Bag A contain 4 white balls and 3 black balls, while Bag B contain 3 white balls and 5 black balls. Two balls are drawn from Bag A and placed in bag B. Then, what is the probability of drawing a white ball from bag B ?

Answer

To solve the problem step by step, we will analyze the situation and calculate the probability of drawing a white ball from Bag B after transferring two balls from Bag A.

### Step 1: Understand the contents of Bag A and Bag B
- Bag A contains 4 white balls and 3 black balls.
- Bag B contains 3 white balls and 5 black balls.

### Step 2: Identify the possible cases for drawing balls from Bag A
When we draw two balls from Bag A, there are three possible cases:
1. Case 1: Draw 2 white balls.
2. Case 2: Draw 2 black balls.
3. Case 3: Draw 1 white ball and 1 black ball.

### Step 3: Calculate the probabilities for each case

#### Case 1: Drawing 2 white balls
- The number of ways to choose 2 white balls from 4 is given by \( \binom{4}{2} \).
- After transferring these 2 white balls to Bag B, Bag B will have \( 3 + 2 = 5 \) white balls and 5 black balls (total = 10 balls).
- The probability of drawing a white ball from Bag B is \( \frac{5}{10} = \frac{1}{2} \).

**Probability for Case 1:**
\[ P_1 = \frac{\binom{4}{2}}{\binom{7}{2}} \cdot \frac{5}{10} = \frac{6}{21} \cdot \frac{1}{2} = \frac{3}{21} = \frac{1}{7} \]

#### Case 2: Drawing 2 black balls
- The number of ways to choose 2 black balls from 3 is given by \( \binom{3}{2} \).
- After transferring these 2 black balls to Bag B, Bag B will have 3 white balls and \( 5 + 2 = 7 \) black balls (total = 10 balls).
- The probability of drawing a white ball from Bag B is \( \frac{3}{10} \).

**Probability for Case 2:**
\[ P_2 = \frac{\binom{3}{2}}{\binom{7}{2}} \cdot \frac{3}{10} = \frac{3}{21} \cdot \frac{3}{10} = \frac{9}{210} = \frac{3}{70} \]

#### Case 3: Drawing 1 white ball and 1 black ball
- The number of ways to choose 1 white ball from 4 and 1 black ball from 3 is given by \( \binom{4}{1} \cdot \binom{3}{1} \).
- After transferring these balls to Bag B, Bag B will have \( 3 + 1 = 4 \) white balls and \( 5 + 1 = 6 \) black balls (total = 10 balls).
- The probability of drawing a white ball from Bag B is \( \frac{4}{10} = \frac{2}{5} \).

**Probability for Case 3:**
\[ P_3 = \frac{\binom{4}{1} \cdot \binom{3}{1}}{\binom{7}{2}} \cdot \frac{4}{10} = \frac{12}{21} \cdot \frac{2}{5} = \frac{24}{105} = \frac{8}{35} \]

### Step 4: Combine the probabilities from all cases
Now, we add the probabilities from all three cases to find the total probability of drawing a white ball from Bag B.

\[ P = P_1 + P_2 + P_3 = \frac{1}{7} + \frac{3}{70} + \frac{8}{35} \]

To add these fractions, we need a common denominator, which is 70.

\[ P = \frac{10}{70} + \frac{3}{70} + \frac{16}{70} = \frac{29}{70} \]

### Final Answer
The probability of drawing a white ball from Bag B is \( \frac{29}{70} \).

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