A bag A has 3 red and 2 black balls, and a bag B had 3 red and 4 black balls. Then one ball is drawn from B and placed in A. if one ball is drawn from A. What is the probability that it is red ?

Answer

To solve the problem step by step, we will calculate the probability of drawing a red ball from bag A after transferring one ball from bag B to bag A.

### Step 1: Identify the initial conditions
- Bag A has 3 red balls and 2 black balls.
- Bag B has 3 red balls and 4 black balls.

### Step 2: Calculate the total number of balls in each bag
- Total balls in Bag A = 3 (red) + 2 (black) = 5 balls.
- Total balls in Bag B = 3 (red) + 4 (black) = 7 balls.

### Step 3: Define events
Let:
- E1 = event that a red ball is transferred from Bag B to Bag A.
- E2 = event that a black ball is transferred from Bag B to Bag A.
- E = event that a red ball is drawn from Bag A after the transfer.

### Step 4: Calculate the probabilities of transferring a ball
- Probability of transferring a red ball (P(E1)):
\[ P(E1) = \frac{3 \text{ (red balls in B)}}{7 \text{ (total balls in B)}} = \frac{3}{7} \]

- Probability of transferring a black ball (P(E2)):
\[ P(E2) = \frac{4 \text{ (black balls in B)}}{7 \text{ (total balls in B)}} = \frac{4}{7} \]

### Step 5: Calculate the probability of drawing a red ball from Bag A given each event
1. If a red ball is transferred (E1):
- Bag A will then have 4 red balls and 2 black balls.
- Total balls in Bag A = 4 + 2 = 6.
- Probability of drawing a red ball (P(E|E1)):
\[ P(E|E1) = \frac{4 \text{ (red balls in A)}}{6 \text{ (total balls in A)}} = \frac{4}{6} = \frac{2}{3} \]

2. If a black ball is transferred (E2):
- Bag A will have 3 red balls and 3 black balls.
- Total balls in Bag A = 3 + 3 = 6.
- Probability of drawing a red ball (P(E|E2)):
\[ P(E|E2) = \frac{3 \text{ (red balls in A)}}{6 \text{ (total balls in A)}} = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Use the law of total probability to find P(E)
Using the law of total probability:
\[ P(E) = P(E1) \cdot P(E|E1) + P(E2) \cdot P(E|E2) \]
Substituting the values:
\[ P(E) = \left(\frac{3}{7} \cdot \frac{2}{3}\right) + \left(\frac{4}{7} \cdot \frac{1}{2}\right) \]
Calculating each term:
- First term:
\[ \frac{3}{7} \cdot \frac{2}{3} = \frac{2}{7} \]
- Second term:
\[ \frac{4}{7} \cdot \frac{1}{2} = \frac{2}{7} \]

Adding both terms:
\[ P(E) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7} \]

### Final Answer
The probability that the ball drawn from Bag A is red is:
\[ \boxed{\frac{4}{7}} \]