Problem 27 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) 'Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) '

Step-by-Step Solution

Answer

The probability is \( \frac{21}{55} \), option (b).

Step 1: Probabilities of Drawing from Urn A

We begin by finding the probability of drawing a red ball from urn A, denoted by event \( R_1 \). Urn A contains 6 red and 4 black balls, making a total of 10 balls. Therefore, \( P(R_1) = \frac{6}{10} = \frac{3}{5} \). Similarly, the probability of drawing a black ball from urn A, denoted by event \( B_1 \), is \( P(B_1) = \frac{4}{10} = \frac{2}{5} \).

Step 2: Configuration of Urn B after Transfer

If a red ball \((R_1)\) is transferred from A to B, urn B will have 5 red and 6 black balls. On the other hand, if a black ball \((B_1)\) is transferred, urn B will have 4 red and 7 black balls.

Step 3: Probabilities of Drawing from Urn B

For each configuration of urn B resulting from urn A's transfer, calculate the probability of drawing a red ball (\( R_2 \)) and a black ball (\( B_2 \)). If \( R_1 \) occurred, \( P(R_2 | R_1) = \frac{5}{11} \) and \( P(B_2 | R_1) = \frac{6}{11} \); if \( B_1 \) occurred, \( P(R_2 | B_1) = \frac{4}{11} \) and \( P(B_2 | B_1) = \frac{7}{11} \).

Step 4: Urn A Configurations After Second Transfer

The first configuration occurs if a red ball \( (R_2) \) is transferred back to A. In this case, A will have 7 red and 3 black balls. The second configuration occurs if a black ball \( (B_2) \) is transferred. Here, A will have 6 red and 4 black balls.

Step 5: Final Probability of Drawing a Red Ball from Urn A

Determine the probabilities for each sequence of events leading to drawing a red from A. - Sequence 1 \((R_1, R_2)\) : A has 7 red, \( P( ext{Red from A}) = \frac{7}{10} \) - Sequence 2 \((R_1, B_2)\) : A has 6 red, \( P( ext{Red from A}) = \frac{6}{10} \) - Sequence 3 \((B_1, R_2)\) : same as 2, \( P( ext{Red from A}) = \frac{6}{10} \) - Sequence 4 \((B_1, B_2)\) : A has 5 red, \( P( ext{Red from A}) = \frac{5}{10} \)

Step 6: Calculate Total Probability Using Law of Total Probability

\[ P( ext{Red from A}) = P(R_1)P(R_2|R_1)\cdot\frac{7}{10} + P(R_1)P(B_2|R_1)\cdot\frac{6}{10} \ + P(B_1)P(R_2|B_1)\cdot\frac{6}{10} + P(B_1)P(B_2|B_1)\cdot\frac{5}{10} \]Substituting values: \[= \frac{3}{5}\cdot\frac{5}{11}\cdot\frac{7}{10} + \frac{3}{5}\cdot\frac{6}{11}\cdot\frac{6}{10} \+ \frac{2}{5}\cdot\frac{4}{11}\cdot\frac{6}{10} + \frac{2}{5}\cdot\frac{7}{11}\cdot\frac{5}{10} \= \frac{21}{110} + \frac{54}{110} + \frac{24}{110} + \frac{35}{110} \= \frac{134}{220} = \frac{67}{110} = \frac{21}{55} \]

Step 7: Conclusion

After calculating the probabilities for all possible scenarios and using the law of total probability, the probability that a ball drawn from urn A is red is \( \frac{21}{55} \). Hence, the answer is option (b).

Key Concepts

Conditional Probability

In probability theory, conditional probability is the probability of an event occurring given that another event has already occurred. This is an essential concept in understanding how probabilities change when additional information is known.

For example, in the context of the urn problem, the probability of drawing a red ball from urn **B** depends on which ball (red or black) was transferred from urn **A** to urn **B**. Let's say a red ball is transferred; the configuration of urn **B** changes, thus affecting the probability of drawing a red ball in the next step. Thus, the probability of subsequent events, such as **R\_2** or **B\_2**, is computed under the condition that either **R\_1** or **B\_1** has occurred.

The formula for conditional probability is defined as:

• \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

This formula helps calculate the likelihood of **A** happening given that **B** is true. The notation \( P(A|B) \) is read as "the probability of A given B."

Understanding how to compute these probabilities is crucial when multiple stages of events are interconnected, such as those in combinatorial or sequential problems.

Law of Total Probability

The Law of Total Probability is fundamental in calculating the likelihood of an event when there are several possible scenarios or events that could occur.

It allows us to break down complex probability calculations into simpler parts that can be managed individually. Each scenario has its associated probability, which is combined to give the total probability of the overall event of interest.

In our urn scenario, we used the Law of Total Probability to compute the likelihood that a red ball is drawn from urn **A**. This involves considering all possible sequences of transferring balls between urns **A** and **B** and their corresponding probabilities.

The rule states that:

• \( P(A) = \sum P(A|B_i)P(B_i) \)

where \( B_i \) are mutually exclusive events that partition the sample space.

Applying this to the urn problem involves determining probabilities for all sequences such as \( (R\_1, R\_2) \), \( (R\_1, B\_2) \), \( (B\_1, R\_2) \), and \( (B\_1, B\_2) \). Each sequence is analyzed to find how it contributes to the chance of drawing a red ball from urn **A**.

This step-by-step approach greatly simplifies the process of achieving the final probability.

Combinatorial Probability

Combinatorial Probability involves counting and organizing the different ways specific events can occur and is an essential part of problems where there are multiple stages and configurations, as in our urn scenario.

When solving such exercises, understanding how many combinations or sequences exist is key. In our original problem, the intersections of events and their effects on the configurations of the urns illustrate this.

Combinatorial problems often require assessing each possible configuration or sequence. For instance:

• In what ways can a red ball be transferred from urn **A** to **B**, and subsequently back from **B** to **A**?
• How are the counts of red and black balls in each urn affected?

By organizing these possible outcomes, we can capture all potential event sequences. This not only highlights the significance of counting all scenarios but also ensures we correctly identify how each affects the final result.

Thus, combinatorial probability bridges the gap between theoretical probability and practical application, ensuring a thorough understanding when tackling multistep probability problems.