A bag contains 7 green, 10 blue and 5 red balls. A ball is drawn at random. Find the probability of this ball being a:
(i) blue ball.
(ii) red ball or a green ball.
(iii) not a green ball.

Answer

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

### Given:
- Number of green balls = 7
- Number of blue balls = 10
- Number of red balls = 5

### Total number of balls:
\[ \text{Total balls} = \text{Green} + \text{Blue} + \text{Red} = 7 + 10 + 5 = 22 \]

### (i) Probability of drawing a blue ball:
- Favorable outcomes for blue balls = 10
- Total outcomes = 22

The probability \( P(\text{Blue}) \) is given by:
\[ P(\text{Blue}) = \frac{\text{Favorable outcomes for blue}}{\text{Total outcomes}} = \frac{10}{22} \]
Simplifying this fraction:
\[ P(\text{Blue}) = \frac{5}{11} \]

### (ii) Probability of drawing a red ball or a green ball:
- Favorable outcomes for red balls = 5
- Favorable outcomes for green balls = 7
- Total favorable outcomes for red or green = 5 + 7 = 12

The probability \( P(\text{Red or Green}) \) is given by:
\[ P(\text{Red or Green}) = \frac{\text{Favorable outcomes for red or green}}{\text{Total outcomes}} = \frac{12}{22} \]
Simplifying this fraction:
\[ P(\text{Red or Green}) = \frac{6}{11} \]

### (iii) Probability of not drawing a green ball:
- Total balls = 22
- Number of green balls = 7
- Therefore, the number of balls that are not green = Total balls - Green balls = 22 - 7 = 15

The probability \( P(\text{Not Green}) \) is given by:
\[ P(\text{Not Green}) = \frac{\text{Favorable outcomes for not green}}{\text{Total outcomes}} = \frac{15}{22} \]

### Final Answers:
1. Probability of drawing a blue ball = \( \frac{5}{11} \)
2. Probability of drawing a red ball or a green ball = \( \frac{6}{11} \)
3. Probability of not drawing a green ball = \( \frac{15}{22} \)