In a bag which contains 60 pens , there are 27 blue pens and some black and red pens . If two pens are picked up from the bag without replacement , then the probability of the first pen being blue and second being black is 9/59.Find the number of red in the bag .

Options

A. 16
B. 13
C. 10
D. 14

Explanation

To solve the problem, we need to find the number of red pens in the bag given the total number of pens, the number of blue pens, and the probability of drawing a blue pen first and a black pen second without replacement.

### Step-by-Step Solution:

1. **Identify the total number of pens and blue pens:**
- Total pens = 60
- Blue pens = 27
- Let the number of black pens be \( b \) and the number of red pens be \( r \).
- Therefore, we have the equation:
\[ 27 + b + r = 60 \]
- This simplifies to:
\[ b + r = 33 \quad \text{(Equation 1)} \]

2. **Understand the probability condition:**
- The probability of picking a blue pen first and a black pen second without replacement is given as \( \frac{9}{59} \).
- The probability can be calculated as:
\[ P(\text{Blue first, Black second}) = P(\text{Blue first}) \times P(\text{Black second | Blue first}) \]
- The probability of picking a blue pen first:
\[ P(\text{Blue first}) = \frac{27}{60} \]
- After picking one blue pen, there are now 59 pens left, including \( b \) black pens. The probability of picking a black pen second:
\[ P(\text{Black second | Blue first}) = \frac{b}{59} \]
- Therefore, we can write:
\[ \frac{27}{60} \times \frac{b}{59} = \frac{9}{59} \]

3. **Simplify the equation:**
- Cross-multiply to eliminate the fractions:
\[ 27b = 60 \times 9 \]
- Calculate \( 60 \times 9 = 540 \):
\[ 27b = 540 \]
- Now, divide both sides by 27:
\[ b = \frac{540}{27} = 20 \]

4. **Find the number of red pens:**
- Substitute \( b = 20 \) back into Equation 1:
\[ 20 + r = 33 \]
- Solving for \( r \):
\[ r = 33 - 20 = 13 \]

### Conclusion:
The number of red pens in the bag is **13**.