JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: 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:

Select Answer:

Progressive Hints

Hint 1

Visualize the experiment as a branching tree. You are standing at the start, and there are three distinct paths the transferred ball can take: Red, Black, or White. Your goal is to identify which path was taken, given that the final outcome is a black ball.

Hint 2

You need to employ Bayes' Theorem. Remember, Bayes' Theorem is the bridge between the 'prior' probability (what we know before the draw) and the 'posterior' probability (what we know after observing the black ball). The core formula is

P (T_{R} ∣ A) = \frac{P ( T _{R} ) P ( A ∣ T _{R} )}{P ( A )}

Hint 3

Begin by calculating the conditional probabilities for each branch. For instance, if a red ball is transferred, Bag II now contains 3 Red, 5 Black, and 2 White balls. The probability of drawing a black ball given this transfer is

P (A ∣ T_{R}) = \frac{5}{10}

. Apply this logic to all three cases and plug them into the Bayes' formula.

Visualized Solution

Visualizing the Initial Setup

Bag I: 3 Red, 4 Black, 3 White (Total = 10)
Bag II: 2 Red, 5 Black, 2 White (Total = 9)

Defining the Experiment

Transfer 1 ball from Bag I to Bag II.
Draw 1 ball from Bag II.
Observed Event $A$ : The drawn ball is black.

Prior Probabilities of Transfer

Let $T_{R}, T_{B}, T_{W}$ be events that the transferred ball is Red, Black, and White.
$P (T_{R}) = \frac{3}{10}$
$P (T_{B}) = \frac{4}{10}$
$P (T_{W}) = \frac{3}{10}$

Case 1: Red Ball Transferred

If $T_{R}$ occurs, Bag II gets +1 Red ball.
Bag II now has: 3 Red, 5 Black, 2 White (Total = 10).
$P (A ∣ T_{R}) = \frac{5}{10}$

Case 2: Black Ball Transferred

If $T_{B}$ occurs, Bag II gets +1 Black ball.
Bag II now has: 2 Red, 6 Black, 2 White (Total = 10).
$P (A ∣ T_{B}) = \frac{6}{10}$

Case 3: White Ball Transferred

If $T_{W}$ occurs, Bag II gets +1 White ball.
Bag II now has: 2 Red, 5 Black, 3 White (Total = 10).
$P (A ∣ T_{W}) = \frac{5}{10}$

Applying Bayes' Theorem

We need to find $P (T_{R} ∣ A)$ .
By Bayes' Theorem:
$P (T_{R} ∣ A) = \frac{P ( T _{R} ) P ( A ∣ T _{R} )}{P ( T _{R} ) P ( A ∣ T _{R} ) + P ( T _{B} ) P ( A ∣ T _{B} ) + P ( T _{W} ) P ( A ∣ T _{W} )}$

Substituting the Values

$P (T_{R} ∣ A) = \frac{( \frac{3}{10} ) ( \frac{5}{10} )}{( \frac{3}{10} ) ( \frac{5}{10} ) + ( \frac{4}{10} ) ( \frac{6}{10} ) + ( \frac{3}{10} ) ( \frac{5}{10} )}$

Simplifying the Expression

Cancel the common denominator $100$ from all terms.
$P (T_{R} ∣ A) = \frac{3 \times 5}{( 3 \times 5 ) + ( 4 \times 6 ) + ( 3 \times 5 )}$
$P (T_{R} ∣ A) = \frac{15}{15 + 24 + 15}$

Final Calculation

$P (T_{R} ∣ A) = \frac{15}{54}$
Divide numerator and denominator by 3:
$P (T_{R} ∣ A) = \frac{5}{18}$

Conclusion

Final Answer: $\frac{5}{18}$
The probability that the transferred ball was red is $\frac{5}{18}$ .

00:00 / 00:00

The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

Welcome, future engineers! Today we are not just solving a probability problem; we are becoming detectives. Imagine you are standing in a lab with two bags.

Bag I contains 3 Red, 4 Black, and 3 White balls, totaling 10. Bag II contains 2 Red, 5 Black, and 2 White balls, totaling 9.

We perform a two-step experiment: first, we transfer one ball from Bag I to Bag II, and then we draw a ball from Bag II. We observe that the final ball drawn is black, and we must work backward to find the probability that the transferred ball was red.

