There 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

### (i) Probability that both balls are black

1. **Calculate the probability of drawing the first black ball:**
- There are 10 black balls out of 18 total balls.
- Probability of first ball being black = \( \frac{10}{18} \)

2. **Calculate the probability of drawing the second black ball:**
- After drawing one black ball, there are now 9 black balls left and a total of 17 balls remaining.
- Probability of second ball being black = \( \frac{9}{17} \)

3. **Calculate the combined probability:**
- Probability that both balls are black = Probability of first black × Probability of second black
- \( P(\text{both black}) = \frac{10}{18} \times \frac{9}{17} = \frac{90}{306} = \frac{5}{17} \)

### (ii) Probability that the first ball is black and the second is white

1. **Calculate the probability of drawing the first black ball:**
- Probability of first ball being black = \( \frac{10}{18} \)

2. **Calculate the probability of drawing the second white ball:**
- After drawing one black ball, there are still 8 white balls left and a total of 17 balls remaining.
- Probability of second ball being white = \( \frac{8}{17} \)

3. **Calculate the combined probability:**
- Probability that first ball is black and second is white = Probability of first black × Probability of second white
- \( P(\text{first black, second white}) = \frac{10}{18} \times \frac{8}{17} = \frac{80}{306} = \frac{40}{153} \)

### (iii) Probability that one ball is black and the other is white

1. **Calculate the probability of the first ball being black and the second being white (already calculated):**
- \( P(\text{first black, second white}) = \frac{40}{153} \)

2. **Calculate the probability of the first ball being white and the second being black:**
- Probability of first ball being white = \( \frac{8}{18} \)
- After drawing one white ball, there are still 10 black balls left and a total of 17 balls remaining.
- Probability of second ball being black = \( \frac{10}{17} \)
- Probability that first ball is white and second is black = \( \frac{8}{18} \times \frac{10}{17} = \frac{80}{306} = \frac{40}{153} \)

3. **Combine the probabilities:**
- Total probability that one ball is black and the other is white = Probability (first black, second white) + Probability (first white, second black)
- \( P(\text{one black, one white}) = \frac{40}{153} + \frac{40}{153} = \frac{80}{153} \)

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