An urn contains 5 blue and unknown number x of red balls. Two balls are drawn at random. If the probability of both of them being blue is `5/14` , then x is

Options

A. A. 1
B. B. 2
C. C. 5
D. D. 3

Answer

To solve the problem step by step, we will use the information given about the urn containing blue and red balls, and the probability of drawing two blue balls.

### Step 1: Define the variables
Let:
- The number of blue balls = 5
- The number of red balls = x
- Total number of balls = 5 + x

### Step 2: Write the probability of drawing 2 blue balls
The probability of drawing 2 blue balls can be expressed as:
\[ P(\text{2 blue}) = \frac{\text{Number of ways to choose 2 blue balls}}{\text{Total ways to choose 2 balls}} \]

The number of ways to choose 2 blue balls from 5 is given by the combination formula:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

The total number of ways to choose 2 balls from (5 + x) is:
\[ \binom{5+x}{2} = \frac{(5+x)(4+x)}{2} \]

### Step 3: Set up the equation
According to the problem, the probability of both balls being blue is given as \( \frac{5}{14} \). Therefore, we can set up the equation:
\[ \frac{10}{\frac{(5+x)(4+x)}{2}} = \frac{5}{14} \]

### Step 4: Simplify the equation
Cross-multiplying gives:
\[ 10 \cdot 14 = 5 \cdot \frac{(5+x)(4+x)}{2} \]
\[ 140 = \frac{5(5+x)(4+x)}{2} \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 280 = 5(5+x)(4+x) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:
\[ 280 = 5(20 + 9x + x^2) \]
\[ 280 = 100 + 45x + 5x^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 5x^2 + 45x + 100 - 280 = 0 \]
\[ 5x^2 + 45x - 180 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 5:
\[ x^2 + 9x - 36 = 0 \]

### Step 8: Factor the quadratic equation
We need to factor the quadratic:
\[ (x + 12)(x - 3) = 0 \]

### Step 9: Solve for x
Setting each factor to zero gives:
\[ x + 12 = 0 \quad \Rightarrow \quad x = -12 \quad (\text{not valid since } x \text{ cannot be negative}) \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Conclusion
The number of red balls \( x \) is 3.