Question

There are 30 marbles in a bag, 10 are green, 8 are blue, $m$ are red and $n$ are yellow, where $m$ and $n$ are integers and $m\ge n$. When two marbles are selected from the bag without replacement the probability they are the same colour is $\frac{107}{435}$. Find the value of $\left(m^2+nm+n^2\right)$.
[Note: $10,8,m$ and $n$ add to 30 .]

Answer 1

Answer: 100

$P($ two drawn marbles are of the same colour $)=\frac{{}^{10}{C}_2+{}^8{C}_2+{}^m{C}_2+{}^n{C}_2}{{}^{30}{C}_2}$
$\therefore \frac{107}{435}=\frac{10\cdot 9+8\cdot 7+m(m-1)+n(n-1)}{30\cdot 29}\Rightarrow \frac{107}{87}=\frac{90+56+\left(m^2+n^2\right)-(m+n)}{6\cdot 29}$
$\Rightarrow 214=146+\left(m^2+n^2\right)+12\Rightarrow 226-146=m^2+n^2\Rightarrow 80=m^2+(12-m{)}^2$
$\Rightarrow 2m^2-24m+144=80\Rightarrow 2m^2-24m+64=0\Rightarrow m^2-12m+32=0\Rightarrow (m-8)(m-4)=0$
$\therefore m=8\text{ and }n=4$
$\therefore m^2+mn+n^2=64+32+4=100$