Ther are 10 black and 8 white balls in a bag. Two balls are drawn without replacement. Find the probability that:
(i) both balls are black
(ii) first ball is black and second is white
(iii) one ball is black and other is white.

Explanation

To solve the problem step by step, we will calculate the probabilities for each part of the question.

### Given:
- Number of black balls = 10
- Number of white balls = 8
- Total number of balls = 10 + 8 = 18

### Total ways to draw 2 balls without replacement:
The total number of ways to choose 2 balls from 18 is given by:
\[ \text{Total ways} = 18 \times 17 \]

### (i) Probability that both balls are black:
1. **Choosing the first black ball**: There are 10 options.
2. **Choosing the second black ball**: After choosing the first black ball, there are 9 black balls left.

So, the number of favorable outcomes for both balls being black is:
\[ \text{Favorable outcomes} = 10 \times 9 \]

3. **Probability**:
\[ P(\text{both black}) = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{10 \times 9}{18 \times 17} \]

Calculating this:
\[ P(\text{both black}) = \frac{90}{306} = \frac{5}{17} \]

### (ii) Probability that the first ball is black and the second is white:
1. **Choosing the first black ball**: There are 10 options.
2. **Choosing the second white ball**: There are 8 options.

So, the number of favorable outcomes for the first ball being black and the second being white is:
\[ \text{Favorable outcomes} = 10 \times 8 \]

3. **Probability**:
\[ P(\text{first black, second white}) = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{10 \times 8}{18 \times 17} \]

Calculating this:
\[ P(\text{first black, second white}) = \frac{80}{306} = \frac{40}{153} \]

### (iii) Probability that one ball is black and the other is white:
In this case, we can have two scenarios:
- Scenario 1: First ball is black and second ball is white (already calculated).
- Scenario 2: First ball is white and second ball is black.

1. **Choosing the first white ball**: There are 8 options.
2. **Choosing the second black ball**: There are 10 options.

So, the number of favorable outcomes for the second scenario is:
\[ \text{Favorable outcomes} = 8 \times 10 \]

3. **Total favorable outcomes for one black and one white**:
\[ \text{Total favorable outcomes} = (10 \times 8) + (8 \times 10) = 80 + 80 = 160 \]

4. **Probability**:
\[ P(\text{one black, one white}) = \frac{\text{Total favorable outcomes}}{\text{Total ways}} = \frac{160}{306} = \frac{80}{153} \]

### Summary of Results:
- (i) Probability that both balls are black: \(\frac{5}{17}\)
- (ii) Probability that the first ball is black and the second is white: \(\frac{40}{153}\)
- (iii) Probability that one ball is black and the other is white: \(\frac{80}{153}\)