A bag contains 5 white, 6 red and 4 green balls. One ball is drawn at random. What is the probability that the ball drawn is (i) green? (ii) white? (iii) non -red?

Answer

To solve the problem step by step, let's break it down into three parts as mentioned in the question.

### Given:
- Number of white balls = 5
- Number of red balls = 6
- Number of green balls = 4

### Total number of balls:
Total = Number of white balls + Number of red balls + Number of green balls
Total = 5 + 6 + 4 = 15

### (i) Probability that the ball drawn is green:
1. **Total outcomes**: The total number of balls is 15.
2. **Favorable outcomes for green**: The number of green balls is 4.
3. **Probability of drawing a green ball**:
\[ P(\text{Green}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{4}{15} \]

### (ii) Probability that the ball drawn is white:
1. **Total outcomes**: The total number of balls is still 15.
2. **Favorable outcomes for white**: The number of white balls is 5.
3. **Probability of drawing a white ball**:
\[ P(\text{White}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{15} = \frac{1}{3} \]

### (iii) Probability that the ball drawn is non-red:
1. **Total outcomes**: The total number of balls is 15.
2. **Favorable outcomes for non-red**: Non-red balls include white and green balls.
- Number of white balls = 5
- Number of green balls = 4
- Total non-red balls = 5 + 4 = 9
3. **Probability of drawing a non-red ball**:
\[ P(\text{Non-red}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{9}{15} = \frac{3}{5} \]

### Summary of the Answers:
- (i) Probability of drawing a green ball = \(\frac{4}{15}\)
- (ii) Probability of drawing a white ball = \(\frac{1}{3}\)
- (iii) Probability of drawing a non-red ball = \(\frac{3}{5}\)