JEE Main 2024 (08 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: There are three bags X, Y and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag Y, is :

Q: Why can we just cancel out the $\frac{1}{3}$ in the final step?

Look at the structure of the equation. The term $\frac{1}{3}$ appears in every single term of the numerator and the denominator. Mathematically, you are factoring it out: $\frac{\frac{1}{3} \cdot X}{\frac{1}{3} \cdot (Y + Z + W)}$. Since $\frac{1}{3}$ is non-zero, it cancels out perfectly. It is a beautiful example of how symmetry simplifies complex problems.

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Progressive Hints

Hint 1

Think of this as a detective story. You are not predicting the future; you are investigating the past. You have the 'effect' (a one-rupee coin), and you need to trace it back to the 'cause' (the specific bag). This is the hallmark of Bayes' Theorem.

Hint 2

Your toolkit must include the definition of conditional probability and the Law of Total Probability. Remember, Bayes' Theorem is just a bridge:

P (E_{i} ∣ A) = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{P ( A )}

. You need to calculate the total probability of drawing a one-rupee coin,

P (A)

, by summing the probabilities of drawing it from each bag.

Hint 3

Start by setting up the ratio:

P (E_{2} ∣ A) = \frac{P ( E _{2} ) \cdot P ( A ∣ E _{2} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} ) + P ( E _{3} ) \cdot P ( A ∣ E _{3} )}

. Substitute the values:

P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}

, and the conditional probabilities based on the coin counts in each bag.

Visualized Solution

Visualizing the Bags

Bag X: 5 (Rs. 1), 4 (Rs. 5) $⟹$ Total = 9
Bag Y: 4 (Rs. 1), 5 (Rs. 5) $⟹$ Total = 9
Bag Z: 3 (Rs. 1), 6 (Rs. 5) $⟹$ Total = 9

Defining Selection Events $E_{1}, E_{2}, E_{3}$

Let $E_{1}, E_{2}, E_{3}$ be the events of selecting bags X, Y, Z respectively.
Since a bag is selected at random:
$P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

Defining the Observed Event $A$

Let $A$ be the event of drawing a Rs. 1 coin.
We need to find the probability that it came from Bag Y, which is $P (E_{2} ∣ A)$ .

Conditional Probabilities $P (A ∣ E_{i})$

$P (A ∣ E_{1}) = \frac{5}{9}$ (5 Rs. 1 coins in Bag X)
$P (A ∣ E_{2}) = \frac{4}{9}$ (4 Rs. 1 coins in Bag Y)
$P (A ∣ E_{3}) = \frac{3}{9}$ (3 Rs. 1 coins in Bag Z)

Applying Bayes' Theorem

We use Bayes' Theorem to find $P (E_{2} ∣ A)$ :
$P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} ) + P ( E _{3} ) P ( A ∣ E _{3} )}$

Substitution of Values

$P (E_{2} ∣ A) = \frac{\frac{1}{3} \cdot \frac{4}{9}}{\frac{1}{3} \cdot \frac{5}{9} + \frac{1}{3} \cdot \frac{4}{9} + \frac{1}{3} \cdot \frac{3}{9}}$

Simplifying the Expression

Factoring out $\frac{1}{3} \cdot \frac{1}{9}$ from numerator and denominator:
$P (E_{2} ∣ A) = \frac{4}{5 + 4 + 3}$

Final Calculation

$P (E_{2} ∣ A) = \frac{4}{12}$
$P (E_{2} ∣ A) = \frac{1}{3}$

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

We are presented with three bags,

X

Y

, and

Z

, each containing a total of 9 coins. The distribution of one-rupee coins and five-rupee coins is as follows:

Bag $X$ ( $E_{1}$ ): 5 one-rupee coins, 4 five-rupee coins. Bag

Y

(

E_{2}

): 4 one-rupee coins, 5 five-rupee coins. * Bag

Z

(

E_{3}

): 3 one-rupee coins, 6 five-rupee coins.

Since the bags are chosen at random, the prior probability of selecting any specific bag is equal:

P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}

The Observed Reality

We define Event

A

as the act of drawing a one-rupee coin. We calculate the conditional probability of drawing a one-rupee coin from each bag:

$P (A ∣ E_{1}) = \frac{5}{9}$

P (A ∣ E_{2}) = \frac{4}{9}

P (A ∣ E_{3}) = \frac{3}{9}

These values represent the likelihood of our observed evidence given each specific hypothesis.

The Master Equation

To find the probability that the coin came from Bag

Y

given that it is a one-rupee coin, we apply Bayes' Theorem:

P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} ) + P ( E _{3} ) P ( A ∣ E _{3} )}

Substituting our known values into the formula, we obtain:

P (E_{2} ∣ A) = \frac{\frac{1}{3} \cdot \frac{4}{9}}{\frac{1}{3} \cdot \frac{5}{9} + \frac{1}{3} \cdot \frac{4}{9} + \frac{1}{3} \cdot \frac{3}{9}}

Final Calculation

Due to the symmetry of the selection process, the common factor of

\frac{1}{3} \cdot \frac{1}{9}

appears in every term. Factoring this out simplifies the expression significantly:

P (E_{2} ∣ A) = \frac{4}{5 + 4 + 3}

Performing the final arithmetic:

P (E_{2} ∣ A) = \frac{4}{12} = \frac{1}{3}

The probability that the one-rupee coin was drawn from Bag

Y

is $\frac{1}{3}$ .

Common Traps & FAQs

Why do we need Bayes' Theorem? Can't we just guess?

Guessing is for the unprepared! Bayes' Theorem is the rigorous mathematical framework for 'inverse probability.' When you know the outcome (the coin is one-rupee) and need to find the source (the bag), you are effectively reversing the flow of time. Bayes' Theorem is the only logical way to quantify this uncertainty.

Why can we just cancel out the

\frac{1}{3}

in the final step?

Look at the structure of the equation. The term

\frac{1}{3}

appears in every single term of the numerator and the denominator. Mathematically, you are factoring it out:

\frac{\frac{1}{3} \cdot X}{\frac{1}{3} \cdot ( Y + Z + W )}

. Since

\frac{1}{3}

is non-zero, it cancels out perfectly. It is a beautiful example of how symmetry simplifies complex problems.

What if the bags had different numbers of coins?

If the bags had different totals, the symmetry would break. You would no longer have a common denominator of 9, and the

\frac{1}{3}

would not cancel out so cleanly. Always check the total count in each container before you start your calculation—it is a common trap!

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Visualized Solution

Visualizing the Bags

Defining Selection Events $E_{1}, E_{2}, E_{3}$

Defining the Observed Event $A$

Conditional Probabilities $P (A ∣ E_{i})$

Applying Bayes' Theorem

Substitution of Values

Simplifying the Expression

Final Calculation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Observed Reality

The Master Equation

Final Calculation

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, 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 $B_{1}$ 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:

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 ____.

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

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 :

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

Select Answer:

Visualized Solution

Visualizing the Bags

Defining Selection Events .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​,E2​,E3​

Defining the Observed Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

Applying Bayes' Theorem

Substitution of Values

Simplifying the Expression

Final Calculation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

The Observed Reality

The Master Equation

Final Calculation

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, 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:

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

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 :

Defining Selection Events $E_{1}, E_{2}, E_{3}$

Defining the Observed Event $A$

Conditional Probabilities $P (A ∣ E_{i})$

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