A bag contians 4 white and 2 black balls and another bag contain 3 white and 5 black balls. If one ball is drawn from each bag, then the probability that one ball is white and one ball is black is

Options

A. `(5)/(24)`
B. `(13)/(24)`
C. `(1)/(4)`
D. `(2)/(3)`

Explanation

To solve the problem, we need to find the probability that one ball drawn from each of the two bags is white and the other is black. We will break this down into steps.

### Step 1: Identify the contents of each bag
- **Bag 1** contains 4 white balls and 2 black balls.
- **Bag 2** contains 3 white balls and 5 black balls.

### Step 2: Calculate the total number of balls in each bag
- Total balls in Bag 1 = 4 (white) + 2 (black) = 6 balls.
- Total balls in Bag 2 = 3 (white) + 5 (black) = 8 balls.

### Step 3: Define the two cases for drawing one white and one black ball
There are two possible cases for drawing one white and one black ball:
1. Draw a white ball from Bag 1 and a black ball from Bag 2.
2. Draw a black ball from Bag 1 and a white ball from Bag 2.

### Step 4: Calculate the probability for each case

**Case 1: White from Bag 1 and Black from Bag 2**
- Probability of drawing a white ball from Bag 1:
\[ P(W_1) = \frac{4}{6} = \frac{2}{3} \]
- Probability of drawing a black ball from Bag 2:
\[ P(B_2) = \frac{5}{8} \]
- Combined probability for Case 1:
\[ P(W_1 \text{ and } B_2) = P(W_1) \times P(B_2) = \frac{2}{3} \times \frac{5}{8} = \frac{10}{24} \]

**Case 2: Black from Bag 1 and White from Bag 2**
- Probability of drawing a black ball from Bag 1:
\[ P(B_1) = \frac{2}{6} = \frac{1}{3} \]
- Probability of drawing a white ball from Bag 2:
\[ P(W_2) = \frac{3}{8} \]
- Combined probability for Case 2:
\[ P(B_1 \text{ and } W_2) = P(B_1) \times P(W_2) = \frac{1}{3} \times \frac{3}{8} = \frac{3}{24} \]

### Step 5: Add the probabilities of both cases
Now we add the probabilities of both cases to find the total probability of drawing one white and one black ball:
\[ P(\text{one white and one black}) = P(W_1 \text{ and } B_2) + P(B_1 \text{ and } W_2 = \frac{10}{24} + \frac{3}{24} = \frac{13}{24} \]

### Final Answer
The probability that one ball is white and one ball is black is:
\[ \frac{13}{24} \]