Step-by-step Solution:

Let's define some events to make this clearer.
Defining Events:
Let \( R \) be the event that a red ball is transferred from Bag I to Bag II.
Let \( B \) be the event that a black ball is transferred from Bag I to Bag II.
Let \( W \) be the event that a white ball is transferred from Bag I to Bag II.
Let \( D \) be the event that a black ball is drawn from Bag II after the transfer.
We need to find \( P(R | D) \), the probability that the transferred ball was red given that a black ball was drawn from Bag II.
Understanding the Problem:
First, we transfer one ball from Bag I to Bag II. Then, we draw one ball from Bag II, and it's black. We need to find the probability that the transferred ball was red.
This is a classic conditional probability problem, and we can use Bayes' Theorem to solve it.
Bayes' Theorem: Bayes' Theorem states that: \[ P(R | D) = \frac{P(D | R) \cdot P(R)}{P(D)} \] where:
\( P(D | R) \) is the probability of drawing a black ball from Bag II given that a red ball was transferred.
\( P(R) \) is the probability of transferring a red ball from Bag I.
\( P(D) \) is the total probability of drawing a black ball from Bag II.
So, we need to find these three probabilities.
Step 1: Find \( P(R) \), the probability of transferring a red ball from Bag I.
Bag I has:
3 red balls
4 black balls
3 white balls
Total balls in Bag I: \( 3 + 4 + 3 = 10 \)
The probability of transferring a red ball: \[ P(R) = \frac{\text{Number of red balls in Bag I}}{\text{Total balls in Bag I}} = \frac{3}{10} \] Step 2: Find \( P(D | R) \), the probability of drawing a black ball from Bag II given that a red ball was transferred.
After transferring a red ball from Bag I to Bag II, Bag II will have:
Original: 2 red, 5 black, 2 white
After transfer: 3 red, 5 black, 2 white
Total balls in Bag II after transfer: \( 3 + 5 + 2 = 10 \)
The probability of drawing a black ball from Bag II: \[ P(D | R) = \frac{\text{Number of black balls in Bag II after transfer}}{\text{Total balls in Bag II after transfer}} = \frac{5}{10} = \frac{1}{2} \] Step 3: Find \( P(D) \), the total probability of drawing a black ball from Bag II.
To find \( P(D) \), we need to consider all possible ways a black ball can be drawn from Bag II, which depends on what type of ball was transferred from Bag I.
There are three scenarios:
1. A red ball was transferred (\( R \))
2. A black ball was transferred (\( B \))
3. A white ball was transferred (\( W \))
So, \[ P(D) = P(D | R) \cdot P(R) + P(D | B) \cdot P(B) + P(D | W) \cdot P(W) \] We already have \( P(R) = \frac{3}{10} \) and \( P(D | R) = \frac{1}{2} \).
Now, let's find \( P(B) \) and \( P(D | B) \), and \( P(W) \) and \( P(D | W) \).
Finding \( P(B) \):
The probability of transferring a black ball from Bag I: \[ P(B) = \frac{\text{Number of black balls in Bag I}}{\text{Total balls in Bag I}} = \frac{4}{10} = \frac{2}{5} \] Finding \( P(D | B) \):
After transferring a black ball from Bag I to Bag II, Bag II will have:
Original: 2 red, 5 black, 2 white
After transfer: 2 red, 6 black, 2 white
Total balls in Bag II after transfer: \( 2 + 6 + 2 = 10 \)
The probability of drawing a black ball from Bag II: \[ P(D | B) = \frac{\text{Number of black balls in Bag II after transfer}}{\text{Total balls in Bag II after transfer}} = \frac{6}{10} = \frac{3}{5} \] Finding \( P(W) \):
The probability of transferring a white ball from Bag I: \[ P(W) = \frac{\text{Number of white balls in Bag I}}{\text{Total balls in Bag I}} = \frac{3}{10} \] Finding \( P(D | W) \):
After transferring a white ball from Bag I to Bag II, Bag II will have:
Original: 2 red, 5 black, 2 white
After transfer: 2 red, 5 black, 3 white
Total balls in Bag II after transfer: \( 2 + 5 + 3 = 10 \)
The probability of drawing a black ball from Bag II: \[ P(D | W) = \frac{\text{Number of black balls in Bag II after transfer}}{\text{Total balls in Bag II after transfer}} = \frac{5}{10} = \frac{1}{2} \] Now, plug all these into the formula for \( P(D) \):
\[ P(D) = P(D | R) \cdot P(R) + P(D | B) \cdot P(B) + P(D | W) \cdot P(W) \] \[ P(D) = \left( \frac{1}{2} \times \frac{3}{10} \right) + \left( \frac{3}{5} \times \frac{2}{5} \right) + \left( \frac{1}{2} \times \frac{3}{10} \right) \] Let's calculate each term:
1. \( \frac{1}{2} \times \frac{3}{10} = \frac{3}{20} \)
2. \( \frac{3}{5} \times \frac{2}{5} = \frac{6}{25} \)
3. \( \frac{1}{2} \times \frac{3}{10} = \frac{3}{20} \)
Now, add them up: \[ P(D) = \frac{3}{20} + \frac{6}{25} + \frac{3}{20} \] To add these fractions, we need a common denominator. The least common multiple of 20 and 25 is 100.
Convert each fraction:
1. \( \frac{3}{20} = \frac{15}{100} \)
2. \( \frac{6}{25} = \frac{24}{100} \)
3. \( \frac{3}{20} = \frac{15}{100} \)
Now, add them: \[ P(D) = \frac{15}{100} + \frac{24}{100} + \frac{15}{100} = \frac{54}{100} = \frac{27}{50} \] Step 4: Apply Bayes' Theorem to find \( P(R | D) \).
We have: \[ P(R | D) = \frac{P(D | R) \cdot P(R)}{P(D)} = \frac{ \left( \frac{1}{2} \times \frac{3}{10} \right) }{ \frac{27}{50} } = \frac{ \frac{3}{20} }{ \frac{27}{50} } \] To divide by a fraction, multiply by its reciprocal: \[ P(R | D) = \frac{3}{20} \times \frac{50}{27} = \frac{3 \times 50}{20 \times 27} = \frac{150}{540} \] Simplify the fraction by dividing numerator and denominator by 30: \[ \frac{150}{540} = \frac{5}{18} \] Final Answer: The probability that the transferred ball was red, given that a black ball was drawn from Bag II, is: \[ {\dfrac{5}{18}} \]