There are 5 red and 5 black balls in first bag and 6 red and 4 black balls is second bag. One -one ball is drawn from each bag. Find the probability that:
(i)both balls are of the same colour,
(ii) one ball is red and other is black.

Explanation

To solve the problem step by step, we will calculate the probabilities for both parts of the question.

### Given:
- **Bag 1**: 5 red balls and 5 black balls (Total = 10 balls)
- **Bag 2**: 6 red balls and 4 black balls (Total = 10 balls)

### Part (i): Probability that both balls are of the same color

1. **Calculate the probability of both balls being black**:
- Probability of drawing a black ball from Bag 1 = Number of black balls in Bag 1 / Total balls in Bag 1 = \( \frac{5}{10} = \frac{1}{2} \)
- Probability of drawing a black ball from Bag 2 = Number of black balls in Bag 2 / Total balls in Bag 2 = \( \frac{4}{10} = \frac{2}{5} \)
- Therefore, the combined probability of both balls being black:
\[ P(\text{Black, Black}) = P(\text{Black from Bag 1}) \times P(\text{Black from Bag 2}) = \frac{1}{2} \times \frac{2}{5} = \frac{2}{10} = \frac{1}{5} \]

2. **Calculate the probability of both balls being red**:
- Probability of drawing a red ball from Bag 1 = Number of red balls in Bag 1 / Total balls in Bag 1 = \( \frac{5}{10} = \frac{1}{2} \)
- Probability of drawing a red ball from Bag 2 = Number of red balls in Bag 2 / Total balls in Bag 2 = \( \frac{6}{10} = \frac{3}{5} \)
- Therefore, the combined probability of both balls being red:
\[ P(\text{Red, Red}) = P(\text{Red from Bag 1}) \times P(\text{Red from Bag 2}) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \]

3. **Total probability of both balls being the same color**:
- Add the probabilities of both scenarios:
\[ P(\text{Same Color}) = P(\text{Black, Black}) + P(\text{Red, Red}) = \frac{1}{5} + \frac{3}{10} \]
- Convert \( \frac{1}{5} \) to a fraction with a denominator of 10:
\[ \frac{1}{5} = \frac{2}{10} \]
- Therefore:
\[ P(\text{Same Color}) = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} \]

### Part (ii): Probability that one ball is red and the other is black

1. **Calculate the probability of drawing a black ball from Bag 1 and a red ball from Bag 2**:
- Probability of drawing a black ball from Bag 1 = \( \frac{5}{10} = \frac{1}{2} \)
- Probability of drawing a red ball from Bag 2 = \( \frac{6}{10} = \frac{3}{5} \)
- Therefore:
\[ P(\text{Black from Bag 1, Red from Bag 2}) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \]

2. **Calculate the probability of drawing a red ball from Bag 1 and a black ball from Bag 2**:
- Probability of drawing a red ball from Bag 1 = \( \frac{5}{10} = \frac{1}{2} \)
- Probability of drawing a black ball from Bag 2 = \( \frac{4}{10} = \frac{2}{5} \)
- Therefore:
\[ P(\text{Red from Bag 1, Black from Bag 2}) = \frac{1}{2} \times \frac{2}{5} = \frac{2}{10} = \frac{1}{5} \]

3. **Total probability of one ball being red and the other being black**:
- Add the probabilities of both scenarios:
\[ P(\text{One Red, One Black}) = P(\text{Black from Bag 1, Red from Bag 2}) + P(\text{Red from Bag 1, Black from Bag 2}) = \frac{3}{10} + \frac{1}{5} \]
- Convert \( \frac{1}{5} \) to a fraction with a denominator of 10:
\[ \frac{1}{5} = \frac{2}{10} \]
- Therefore:
\[ P(\text{One Red, One Black}) = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2} \]

### Final Answers:
- (i) The probability that both balls are of the same color is \( \frac{1}{2} \).
- (ii) The probability that one ball is red and the other is black is \( \frac{1}{2} \).