum possible value, which is zero.

Why does the inequality sign flip when we evaluate Option D?

Think of the function

f (x) = 1 - x

. This is a strictly decreasing function. When you have an inequality

x \leq A

, applying a decreasing function reverses the inequality. Since

P (E \cup F \cup G) \leq \frac{13}{24}

, subtracting this from

1

gives

1 - P (E \cup F \cup G) \geq 1 - \frac{13}{24}

. It's a classic trap, so stay alert!

Can we use the Inclusion-Exclusion Principle for Option C?

You could, but it would be overkill and potentially confusing. The Inclusion-Exclusion Principle gives you an exact equality for the union. Boole's Inequality (the Union Bound) is much faster here because the question asks for an inequality, not an exact value. Always choose the tool that fits the specific question asked.

Select Answer:

Visualized Solution

Visual Anchor: The Venn Diagram Setup

Logic Bridge: Analyzing Option A

Raw Setup: Non-Negativity Constraint

Atomic Compute: Solving for Option A

Logic Bridge: Analyzing Option B

Raw Setup: Constraint for Circle F

Atomic Compute: Solving for Option B

Logic Bridge: Boole's Inequality for Option C

Raw Setup: Substituting Probabilities

Atomic Compute: Solving for Option C

Logic Bridge: De Morgan's Laws for Option D

Raw Setup: Bounding the Outside Region

Atomic Compute: Disproving Option D

The Way Forward

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Non-Negativity Constraint

The Power of Boole's Inequality

The De Morgan Trap

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Select Answer:

Visualized Solution

Visual Anchor: The Venn Diagram Setup

Logic Bridge: Analyzing Option A

Raw Setup: Non-Negativity Constraint

Atomic Compute: Solving for Option A

Logic Bridge: Analyzing Option B

Raw Setup: Constraint for Circle F

Atomic Compute: Solving for Option B

Logic Bridge: Boole's Inequality for Option C

Raw Setup: Substituting Probabilities

Atomic Compute: Solving for Option C

Logic Bridge: De Morgan's Laws for Option D

Raw Setup: Bounding the Outside Region

Atomic Compute: Disproving Option D

The Way Forward

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Non-Negativity Constraint

The Power of Boole's Inequality

The De Morgan Trap

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