JEE Main 2023 (24 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Probability: The urns A, B and C contain 4 red, 6 black; 5 red, 5 black and $λ$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^{2} = λ x$ with one vertex at the vertex of the parabola is

Q: How do I know the equilateral triangle is symmetric about the x-axis?

Great question. The parabola $y^2 = 6x$ is symmetric about the x-axis. If you were to place the equilateral triangle such that it is not symmetric, the vertices would not lie on the parabola in a way that satisfies the equilateral condition while keeping one vertex at the origin. By placing the triangle symmetrically, we ensure that the two vertices on the parabola have the same x-coordinate and opposite y-coordinates, which is the only way to maintain the equilateral property given the symmetry of the parabola itself.

Q: Is there a faster way to find the side length without finding the parameter $t$?

You could use the intersection of the line $y = \frac{1}{\sqrt{3}}x$ and the parabola $y^2 = 6x$. By substituting $y = \frac{x}{\sqrt{3}}$ into the parabola equation, you get $\frac{x^2}{3} = 6x$, which simplifies to $x^2 = 18x$. Since $x=0$ is the origin, the other vertex has $x=18$. Then, $y = \frac{18}{\sqrt{3}} = 6\sqrt{3}$. The side length $L$ is the distance from $(0,0)$ to $(18, 6\sqrt{3})$, so $L^2 = 18^2 + (6\sqrt{3})^2 = 324 + 108 = 432$. Both methods are valid, but understanding the parameter $t$ is crucial for more complex problems!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a classic Bayes' Theorem problem. The unknown

λ

is hidden inside the probability of selecting a red ball from Urn C. Visualize the urns as three distinct branches of a probability tree. The key is to express the total probability of drawing a red ball as a weighted sum of the probabilities from each urn.

Hint 2

Hint 2: The Arsenal. Once you have the probability equation, you will find that the prior probability of selecting any urn, which is

\frac{1}{3}

, appears in both the numerator and the denominator. This allows you to simplify the expression significantly. Focus on solving the resulting linear equation for

λ

Hint 3

Hint 3: The Execution. With

λ

determined, you have the equation of the parabola

y^{2} = 6 x

. For an equilateral triangle inscribed with one vertex at the origin, symmetry is your best friend. If the triangle is symmetric about the x-axis, the line connecting the origin to one of the other vertices must make an angle of

3 0^{\circ}

with the x-axis. Use the slope

m = tan (3 0^{\circ})

to find the coordinates of the vertex.

Visualized Solution

Defining the Events

Let $E_{1}, E_{2}, E_{3}$ be events of selecting Urns $A, B, C$ .
$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$
Let $R$ be the event that the ball drawn is red.

Conditional Probabilities

$P (R ∣ E_{1}) = \frac{4}{10} = 0.4$
$P (R ∣ E_{2}) = \frac{5}{10} = 0.5$
$P (R ∣ E_{3}) = \frac{λ}{λ + 4}$

Applying Bayes' Theorem

Given $P (E_{3} ∣ R) = 0.4 = \frac{2}{5}$
By Bayes' Theorem: $P (E_{3} ∣ R) = \frac{P ( E _{3} ) P ( R ∣ E _{3} )}{\sum _{i = 1}^{3} P ( E _{i} ) P ( R ∣ E _{i} )}$

Substituting Values

$\frac{\frac{1}{3} \cdot \frac{λ}{λ + 4}}{\frac{1}{3} ( 0.4 + 0.5 + \frac{λ}{λ + 4} )} = 0.4$
Canceling $\frac{1}{3}$ : $\frac{\frac{λ}{λ + 4}}{0.9 + \frac{λ}{λ + 4}} = 0.4$

Solving for $λ$

$\frac{λ}{0.9 ( λ + 4 ) + λ} = 0.4$
$λ = 0.4 (1.9 λ + 3.6)$
$λ = 0.76 λ + 1.44$
$0.24 λ = 1.44 ⟹ λ = 6$

The Parabola Equation

Parabola: $y^{2} = λ x ⟹ y^{2} = 6 x$
Standard form: $y^{2} = 4 a x ⟹ 4 a = 6 ⟹ a = 1.5$

Geometry of the Inscribed Triangle

Let vertices be $O (0, 0)$ , $P (a t^{2}, 2 a t)$ , and $Q (a t^{2}, - 2 a t)$ .
By symmetry, the line $O P$ makes an angle of $3 0^{\circ}$ with the x-axis.

