JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $29/45$ then n is equal to:

Q: Why does the denominator in Bag 2 change to $n+4$?

Think about the physical process. Bag 2 originally had $n$ white balls and $3$ black balls, for a total of $n+3$ balls. When we transfer one ball from Bag 1 to Bag 2, the total count in Bag 2 must increase by one. So, $(n+3) + 1 = n+4$. It is a common trap to forget this! Always track the total number of items in your sample space carefully.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the experiment as a branching tree. The first event is the transfer of a ball from Bag 1 to Bag 2. Since Bag 1 has 4 white and 5 black balls, there are two distinct, mutually exclusive paths: either a white ball is transferred or a black ball is transferred. Your first task is to calculate the probability of each path.

Hint 2

The Arsenal: Once you have the probabilities of the two paths, you must determine the new composition of Bag 2 for each case. Remember, the total number of balls in Bag 2 increases by one after the transfer. Use the Law of Total Probability to combine these conditional outcomes into a single expression for the probability of drawing a white ball from Bag 2.

Hint 3

The Execution: Set up the equation using the Law of Total Probability:

P (W_{2}) = P (W_{1}) \cdot P (W_{2} ∣ W_{1}) + P (B_{1}) \cdot P (W_{2} ∣ B_{1})

. Substitute the values

P (W_{1}) = \frac{4}{9}

P (B_{1}) = \frac{5}{9}

P (W_{2} ∣ W_{1}) = \frac{n + 1}{n + 4}

, and

P (W_{2} ∣ B_{1}) = \frac{n}{n + 4}

into this equation and equate it to

\frac{29}{45}

to solve for

n

Visualized Solution

Understanding the Setup

Bag 1: $4$ White ( $W$ ), $5$ Black ( $B$ )
Bag 2: $n$ White ( $W$ ), $3$ Black ( $B$ )
Process: Transfer $1$ ball from Bag 1 to Bag 2, then draw $1$ ball from Bag 2.
Given: $P (W_{drawn from Bag 2}) = \frac{29}{45}$

Identifying the Two Cases

Let $W_{1}$ be the event: White ball transferred from Bag 1.
Let $B_{1}$ be the event: Black ball transferred from Bag 1.
These are the only two mutually exclusive cases for the transfer.

Probabilities of Transfer

$P (W_{1}) = \frac{4}{4 + 5} = \frac{4}{9}$
$P (B_{1}) = \frac{5}{4 + 5} = \frac{5}{9}$

Case 1: White Ball Transferred

Case 1: If $W_{1}$ occurs:
Bag 2 now contains $(n + 1)$ White and $3$ Black balls.
Total balls in Bag 2 = $(n + 1) + 3 = n + 4$

Conditional Probability: $P (W_{2} ∣ W_{1})$

Probability of drawing White from Bag 2 given $W_{1}$ :
$P (W_{2} ∣ W_{1}) = \frac{n + 1}{n + 4}$

Case 2: Black Ball Transferred

Case 2: If $B_{1}$ occurs:
Bag 2 now contains $n$ White and $4$ Black balls.
Total balls in Bag 2 = $n + 4$

Conditional Probability: $P (W_{2} ∣ B_{1})$

Probability of drawing White from Bag 2 given $B_{1}$ :
$P (W_{2} ∣ B_{1}) = \frac{n}{n + 4}$

Applying Law of Total Probability

By the Law of Total Probability:
$P (W_{2}) = P (W_{1}) \cdot P (W_{2} ∣ W_{1}) + P (B_{1}) \cdot P (W_{2} ∣ B_{1})$

Substituting the Values

Substitute the values into the formula:
$P (W_{2}) = (\frac{4}{9}) (\frac{n + 1}{n + 4}) + (\frac{5}{9}) (\frac{n}{n + 4})$

Simplifying the Expression

Combine the fractions over the common denominator $9 (n + 4)$ :
$P (W_{2}) = \frac{4 ( n + 1 ) + 5 n}{9 ( n + 4 )}$
$P (W_{2}) = \frac{4 n + 4 + 5 n}{9 ( n + 4 )} = \frac{9 n + 4}{9 ( n + 4 )}$

Setting up the Final Equation

Given $P (W_{2}) = \frac{29}{45}$ :
$\frac{9 n + 4}{9 ( n + 4 )} = \frac{29}{45}$
Multiply both sides by $9$ :
$\frac{9 n + 4}{n + 4} = \frac{29}{5}$

Cross-Multiplication

Cross-multiply to solve for $n$ :
$5 (9 n + 4) = 29 (n + 4)$
$45 n + 20 = 29 n + 116$

Solving for $n$

Rearrange the terms:
$45 n - 29 n = 116 - 20$
$16 n = 96$
$n = \frac{96}{16} = 6$

Conclusion \& Key Takeaway

Final Answer: $n = 6$
Key Takeaway: The Law of Total Probability is essential when an event depends on several preceding mutually exclusive events.
Challenge: How would the equation change if a black ball was drawn from Bag 2 instead?

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The Sigma Insight: Total Probability Theorem

Solution Diagram

Analyzing the Setup

Imagine you are standing in a laboratory, tasked with a simple yet profound experiment. You have two bags. Bag 1 contains

4

white balls and

5

black balls. Bag 2 is a mystery, containing

n

white balls and

3

black balls.

