A bag contains 150 chips. There are 60 blue chips, x red chips and y green chips in the bag. The probability of drawing a red chip is $\frac{7}{30}$.

1. Use the probability of drawing a red chip to find $x$.

$\frac{x}{150}=\frac{7}{30}$
$x=150\times \frac{7}{30}=35$

2. Use the total number of chips to find $y$.

There are 60 blue chips, 35 red chips and $y$ green chips.

$60+35+y=150$
$y=150-60-35=55$

1. Write the probability of drawing one yellow chip after adding $n$ yellow chips.

After adding $n$ yellow chips,

Total number of chips $=150+n$.

Number of yellow chips $=n$.

So,
$P(\text{yellow})=\frac{n}{150+n}.$

2. Use replacement to form the equation for choosing two yellow chips.

Since the chips are chosen with replacement, the probability remains the same for each draw.

$P(\text{two yellow chips})=\frac{n}{150+n}\times \frac{n}{150+n}$

Given that this probability is $\frac{1}{256}$,

$\left(\frac{n}{150+n}\right)^2=\frac{1}{256}.$

3. Simplify the equation to the required quadratic form.

$\left(\frac{n}{150+n}\right)^2=\frac{1}{256}$

$\frac{n^2}{(150+n)^2}=\frac{1}{256}$

$256n^2=(150+n)^2$

$256n^2=n^2+300n+22500$

$255n^2-300n-22500=0$

Divide throughout by $15$:

$17n^2-20n-1500=0.$

1. Factorise the quadratic equation.

$17n^2-20n-1500=0$

$17n^2-170n+150n-1500=0$

$17n(n-10)+150(n-10)=0$

$(17n+150)(n-10)=0$

2. Find the possible values of $n$ and choose the valid number of chips.

$(17n+150)(n-10)=0$

So,
$17n+150=0 \quad \text{or} \quad n-10=0$

$n=-\frac{150}{17} \quad \text{or} \quad n=10$

Since $n$ is the number of yellow chips added, $n$ cannot be negative.

$n=10$