Question

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.

Work out the number of green pens in the box

Answer 1

SoluTion :-

Let us assume there are x green pens.

Then,

Blue pens = x+3

Total = 2x+3 pens

The probability of taking two green pens out of the box is :-

$\sf {\frac{x}{2x+3} \times \frac{x-1}{2x+2} = \frac{x^{2}-x}{2x+3(2x+2)} }$

The probability of taking two blue pens out of the box is :-

$\sf {\frac{(x+3)(x+2)}{(2x+3)(2x+2)}}$

Therefore, the total probability of taking two pens of the same colour is :-

$\sf {\frac{ (2x^2+4x+6)}{(4x^2+10x+6)} = \frac{(x^2+2x+3)}{(2x^2+5x+3)} =\frac{27}{55} }$

$\sf {= 55x^2 + 110x + 165 = 54x^2 + 135x + 81}$

$\sf {= x^2 - 25x + 84 = 0}$

$The\ solutions\ are=\begin{cases}x=4\\x=21\end{cases}$

Now we know that 2 x 4 + 3 = 11, which does not fit the "more than 12 pens" rule, so we conclude that x = 21.