Bag A contains 1 white, 2 blue and 3 red balls. Bag B contains 3 white, 3 blue and 4 red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red.

Answer

To find the probability that the balls drawn are one white and one red, we will follow these steps:

### Step 1: Identify the contents of each bag
- **Bag A** contains:
- 1 white ball
- 2 blue balls
- 3 red balls
- **Total = 1 + 2 + 3 = 6 balls**

- **Bag B** contains:
- 3 white balls
- 3 blue balls
- 4 red balls
- **Total = 3 + 3 + 4 = 10 balls**

### Step 2: Determine the probabilities of selecting each bag
Since one bag is selected at random:
- Probability of selecting Bag A, \( P(E_1) = \frac{1}{2} \)
- Probability of selecting Bag B, \( P(E_2) = \frac{1}{2} \)

### Step 3: Calculate the probability of drawing one white and one red ball from Bag A
To find the probability of drawing one white and one red ball from Bag A:
- The number of ways to choose 1 white from 1 white: \( \binom{1}{1} = 1 \)
- The number of ways to choose 1 red from 3 red: \( \binom{3}{1} = 3 \)
- Total ways to choose 2 balls from Bag A: \( \binom{6}{2} = 15 \)

Thus, the probability of drawing one white and one red ball from Bag A:
\[ P(F | E_1) = \frac{\binom{1}{1} \cdot \binom{3}{1}}{\binom{6}{2}} = \frac{1 \cdot 3}{15} = \frac{3}{15} = \frac{1}{5} \]

### Step 4: Calculate the probability of drawing one white and one red ball from Bag B
To find the probability of drawing one white and one red ball from Bag B:
- The number of ways to choose 1 white from 3 white: \( \binom{3}{1} = 3 \)
- The number of ways to choose 1 red from 4 red: \( \binom{4}{1} = 4 \)
- Total ways to choose 2 balls from Bag B: \( \binom{10}{2} = 45 \)

Thus, the probability of drawing one white and one red ball from Bag B:
\[ P(F | E_2) = \frac{\binom{3}{1} \cdot \binom{4}{1}}{\binom{10}{2}} = \frac{3 \cdot 4}{45} = \frac{12}{45} = \frac{4}{15} \]

### Step 5: Use the law of total probability to find \( P(F) \)
Now, we can find the total probability of drawing one white and one red ball:
\[ P(F) = P(E_1) \cdot P(F | E_1) + P(E_2) \cdot P(F | E_2) \]
Substituting the values:
\[ P(F) = \left(\frac{1}{2} \cdot \frac{1}{5}\right) + \left(\frac{1}{2} \cdot \frac{4}{15}\right) \]
\[ P(F) = \frac{1}{10} + \frac{4}{30} \]
To add these fractions, convert \( \frac{1}{10} \) to a common denominator of 30:
\[ P(F) = \frac{3}{30} + \frac{4}{30} = \frac{7}{30} \]

### Final Answer
Thus, the probability that the balls drawn are one white and one red is:
\[ \boxed{\frac{7}{30}} \]