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 ratio of number of red balls and white balls is 7 : 5, then what is the probability of drawing 2 balls such that one is red and other is white?

Options

A. `5//29`
B. `14//87`
C. `15//58`
D. `28//87`

Explanation

To solve the problem step by step, let's break down the information given and derive the required probability.

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

### Step 2: Use the Given Ratio
From the problem, we know that the ratio of the number of red balls to white balls is \( 7:5 \). We can express this as:
\[ R = 7x \quad \text{and} \quad W = 5x \]
where \( x \) is a common multiplier.

### Step 3: Calculate Total Balls
The total number of balls in the bag is:
\[ \text{Total} = R + G + W = 7x + G + 5x = 12x + G \]

### Step 4: Use the Probability of Drawing a White Ball
The probability of drawing one white ball is given as \( \frac{1}{3} \):
\[ \frac{W}{\text{Total}} = \frac{5x}{12x + G} = \frac{1}{3} \]
Cross-multiplying gives:
\[ 5x \cdot 3 = 1 \cdot (12x + G) \implies 15x = 12x + G \implies G = 3x \]

### Step 5: Total Number of Balls
Now substituting \( G \) back into the total:
\[ \text{Total} = 7x + 3x + 5x = 15x \]

### Step 6: Probability of Drawing a Green Ball
The probability of drawing a green ball initially is:
\[ \frac{G}{\text{Total}} = \frac{3x}{15x} = \frac{1}{5} \]

### Step 7: Adjusting for Additional Red Balls
If there were 6 more red balls, the new number of red balls would be \( R + 6 = 7x + 6 \). The new total number of balls would be:
\[ (7x + 6) + 3x + 5x = 15x + 6 \]
The new probability of drawing a green ball would then be:
\[ \frac{G}{\text{New Total}} = \frac{3x}{15x + 6} \]

### Step 8: Set Up the Equation
According to the problem, the difference in probabilities is \( \frac{1}{30} \):
\[ \frac{1}{5} - \frac{3x}{15x + 6} = \frac{1}{30} \]
To solve this, we first find a common denominator and simplify:
\[ \frac{6(15x + 6) - 15x \cdot 3x}{5(15x + 6)} = \frac{1}{30} \]
Cross-multiplying gives:
\[ 6(15x + 6) - 15x \cdot 3x = \frac{1}{30} \cdot 5(15x + 6) \]

### Step 9: Solve for \( x \)
After simplifying and solving for \( x \), we find:
\[ x = 2 \]

### Step 10: Find Number of Balls
Substituting \( x \) back:
- Red balls \( R = 7x = 14 \)
- Green balls \( G = 3x = 6 \)
- White balls \( W = 5x = 10 \)

### Step 11: Calculate Probability of Drawing One Red and One White Ball
The total number of balls is \( 30 \). The probability of drawing one red and one white can occur in two ways:
1. Red first, then white:
\[ P(RW) = \frac{14}{30} \cdot \frac{10}{29} \]
2. White first, then red:
\[ P(WR) = \frac{10}{30} \cdot \frac{14}{29} \]

### Step 12: Combine Probabilities
Adding these probabilities gives:
\[ P(RW) + P(WR) = \frac{14}{30} \cdot \frac{10}{29} + \frac{10}{30} \cdot \frac{14}{29} = 2 \cdot \frac{14 \cdot 10}{30 \cdot 29} = \frac{280}{870} = \frac{28}{87} \]

### Final Answer
The probability of drawing one red ball and one white ball is:
\[ \frac{28}{87} \]