Let's solve the problem step by step.

Let:

• $g$ represent the number of green pens in the box.
• $b$ represent the number of blue pens in the box.

Given Information:

1. There are only green pens and blue pens.
2. There are three more blue pens than green pens: $b = g + 3$
3. There are more than 12 pens in total: $g + b > 12$
4. Simon takes two pens at random, and the probability that both are of the same color is $\frac{27}{55}$.

Probability Analysis:

The total number of ways to pick any 2 pens from the box is: $\text{Total ways} = \binom{g + b}{2} = \frac{(g + b)(g + b - 1)}{2}$

The ways to pick 2 green pens are: $\binom{g}{2} = \frac{g(g - 1)}{2}$

The ways to pick 2 blue pens are: $\binom{b}{2} = \frac{b(b - 1)}{2}$

The total number of favorable outcomes (picking 2 pens of the same color) is: $\text{Favorable ways} = \frac{g(g - 1)}{2} + \frac{b(b - 1)}{2}$

The probability that Simon picks two pens of the same color is: $\frac{\frac{g(g - 1)}{2} + \frac{b(b - 1)}{2}}{\frac{(g + b)(g + b - 1)}{2}} = \frac{27}{55}$

Simplifying the Equation:

Multiply both the numerator and the denominator by 2 to clear the fractions: $\frac{g(g - 1) + b(b - 1)}{(g + b)(g + b - 1)} = \frac{27}{55}$

Substitute $b = g + 3$ into the equation: $\frac{g(g - 1) + (g + 3)(g + 2)}{(2g + 3)(2g + 2)} = \frac{27}{55}$

Solving for $g$:

Expand and simplify: $g(g - 1) + (g + 3)(g + 2) = g^2 - g + g^2 + 5g + 6 = 2g^2 + 4g + 6$ $(2g + 3)(2g + 2) = 4g^2 + 10g + 6$ So: $\frac{2g^2 + 4g + 6}{4g^2 + 10g + 6} = \frac{27}{55}$

Cross-multiply: $55(2g^2 + 4g + 6) = 27(4g^2 + 10g + 6)$ $110g^2 + 220g + 330 = 108g^2 + 270g + 162$ Simplify: $2g^2 - 50g + 168 = 0$ Divide by 2: $g^2 - 25g + 84 = 0$

Factoring:

$(g - 21)(g - 4) = 0$

Thus, $g = 21$ or $g = 4$. Since the total number of pens must be more than 12, we discard $g = 4$.

Solution:

The number of green pens in the box is: $\boxed{21}$

Verification:

If $g = 21$, then $b = 21 + 3 = 24$, and the total number of pens is $45$. The total ways to pick any 2 pens are: $\binom{45}{2} = \frac{45 \times 44}{2} = 990$

The favorable ways: $\binom{21}{2} + \binom{24}{2} = \frac{21 \times 20}{2} + \frac{24 \times 23}{2} = 210 + 276 = 486$

The probability: $\frac{486}{990} = \frac{27}{55}$

The solution is correct.

Would you like more details on any part of this solution or have any other questions?

Related Questions:

1. How would the problem change if there were fewer than 12 pens in the box?
2. What would the calculation be if the probability were different, say $\frac{1}{2}$?
3. Can you explain how to derive the probability formulas used in this problem?
4. What would happen if the number of blue pens was not a fixed amount more than the green pens?
5. How would you approach the problem if there were three colors of pens?

Tip:

When dealing with probability problems involving combinations, always double-check the total number of outcomes and favorable outcomes to ensure accuracy.