A bag contains b blue balls and r red balls. If two balls are drawn at random, the probability drawing two red balls is five times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. Then

Options

A. b + r = 9
B. br = 18
C. `|b - r| = 4`
D. `b//r = 2`

Explanation

To solve the problem, we need to find the values of \( b \) (the number of blue balls) and \( r \) (the number of red balls) in a bag such that the following conditions hold:

1. The probability of drawing two red balls is five times the probability of drawing two blue balls.
2. The probability of drawing one ball of each color is six times the probability of drawing two blue balls.

### Step 1: Define the probabilities

- The total number of balls is \( r + b \).

**Probability of drawing two red balls (P1):**
\[ P1 = \frac{rC2}{(r+b)C2} = \frac{\frac{r(r-1)}{2}}{\frac{(r+b)(r+b-1)}{2}} = \frac{r(r-1)}{(r+b)(r+b-1)} \]

**Probability of drawing two blue balls (P3):**
\[ P3 = \frac{bC2}{(r+b)C2} = \frac{\frac{b(b-1)}{2}}{\frac{(r+b)(r+b-1)}{2}} = \frac{b(b-1)}{(r+b)(r+b-1)} \]

**Probability of drawing one ball of each color (P2):**
\[ P2 = \frac{rC1 \cdot bC1}{(r+b)C2} = \frac{rb}{\frac{(r+b)(r+b-1)}{2}} = \frac{2rb}{(r+b)(r+b-1)} \]

### Step 2: Set up the equations based on the conditions

From the problem statement, we have:
1. \( P1 = 5 \cdot P3 \)
2. \( P2 = 6 \cdot P3 \)

### Step 3: Substitute the probabilities into the equations

**From the first condition:**
\[ \frac{r(r-1)}{(r+b)(r+b-1)} = 5 \cdot \frac{b(b-1)}{(r+b)(r+b-1)} \]
Cancelling the common terms:
\[ r(r-1) = 5b(b-1) \tag{1} \]

**From the second condition:**
\[ \frac{2rb}{(r+b)(r+b-1)} = 6 \cdot \frac{b(b-1)}{(r+b)(r+b-1)} \]
Cancelling the common terms:
\[ 2rb = 6b(b-1) \tag{2} \]

### Step 4: Solve the equations

From equation (2):
\[ 2r = 6(b-1) \]
\[ r = 3(b-1) \tag{3} \]

Substituting equation (3) into equation (1):
\[ 3(b-1)(3(b-1)-1) = 5b(b-1) \]
Expanding this:
\[ 3(b-1)(3b-3-1) = 5b(b-1) \]
\[ 3(b-1)(3b-4) = 5b(b-1) \]
Expanding both sides:
\[ 9b^2 - 12b - 3b + 4 = 5b^2 - 5b \]
Combining like terms:
\[ 9b^2 - 15b + 4 = 5b^2 - 5b \]
Rearranging gives:
\[ 4b^2 - 10b + 4 = 0 \]
Dividing by 2:
\[ 2b^2 - 5b + 2 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
\[ b = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ b = \frac{5 \pm \sqrt{25 - 16}}{4} \]
\[ b = \frac{5 \pm 3}{4} \]
This gives:
\[ b = 2 \quad \text{or} \quad b = \frac{1}{2} \text{ (not possible since b must be an integer)} \]
So, \( b = 2 \).

### Step 6: Find \( r \)

Using equation (3):
\[ r = 3(2-1) = 3 \]

### Conclusion

The number of blue balls \( b = 2 \) and the number of red balls \( r = 3 \). Therefore, the total number of balls is:
\[ b + r = 2 + 3 = 5 \]