A box contains x red balls and 10 black balls. 3 balls are drawn one by one without replacement. If the probability of choosing 3 red balls is equal to the probability of choosing 2 red and 1 black ball, then the possible value of x can be

Options

A. 1
B. 32
C. 53
D. 40

Answer

To solve the problem, we need to find the possible value of \( x \) (the number of red balls) such that the probability of drawing 3 red balls is equal to the probability of drawing 2 red balls and 1 black ball from a box containing \( x \) red balls and 10 black balls.

### Step-by-Step Solution:

1. **Understanding the Total Number of Balls**:
The total number of balls in the box is \( x + 10 \) (where \( x \) is the number of red balls and 10 is the number of black balls).

2. **Calculating the Probability of Drawing 3 Red Balls**:
The probability of drawing 3 red balls can be calculated using combinations:
\[ P(\text{3 red}) = \frac{\binom{x}{3}}{\binom{x + 10}{3}} \]

3. **Calculating the Probability of Drawing 2 Red Balls and 1 Black Ball**:
The probability of drawing 2 red balls and 1 black ball is:
\[ P(\text{2 red, 1 black}) = \frac{\binom{x}{2} \cdot \binom{10}{1}}{\binom{x + 10}{3}} \]
Here, \( \binom{10}{1} = 10 \).

4. **Setting the Probabilities Equal**:
According to the problem, these two probabilities are equal:
\[ \frac{\binom{x}{3}}{\binom{x + 10}{3}} = \frac{\binom{x}{2} \cdot 10}{\binom{x + 10}{3}} \]
Since \( \binom{x + 10}{3} \) appears in both denominators, we can cancel it out:
\[ \binom{x}{3} = 10 \cdot \binom{x}{2} \]

5. **Expanding the Combinations**:
The combinations can be expressed as:
\[ \binom{x}{3} = \frac{x!}{3!(x-3)!} = \frac{x(x-1)(x-2)}{6} \]
\[ \binom{x}{2} = \frac{x!}{2!(x-2)!} = \frac{x(x-1)}{2} \]
Substituting these into the equation gives:
\[ \frac{x(x-1)(x-2)}{6} = 10 \cdot \frac{x(x-1)}{2} \]

6. **Simplifying the Equation**:
We can simplify the equation:
\[ \frac{x(x-1)(x-2)}{6} = 5x(x-1) \]
Dividing both sides by \( x(x-1) \) (assuming \( x \neq 0 \) and \( x \neq 1 \)):
\[ \frac{x-2}{6} = 5 \]

7. **Solving for \( x \)**:
Multiplying both sides by 6:
\[ x - 2 = 30 \]
Adding 2 to both sides:
\[ x = 32 \]

### Conclusion:
The possible value of \( x \) is \( 32 \).