One bag contains 5 white and 3 red balls, and a second bag contains 4 white and 5 red balls. From one of them chosen at random two balls are drawn, find the chance that they are of different colours.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two bags:
- Bag 1: 5 white balls and 3 red balls
- Bag 2: 4 white balls and 5 red balls

We need to find the probability that when we randomly choose one of the bags and then draw two balls from it, the balls will be of different colors.

### Step 2: Calculate the Probability of Choosing Each Bag
Since we are choosing one of the two bags at random, the probability of choosing either bag is:
- \( P(B1) = \frac{1}{2} \)
- \( P(B2) = \frac{1}{2} \)

### Step 3: Calculate the Probability of Drawing Two Balls of Different Colors from Bag 1
From Bag 1:
- Total balls = 5 white + 3 red = 8 balls
- The number of ways to choose 2 balls from 8 is \( \binom{8}{2} = \frac{8 \times 7}{2} = 28 \).

Now, we need to find the favorable outcomes for drawing one white and one red ball:
- The number of ways to choose 1 white ball from 5 is \( \binom{5}{1} = 5 \).
- The number of ways to choose 1 red ball from 3 is \( \binom{3}{1} = 3 \).
- Therefore, the total favorable outcomes for Bag 1 is \( 5 \times 3 = 15 \).

Thus, the probability of drawing two balls of different colors from Bag 1 is:
\[ P(Different | B1) = \frac{15}{28} \]

### Step 4: Calculate the Probability of Drawing Two Balls of Different Colors from Bag 2
From Bag 2:
- Total balls = 4 white + 5 red = 9 balls
- The number of ways to choose 2 balls from 9 is \( \binom{9}{2} = \frac{9 \times 8}{2} = 36 \).

Now, we need to find the favorable outcomes for drawing one white and one red ball:
- The number of ways to choose 1 white ball from 4 is \( \binom{4}{1} = 4 \).
- The number of ways to choose 1 red ball from 5 is \( \binom{5}{1} = 5 \).
- Therefore, the total favorable outcomes for Bag 2 is \( 4 \times 5 = 20 \).

Thus, the probability of drawing two balls of different colors from Bag 2 is:
\[ P(Different | B2) = \frac{20}{36} = \frac{5}{9} \]

### Step 5: Calculate the Total Probability of Drawing Two Balls of Different Colors
Using the law of total probability:
\[ P(Different) = P(B1) \cdot P(Different | B1) + P(B2) \cdot P(Different | B2) \]
Substituting the values we calculated:
\[ P(Different) = \frac{1}{2} \cdot \frac{15}{28} + \frac{1}{2} \cdot \frac{5}{9} \]
\[ = \frac{15}{56} + \frac{5}{18} \]

### Step 6: Find a Common Denominator and Add the Fractions
The least common multiple of 56 and 18 is 504. We convert each fraction:
\[ \frac{15}{56} = \frac{15 \times 9}{56 \times 9} = \frac{135}{504} \]
\[ \frac{5}{18} = \frac{5 \times 28}{18 \times 28} = \frac{140}{504} \]
Now, adding these fractions:
\[ P(Different) = \frac{135}{504} + \frac{140}{504} = \frac{275}{504} \]

### Final Answer
The probability that the two balls drawn are of different colors is:
\[ \boxed{\frac{275}{504}} \]