Study the following information carefully and answer the given questions.
There were some red, some green and some white balls in a bag. Probability of drawing 1 white ball from the bag is . `(1)/(3)` If there were 6 more red balls in the bag, the probability of getting 1 green ball would have been `(1)/(30)` less.
If Probability of drawing 1 green ball from the bag is `(1)/(5)` , then what is the probability of drawing 2 balls such that none of them is red?

Options

A. `3//29`
B. `4//29`
C. `8//29`
D. `8//87`

Explanation

To solve the problem, we need to find the probability of drawing 2 balls such that none of them is red. Let's break down the information provided step by step.

### Step 1: Define Variables
Let:
- \( R \) = number of red balls
- \( G \) = number of green balls
- \( W \) = number of white balls

### Step 2: Use the Given Probability of White Balls
The probability of drawing 1 white ball from the bag is given as \( \frac{1}{3} \). This can be expressed as:
\[ \frac{W}{R + G + W} = \frac{1}{3} \]
Cross-multiplying gives us:
\[ 3W = R + G + W \]
Simplifying this, we get:
\[ 2W = R + G \quad \text{(Equation 1)} \]

### Step 3: Use the Given Probability of Green Balls
If there were 6 more red balls, the probability of getting 1 green ball would have been \( \frac{1}{30} \). This can be expressed as:
\[ \frac{G}{R + 6 + G + W} = \frac{1}{30} \]
Cross-multiplying gives us:
\[ 30G = R + 6 + G + W \]
Rearranging this, we get:
\[ 29G = R + W + 6 \quad \text{(Equation 2)} \]

### Step 4: Use the Given Probability of Green Balls Again
We are also given that the probability of drawing 1 green ball from the bag is \( \frac{1}{5} \):
\[ \frac{G}{R + G + W} = \frac{1}{5} \]
Cross-multiplying gives us:
\[ 5G = R + G + W \]
Rearranging gives us:
\[ 4G = R + W \quad \text{(Equation 3)} \]

### Step 5: Solve the Equations
Now we have three equations:
1. \( 2W = R + G \) (Equation 1)
2. \( 29G = R + W + 6 \) (Equation 2)
3. \( 4G = R + W \) (Equation 3)

From Equation 3, we can express \( R \) in terms of \( G \) and \( W \):
\[ R = 4G - W \quad \text{(Substituting into Equation 1)} \]
Substituting \( R \) into Equation 1:
\[ 2W = (4G - W) + G \]
This simplifies to:
\[ 2W = 5G - W \]
Adding \( W \) to both sides gives:
\[ 3W = 5G \quad \Rightarrow \quad W = \frac{5}{3}G \]

### Step 6: Substitute Back to Find R
Substituting \( W \) back into Equation 3:
\[ 4G = R + \frac{5}{3}G \]
This gives:
\[ R = 4G - \frac{5}{3}G = \frac{12G - 5G}{3} = \frac{7G}{3} \]

### Step 7: Calculate the Total Number of Balls
Now we can express the total number of balls:
\[ Total = R + G + W = \frac{7G}{3} + G + \frac{5G}{3} = \frac{7G + 3G + 5G}{3} = \frac{15G}{3} = 5G \]

### Step 8: Find the Probability of Drawing 2 Balls that are Not Red
The probability of drawing 2 balls such that none of them is red is given by:
\[ P(\text{not red}) = \frac{G + W}{Total} = \frac{G + \frac{5}{3}G}{5G} = \frac{\frac{8G}{3}}{5G} = \frac{8}{15} \]

### Final Answer
The probability of drawing 2 balls such that none of them is red is:
\[ \frac{8}{15} \]