You are asked to transfer one ball from Bag 1 to Bag 2 and then draw a ball from Bag 2. The final probability of drawing a white ball is given as

\frac{29}{45}

. Our mission is to uncover the value of

n

The Anatomy of the Transfer

The first step in our journey is to recognize that the experiment has two distinct, mutually exclusive paths. When you reach into Bag 1 to pull out a ball, you are essentially splitting the universe into two possibilities.

Let

W_{1}

be the event that a white ball is transferred, and

B_{1}

be the event that a black ball is transferred. The probabilities are:

P (W_{1}) = \frac{4}{4 + 5} = \frac{4}{9}

P (B_{1}) = \frac{5}{4 + 5} = \frac{5}{9}

The Two Worlds of Bag 2

Now, let us look at what happens to Bag 2 in these two scenarios. If

W_{1}

occurs, Bag 2 receives a white ball, changing its composition to

(n + 1)

white and

3

black balls. The total number of balls is now

n + 4

The conditional probability of drawing a white ball from this updated bag is:

P (W_{2} ∣ W_{1}) = \frac{n + 1}{n + 4}

Conversely, if

B_{1}

occurs, Bag 2 receives a black ball. The composition becomes

n

white and

4

black balls, with a total of

n + 4

balls. The conditional probability here is:

P (W_{2} ∣ B_{1}) = \frac{n}{n + 4}

The Law of Total Probability

The Bridge

To find the overall probability of drawing a white ball, we must bridge these two worlds. The Law of Total Probability is our master key:

P (W_{2}) = P (W_{1}) \cdot P (W_{2} ∣ W_{1}) + P (B_{1}) \cdot P (W_{2} ∣ B_{1})

Substituting our values, we get:

P (W_{2}) = (\frac{4}{9}) (\frac{n + 1}{n + 4}) + (\frac{5}{9}) (\frac{n}{n + 4})

The Algebraic Resolution

Both terms share a common denominator of

9 (n + 4)

. Combining them, we simplify the expression:

P (W_{2}) = \frac{4 ( n + 1 ) + 5 n}{9 ( n + 4 )} = \frac{9 n + 4}{9 ( n + 4 )}

We are given that

P (W_{2}) = \frac{29}{45}

. Equating our expression to this value:

\frac{9 n + 4}{9 ( n + 4 )} = \frac{29}{45}

Multiplying both sides by

9

, we simplify to:

\frac{9 n + 4}{n + 4} = \frac{29}{5}

Cross-multiplying gives us:

5 (9 n + 4) = 29 (n + 4)

45 n + 20 = 29 n + 116

Subtracting

29 n

from

45 n

leaves

16 n

, and subtracting

20

from

116

gives

96

. Thus,

16 n = 96

, leading us to the final result: $n = 6$

Common Traps & FAQs

Wait, why do we use the Law of Total Probability here? Can't we just use simple probability?

Great question! We use the Law of Total Probability because the outcome of the second draw depends entirely on the outcome of the first transfer. Since we don't know which ball was transferred, we must account for both possibilities (transferring white or black) and weight them by their respective probabilities. It is the only way to cover all possible scenarios.

Why does the denominator in Bag 2 change to

n + 4

Think about the physical process. Bag 2 originally had

n

white balls and

3

black balls, for a total of

n + 3

balls. When we transfer one ball from Bag 1 to Bag 2, the total count in Bag 2 must increase by one. So,

(n + 3) + 1 = n + 4

. It is a common trap to forget this! Always track the total number of items in your sample space carefully.

Is there a shortcut to solve for

n

without all this algebra?

While you could potentially test the options provided, the algebraic approach is much more robust and less prone to errors. By setting up the equation

\frac{9 n + 4}{9 ( n + 4 )} = \frac{29}{45}

, you are building a solid mathematical model. Once you simplify it to

\frac{9 n + 4}{n + 4} = \frac{29}{5}

, the cross-multiplication is straightforward and guarantees the correct answer.

Select Answer:

Visualized Solution

Understanding the Setup

Identifying the Two Cases

Probabilities of Transfer

Case 1: White Ball Transferred

Conditional Probability: $P (W_{2} ∣ W_{1})$

Case 2: Black Ball Transferred

Conditional Probability: $P (W_{2} ∣ B_{1})$

Applying Law of Total Probability

Substituting the Values

Simplifying the Expression

Setting up the Final Equation

Cross-Multiplication

Solving for $n$

Conclusion \& Key Takeaway

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The Anatomy of the Transfer

The Two Worlds of Bag 2

The Law of Total Probability

The Algebraic Resolution

Similar Questions

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$ where $g cd (m, n) = 1$ , then $m + n$ is equal to:

A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)

Select Answer:

Visualized Solution

Understanding the Setup

Identifying the Two Cases

Probabilities of Transfer

Case 1: White Ball Transferred

Conditional Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(W2​∣W1​)

Case 2: Black Ball Transferred

Conditional Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(W2​∣B1​)

Applying Law of Total Probability

Substituting the Values

Simplifying the Expression

Setting up the Final Equation

Cross-Multiplication

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

Conclusion \& Key Takeaway

The Sigma Insight: Total Probability Theorem

Analyzing the Setup

The Anatomy of the Transfer

The Two Worlds of Bag 2

The Law of Total Probability

The Algebraic Resolution

Similar Questions

A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)

Conditional Probability: $P (W_{2} ∣ W_{1})$

Conditional Probability: $P (W_{2} ∣ B_{1})$

Solving for $n$