Finding the Parameter $t$

Slope of $O P = tan 3 0^{\circ} = \frac{2 a t}{a t ^{2}} = \frac{2}{t}$
$\frac{1}{3} = \frac{2}{t} ⟹ t = 23$

Calculating Side Length Squared

Side length $L = O P = (a t^{2})^{2} + (2 a t)^{2}$
$L^{2} = a^{2} t^{4} + 4 a^{2} t^{2} = a^{2} (t^{4} + 4 t^{2})$

Final Computation

Substitute $a = 1.5$ , $t^{2} = 12$ , $t^{4} = 144$ :
$L^{2} = (1.5)^{2} (144 + 4 (12))$
$L^{2} = 2.25 (144 + 48) = 2.25 \times 192 = 432$

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

Welcome, future engineer! Today, we are embarking on a journey that bridges two seemingly disparate worlds: the unpredictable realm of Probability and the rigid, elegant structure of Coordinate Geometry.

This problem is a classic JEE Advanced challenge because it tests your ability to pivot your mindset. You start as a statistician, unraveling a mystery of urns, and finish as a geometer, carving an equilateral triangle out of a parabola.

The Urn Mystery

Imagine you are standing before three urns,

A

B

, and

C

. You are told that one is selected at random, meaning the probability of choosing any one of them is exactly

\frac{1}{3}

We define events

E_{1}, E_{2}, E_{3}

as selecting Urns

A

B

, and

C

, respectively. Our goal is to find the probability of drawing a red ball, which we call event

R

The conditional probabilities are:

P (R ∣ E_{1}) = 0.4

P (R ∣ E_{2}) = 0.5

P (R ∣ E_{3}) = \frac{λ}{λ + 4}

The Master Equation

We are given that if the ball is red, the probability it came from Urn

C

0.4

. This is the perfect stage for Bayes' Theorem.

The formula is:

P (E_{3} ∣ R) = \frac{P ( E _{3} ) P ( R ∣ E _{3} )}{\sum _{i = 1}^{3} P ( E _{i} ) P ( R ∣ E _{i} )}

When you plug in the values, the

\frac{1}{3}

factor appears in every term and cancels out. The equation simplifies to:

\frac{\frac{λ}{λ + 4}}{0.4 + 0.5 + \frac{λ}{λ + 4}} = 0.4

Solving this algebraic puzzle, we find that

λ = 6

. The mystery is solved, and the path to the next phase is clear.

The Parabola's Secret

Now, we shift gears. With

λ = 6

, our parabola is defined by

y^{2} = 6 x

Comparing this to the form

y^{2} = 4 a x

, we identify

4 a = 6

, which gives us the focal parameter

a = 1.5

. This value is the heartbeat of our geometric construction.

The Geometric Elegance

Visualize the parabola. To fit an equilateral triangle with a vertex at the origin

(0, 0)

, symmetry is essential. The

x

-axis acts as our line of symmetry.

If the triangle is equilateral, the angle at the origin must be

6 0^{\circ}

. Because of the symmetry, the line segment from the origin to one of the vertices on the parabola must make an angle of

3 0^{\circ}

with the

x

-axis.

The slope of this line is

m = tan (3 0^{\circ}) = \frac{1}{3}

Final Calculation

Using the equation of the line

y = \frac{1}{3} x

, we find the intersection with the parabola

y^{2} = 6 x

. Substituting

y

, we get:

\frac{x ^{2}}{3} = 6 x \Rightarrow x^{2} = 18 x

Since the vertex is not at the origin, we take

x = 18

. Consequently,

y = \frac{18}{3} = 63

The side length

L

of our triangle is the distance from the origin to this point

(18, 63)

. Calculating

L^{2}

L^{2} = 1 8^{2} + (63)^{2} = 324 + 108 = 432

The final result for the square of the side length is 432.

