JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Probability: If A and B are two events such that $P (A \cap B) = 0.1,$ and $P (A ∣ B)$ and $P (B ∣ A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P ( A \cup B )}{P ( A \cap B )}$ is:

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the sample space as a Venn diagram. The intersection

P (A \cap B) = 0.1

is the shared region. The conditional probabilities

P (A ∣ B)

and

P (B ∣ A)

are simply ratios of this intersection to the individual probabilities

P (B)

and

P (A)

respectively. The quadratic equation is just a way to find these two conditional values.

Hint 2

Hint 2 (The Arsenal): Recall the definition of conditional probability:

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

and

P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}

. Also, remember De Morgan's Laws:

\overline{A} \cup \overline{B} = \overline{A \cap B}

and

\overline{A} \cap \overline{B} = \overline{A \cup B}

Hint 3

Hint 3 (The Execution): Start by solving

12 x^{2} - 7 x + 1 = 0

to find the two possible values for the conditional probabilities. Once you have

P (A ∣ B)

and

P (B ∣ A)

, use the intersection value

0.1

to calculate

P (A)

and

P (B)

. Finally, use the addition theorem

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

to find the union, and apply De Morgan's Laws to compute the required ratio.

Visualized Solution

Visualizing the Problem

Let $S$ be the sample space containing events $A$ and $B$ .
Given: $P (A \cap B) = 0.1$

Analyzing the Quadratic Equation

The roots of $12 x^{2} - 7 x + 1 = 0$ are $P (A ∣ B)$ and $P (B ∣ A)$ .
We must solve this quadratic equation to find their values.

Solving the Quadratic Equation

$12 x^{2} - 4 x - 3 x + 1 = 0$
$4 x (3 x - 1) - 1 (3 x - 1) = 0$
$(4 x - 1) (3 x - 1) = 0 ⟹ x = \frac{1}{4}, \frac{1}{3}$

Using Conditional Probability Formula

Recall: $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$
Recall: $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$

Calculating $P (B)$

Let $P (A ∣ B) = \frac{1}{4}$
$\frac{1}{4} = \frac{0.1}{P ( B )} ⟹ P (B) = 0.4$

Calculating $P (A)$

Let $P (B ∣ A) = \frac{1}{3}$
$\frac{1}{3} = \frac{0.1}{P ( A )} ⟹ P (A) = 0.3$

Finding the Union $P (A \cup B)$

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$P (A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$

De Morgan's Law: Numerator

By De Morgan's Law: $\overline{A} \cup \overline{B} = \overline{A \cap B}$
$P (\overline{A} \cup \overline{B}) = 1 - P (A \cap B) = 1 - 0.1 = 0.9$

De Morgan's Law: Denominator

By De Morgan's Law: $\overline{A} \cap \overline{B} = \overline{A \cup B}$
$P (\overline{A} \cap \overline{B}) = 1 - P (A \cup B) = 1 - 0.6 = 0.4$

Final Calculation

Ratio $= \frac{P ( A \cup B )}{P ( A \cap B )}$
Ratio $= \frac{0.9}{0.4} = \frac{9}{4}$

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The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Quadratic Gatekeeper

We begin with the quadratic equation:

12 x^{2} - 7 x + 1 = 0

To solve this, we split the middle term:

12 x^{2} - 4 x - 3 x + 1 = 0

Factoring the expression, we obtain:

(4 x - 1) (3 x - 1) = 0

Thus, the roots are

x = \frac{1}{4}

and

x = \frac{1}{3}

. These values represent our conditional probabilities,

P (A ∣ B)

and

P (B ∣ A)

, which define the influence between events.

The Conditional Bridge

We utilize the definitions of conditional probability:

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} and P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}

Given

P (A \cap B) = 0.1

, we isolate the individual probabilities. If

P (A ∣ B) = \frac{1}{4}

, then:

\frac{0.1}{P ( B )} = \frac{1}{4} \Rightarrow P (B) = 0.4

Similarly, if

P (B ∣ A) = \frac{1}{3}

, then:

\frac{0.1}{P ( A )} = \frac{1}{3} \Rightarrow P (A) = 0.3

The De Morgan Revelation

We now evaluate the ratio

\frac{P ( A \cup B )}{P ( A \cap B )}

. Applying De Morgan's Laws, the numerator is equivalent to

P (\overline{A \cap B})

Since

P (A \cap B) = 0.1

, the probability of its complement is:

P (\overline{A \cap B}) = 1 - 0.1 = 0.9

The denominator is equivalent to

P (\overline{A \cup B})

. First, we calculate the union:

P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.3 + 0.4 - 0.1 = 0.6

Therefore, the probability of the complement of the union is:

P (\overline{A \cup B}) = 1 - 0.6 = 0.4

Final Synthesis

We arrive at the final calculation by dividing the numerator by the denominator:

\frac{P ( A \cup B )}{P ( A \cap B )} = \frac{0.9}{0.4} = \frac{9}{4}

The final result is $\frac{9}{4}$ (or $2.25$ ).

Common Traps & FAQs

Wait, does it matter which root of the quadratic equation I assign to

P (A ∣ B)

and which to

P (B ∣ A)

Great question! In this specific problem, it actually does not matter. Because the intersection

P (A \cap B)

is symmetric, assigning

P (A ∣ B) = 1/4

and

P (B ∣ A) = 1/3

will yield

P (B) = 0.4

and

P (A) = 0.3

. If you swapped them, you would simply get

P (A) = 0.4

and

P (B) = 0.3

. Since the final ratio depends on

P (A \cup B)

and

P (A \cap B)

, and the union is commutative (

P (A) + P (B) - P (A \cap B)

), the final result remains identical.

How do I remember De Morgan's Laws correctly? I always get the union and intersection signs mixed up.

Think of it as a 'flipping' rule. When you take the complement of a union, the union symbol flips to an intersection, and vice versa. So,

\overline{A \cup B}

becomes

\overline{A} \cap \overline{B}

, and

\overline{A \cap B}

becomes

\overline{A} \cup \overline{B}

. Just remember: 'Complement of the union is the intersection of the complements'.

Why can't we just use energy conservation or some other physics principle here?

Haha, I love the enthusiasm! But remember, this is a pure probability problem, not mechanics. We are dealing with set theory and the logic of events, not forces or energy. Stick to the axioms of probability—they are the 'laws of motion' for this specific universe!

Animated Solution for Mathematics - Probability: If A and B are two events such that $P (A \cap B) = 0.1,$ and $P (A ∣ B)$ and $P (B ∣ A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P ( A \cup B )}{P ( A \cap B )}$ is:

Select Answer:

Visualized Solution

Visualizing the Problem

Analyzing the Quadratic Equation

Solving the Quadratic Equation

Using Conditional Probability Formula

Calculating $P (B)$

Calculating $P (A)$

Finding the Union $P (A \cup B)$

De Morgan's Law: Numerator

De Morgan's Law: Denominator

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Quadratic Gatekeeper

The Conditional Bridge

The De Morgan Revelation

Final Synthesis

Similar Questions

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If $A$ and $B$ are two events such that $P (A) > 0$ , and $P (B) \neq = 1$ , then $P (\frac{A ˉ}{B ˉ})$ is equal to

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Select Answer:

Visualized Solution

Visualizing the Problem

Analyzing the Quadratic Equation

Solving the Quadratic Equation

Using Conditional Probability Formula

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

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

Finding the Union .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∪B)

De Morgan's Law: Numerator

De Morgan's Law: Denominator

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Quadratic Gatekeeper

The Conditional Bridge

The De Morgan Revelation

Final Synthesis

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