There are only green pens and blue pens in a box. There are three more blue pens than green pens in the box. There are more than 12 pens in the box. Simon is going to take at random two pens from the box. The probability that Simon will take two pens of the same colour is 27/55. b= blue pens // g= green pens // x= total pens P(two of same colour) = P(green, green) + P(blue, blue) P(two of same colour) = (g/x)(g-1/x-1) + (b/x)(b-1/x-1)From Q: b = g + 3x = b + gx = (g + 3) + g = 2g + 3P(two of same colour) = (g/2g+3)(g-1/2g+2) + (g+3/2g+3)(g+2/2g+2) = 27/55Expanding + Solving: (g2-g)/(4g2+10g+6) + (g2+5g +6)/(4g2+10g+6) = 27/552g2+ 4g + 6 = 27/55 (4g2+10g+6) g2/55 - 5g/11 + 84/55 = 0g2 -25g +84 = 0 (g-21)(g-4) = 0 g= 21 g= 4BUT - Q states g > 12 Therefore g = 21