Common Traps & FAQs

Wait, why can we just cancel the

\frac{1}{3}

in Bayes' Theorem?

That is a fantastic observation! In Bayes' Theorem, the probability of selecting any urn is

P (E_{i}) = \frac{1}{3}

for all

i

. Since this factor appears in every term of the numerator and the denominator, it acts as a common multiplier. Mathematically, you are factoring out

\frac{1}{3}

from the entire denominator, which then cancels with the

\frac{1}{3}

in the numerator. It is a beautiful simplification that saves you from unnecessary arithmetic!

How do I know the equilateral triangle is symmetric about the x-axis?

Great question. The parabola

y^{2} = 6 x

is symmetric about the x-axis. If you were to place the equilateral triangle such that it is not symmetric, the vertices would not lie on the parabola in a way that satisfies the equilateral condition while keeping one vertex at the origin. By placing the triangle symmetrically, we ensure that the two vertices on the parabola have the same x-coordinate and opposite y-coordinates, which is the only way to maintain the equilateral property given the symmetry of the parabola itself.

Is there a faster way to find the side length without finding the parameter

t

You could use the intersection of the line

y = \frac{1}{3} x

and the parabola

y^{2} = 6 x

. By substituting

y = \frac{x}{3}

into the parabola equation, you get

\frac{x ^{2}}{3} = 6 x

, which simplifies to

x^{2} = 18 x

. Since

x = 0

is the origin, the other vertex has

x = 18

. Then,

y = \frac{18}{3} = 63

. The side length

L

is the distance from

(0, 0)

(18, 63)

, so

L^{2} = 1 8^{2} + (63)^{2} = 324 + 108 = 432

. Both methods are valid, but understanding the parameter

t

is crucial for more complex problems!

Enter Numerical Value:

Visualized Solution

Defining the Events

Conditional Probabilities

Applying Bayes' Theorem

Substituting Values

Solving for $λ$

The Parabola Equation

Geometry of the Inscribed Triangle

Finding the Parameter $t$

Calculating Side Length Squared

Final Computation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Urn Mystery

The Master Equation

The Parabola's Secret

The Geometric Elegance

Final Calculation

Similar Questions

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Comprehension Passage

If $P (u_{i}) \propto i$ , where $i = 1, 2, 3, \dots, n$ , then $lim_{n \to \infty} P (w) =$

If $P (u_{i}) = c$ , (a constant) then $P (u_{n} / w) =$

Let $P (u_{i}) = \frac{1}{n}$ , if $n$ is even and $E$ denotes the event of choosing even numbered urn, then the value of $P (w / E)$ is

Comprehension Passage

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

Comprehension Passage

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_{2}$ is

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ is:

Enter Numerical Value:

Visualized Solution

Defining the Events

Conditional Probabilities

Applying Bayes' Theorem

Substituting Values

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

The Parabola Equation

Geometry of the Inscribed Triangle

Finding the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Calculating Side Length Squared

Final Computation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Urn Mystery

The Master Equation

The Parabola's Secret

The Geometric Elegance

Final Calculation

Similar Questions

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)∝i, where i=1,2,3,…,n, then limn→∞​P(w)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)=c, (a constant) then P(un​/w)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P(ui​)=n1​, if n is even and E denotes the event of choosing even numbered urn, then the value of P(w/E) is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability of the drawn ball from U2​ being white is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given that the drawn ball from U2​ is white, the probability that head appeared on the coin is

Comprehension Passage

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1 ball is drawn from each of the boxes B1​,B2​ and B3​, the probability that all 3 drawn balls are of the same colour is

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

Solving for $λ$

Finding the Parameter $t$

If $P (u_{i}) \propto i$ , where $i = 1, 2, 3, \dots, n$ , then $lim_{n \to \infty} P (w) =$

If $P (u_{i}) = c$ , (a constant) then $P (u_{n} / w) =$

Let $P (u_{i}) = \frac{1}{n}$ , if $n$ is even and $E$ denotes the event of choosing even numbered urn, then the value of $P (w / E)$ is

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is