A bag contains 15 white balls and some black balls. If the probability of drawing a black ball is thrice that of a white ball, find the number of black balls in the bag.

Options

A. `40`
B. `45`
C. `15`
D. `60`

Explanation

To solve the problem step by step, we need to find the number of black balls in the bag given that the probability of drawing a black ball is three times that of drawing a white ball.

### Step 1: Define the variables
Let:
- The number of white balls = 15 (given)
- The number of black balls = \( x \) (unknown)

### Step 2: Calculate the total number of balls
The total number of balls in the bag is the sum of white and black balls:
\[ \text{Total number of balls} = 15 + x \]

### Step 3: Calculate the probability of drawing a white ball
The probability of drawing a white ball is given by the formula:
\[ P(\text{White}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{15}{15 + x} \]

### Step 4: Calculate the probability of drawing a black ball
The probability of drawing a black ball is:
\[ P(\text{Black}) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{x}{15 + x} \]

### Step 5: Set up the equation based on the given condition
According to the problem, the probability of drawing a black ball is three times that of drawing a white ball:
\[ P(\text{Black}) = 3 \times P(\text{White}) \]
Substituting the probabilities we calculated:
\[ \frac{x}{15 + x} = 3 \times \frac{15}{15 + x} \]

### Step 6: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ x(15 + x) = 3 \times 15 \]
This simplifies to:
\[ x(15 + x) = 45 \]

### Step 7: Rearrange the equation
Expanding the left side:
\[ 15x + x^2 = 45 \]
Rearranging gives us a standard quadratic equation:
\[ x^2 + 15x - 45 = 0 \]

### Step 8: Solve the quadratic equation using the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 15 \), and \( c = -45 \):
\[ x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 15^2 = 225 \]
\[ -4 \cdot 1 \cdot (-45) = 180 \]
\[ b^2 - 4ac = 225 + 180 = 405 \]
Now substituting back into the formula:
\[ x = \frac{-15 \pm \sqrt{405}}{2} \]
Calculating \( \sqrt{405} \):
\[ \sqrt{405} = 3\sqrt{45} = 3 \cdot 3\sqrt{5} = 9\sqrt{5} \]
So,
\[ x = \frac{-15 \pm 9\sqrt{5}}{2} \]

### Step 9: Determine the positive solution
Since \( x \) must be a positive number (number of black balls), we take the positive root:
\[ x = \frac{-15 + 9\sqrt{5}}{2} \]

### Step 10: Approximate the value of \( x \)
Calculating \( \sqrt{5} \approx 2.236 \):
\[ 9\sqrt{5} \approx 20.124 \]
Thus,
\[ x \approx \frac{-15 + 20.124}{2} \approx \frac{5.124}{2} \approx 2.562 \]
Since \( x \) must be a whole number, we round it to the nearest whole number, which is 3.

### Conclusion
The number of black balls in the bag is approximately 3.

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