There are two bage. One bag contains 4 white and 2 black balls. Second bag contains 5 white and 4 black balls. Two balls are transferred from first bag to second bag. Then one ball is taken from the second bag. Find the probability that it is a white ball.

Answer

To solve the problem step-by-step, we will analyze the situation and calculate the probability that a ball drawn from the second bag is white after transferring two balls from the first bag.

### Step-by-Step Solution:

1. **Identify the contents of each bag:**
- Bag 1: 4 white balls and 2 black balls (Total = 6 balls)
- Bag 2: 5 white balls and 4 black balls (Total = 9 balls)

2. **Transfer two balls from Bag 1 to Bag 2:**
We need to consider three possible cases for the transfer of the two balls:
- Case 1: Transfer 2 black balls.
- Case 2: Transfer 2 white balls.
- Case 3: Transfer 1 white ball and 1 black ball.

3. **Calculate the probabilities for each case:**

**Case 1: Transfer 2 black balls**
- The number of ways to choose 2 black balls from 2 black balls:
\[ \text{Ways} = \binom{2}{2} = 1 \]
- The total ways to choose any 2 balls from 6:
\[ \text{Total ways} = \binom{6}{2} = 15 \]
- Probability of transferring 2 black balls:
\[ P(BB) = \frac{1}{15} \]
- After transferring, Bag 2 will have:
- White: 5
- Black: 6 (4 original + 2 transferred)
- Probability of drawing a white ball from Bag 2:
\[ P(\text{White} | BB) = \frac{5}{11} \]

**Case 2: Transfer 2 white balls**
- The number of ways to choose 2 white balls from 4 white balls:
\[ \text{Ways} = \binom{4}{2} = 6 \]
- Probability of transferring 2 white balls:
\[ P(WW) = \frac{6}{15} = \frac{2}{5} \]
- After transferring, Bag 2 will have:
- White: 7 (5 original + 2 transferred)
- Black: 4
- Probability of drawing a white ball from Bag 2:
\[ P(\text{White} | WW) = \frac{7}{11} \]

**Case 3: Transfer 1 white ball and 1 black ball**
- The number of ways to choose 1 white from 4 and 1 black from 2:
\[ \text{Ways} = \binom{4}{1} \cdot \binom{2}{1} = 4 \cdot 2 = 8 \]
- Probability of transferring 1 white and 1 black:
\[ P(WB) = \frac{8}{15} \]
- After transferring, Bag 2 will have:
- White: 6 (5 original + 1 transferred)
- Black: 5 (4 original + 1 transferred)
- Probability of drawing a white ball from Bag 2:
\[ P(\text{White} | WB) = \frac{6}{11} \]

4. **Combine the probabilities using the law of total probability:**
\[ P(\text{White}) = P(\text{White} | BB) \cdot P(BB) + P(\text{White} | WW) \cdot P(WW) + P(\text{White} | WB) \cdot P(WB) \]
Substituting the values:
\[ P(\text{White}) = \left(\frac{5}{11} \cdot \frac{1}{15}\right) + \left(\frac{7}{11} \cdot \frac{2}{5}\right) + \left(\frac{6}{11} \cdot \frac{8}{15}\right) \]
\[ = \frac{5}{165} + \frac{14}{55} + \frac{48}{165} \]
Convert \(\frac{14}{55}\) to a common denominator:
\[ \frac{14}{55} = \frac{42}{165} \]
Now combine:
\[ P(\text{White}) = \frac{5 + 42 + 48}{165} = \frac{95}{165} = \frac{19}{33} \]

### Final Answer:
The probability that the ball drawn from the second bag is white is \(\frac{19}{33}\).