Mapping the Branches

Before we touch any math, we must visualize the flow. We define three possible scenarios for the transferred ball:

T_{R}

(Red),

T_{B}

(Black), and

T_{W}

(White).

Since Bag I has 10 balls, the prior probabilities are:

P (T_{R}) = \frac{3}{10}, P (T_{B}) = \frac{4}{10}, P (T_{W}) = \frac{3}{10}

The Conditional Reality

The transfer changes the composition of Bag II. If we transferred a Red ball (

T_{R}

), Bag II now contains 3 Red, 5 Black, and 2 White balls (total 10). The probability of drawing a black ball (

A

) given this transfer is:

P (A ∣ T_{R}) = \frac{5}{10}

If we transferred a Black ball (

T_{B}

), Bag II now contains 2 Red, 6 Black, and 2 White balls. The probability of drawing a black ball is:

P (A ∣ T_{B}) = \frac{6}{10}

If we transferred a White ball (

T_{W}

), Bag II contains 2 Red, 5 Black, and 3 White balls. The probability of drawing a black ball is:

P (A ∣ T_{W}) = \frac{5}{10}

The Bayesian Engine

We want to find

P (T_{R} ∣ A)

, the probability that the transferred ball was red given that the drawn ball was black. Bayes' Theorem provides the engine for this calculation:

P (T_{R} ∣ A) = \frac{P ( T _{R} ) P ( A ∣ T _{R} )}{P ( T _{R} ) P ( A ∣ T _{R} ) + P ( T _{B} ) P ( A ∣ T _{B} ) + P ( T _{W} ) P ( A ∣ T _{W} )}

This formula takes the specific path we care about (Red transfer leading to Black draw) and divides it by the sum of all possible paths that could lead to a Black draw.

Final Calculation

Plugging in our values, we get:

P (T_{R} ∣ A) = \frac{( \frac{3}{10} ) ( \frac{5}{10} )}{( \frac{3}{10} ) ( \frac{5}{10} ) + ( \frac{4}{10} ) ( \frac{6}{10} ) + ( \frac{3}{10} ) ( \frac{5}{10} )}

Since every term has a denominator of 100, we can cancel them out:

P (T_{R} ∣ A) = \frac{3 \times 5}{( 3 \times 5 ) + ( 4 \times 6 ) + ( 3 \times 5 )}

Simplifying the arithmetic:

P (T_{R} ∣ A) = \frac{15}{15 + 24 + 15} = \frac{15}{54}

Dividing both the numerator and denominator by 3, we arrive at the final result:

$P (T_{R} ∣ A) = \frac{5}{18}$

Common Traps & FAQs

Wait, why does the total number of balls in Bag II change? I thought it was always 9.

That is a classic trap! When you transfer a ball from Bag I to Bag II, you are physically adding an object to Bag II. The total count increases from 9 to 10. Always update your sample space after the transfer!

Do I really need to calculate the probability of the white ball transfer? It doesn't add a black ball.

Yes, you absolutely do! Even though a white ball doesn't increase the count of black balls, it increases the total number of balls in the bag. This changes the denominator, which affects the overall probability. Never ignore a branch in your tree diagram.

Why can't we just use simple probability?

Simple probability works for forward-moving events. Here, we are moving backward in time—we know the result (a black ball) and want to infer the cause (the transferred ball). That is the exact definition of a Bayesian problem.

Select Answer:

Visualized Solution

Visualizing the Initial Setup

Defining the Experiment

Prior Probabilities of Transfer

Case 1: Red Ball Transferred

Case 2: Black Ball Transferred

Case 3: White Ball Transferred

Applying Bayes' Theorem

Substituting the Values

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Mapping the Branches

The Conditional Reality

The Bayesian Engine

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 $B_{1}$ 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:

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)

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:

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

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

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Select Answer:

Visualized Solution

Visualizing the Initial Setup

Defining the Experiment

Prior Probabilities of Transfer

Case 1: Red Ball Transferred

Case 2: Black Ball Transferred

Case 3: White Ball Transferred

Applying Bayes' Theorem

Substituting the Values

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Mapping the Branches

The Conditional Reality

The Bayesian Engine

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 :

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

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls 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