A bag contains 12 balls.
2 of the balls are red, n are green balls and the rest are blue balls.
Eunice takes two balls from the bag, at random, without replacement.

1. Write the first-draw probabilities for each colour.

There are 12 balls in total.

Red balls: $2$

Green balls: $n$

Blue balls: $12-2-n=10-n$

So the first-draw probabilities are
$P(R)=\frac{2}{12},\quad P(G)=\frac{n}{12},\quad P(B)=\frac{10-n}{12}.$

2. Fill in the second-draw conditional probabilities after each possible first draw.

Since the balls are taken without replacement, there are $11$ balls left for the second draw.

If the first ball is Red:
$P(R)=\frac{1}{11},\quad P(G)=\frac{n}{11},\quad P(B)=\frac{10-n}{11}.$

If the first ball is Green:
$P(R)=\frac{2}{11},\quad P(G)=\frac{n-1}{11},\quad P(B)=\frac{10-n}{11}.$

If the first ball is Blue:
$P(R)=\frac{2}{11},\quad P(G)=\frac{n}{11},\quad P(B)=\frac{9-n}{11}.$

1. Write the probability of taking two blue balls without replacement.

From the tree diagram,

$P(\text{Blue then Blue})=\frac{10-n}{12}\times\frac{9-n}{11}$

Given that this probability is $\frac{5}{22}$,

$\frac{10-n}{12}\times\frac{9-n}{11}=\frac{5}{22}$

2. Clear denominators to remove fractions.

$\frac{(10-n)(9-n)}{132}=\frac{5}{22}$

Multiply both sides by $132$:

$(10-n)(9-n)=30$

3. Expand and rearrange into the required quadratic form.

$(10-n)(9-n)=30$

$90-10n-9n+n^2=30$

$n^2-19n+90=30$

$n^2-19n+60=0$

1. Factorise the quadratic expression.

Find two numbers with product $60$ and sum $19$: $4$ and $15$.

$n^2-19n+60=0$
$(n-4)(n-15)=0$

2. Use the zero product property to find the two values of $n$.

$(n-4)(n-15)=0$

So,
$n-4=0 \quad \text{or} \quad n-15=0$

$n=4 \quad \text{or} \quad n=15$

1. Identify the impossible solution using the total number of balls.

From part (c), $n=4$ or $n=15$.

But the bag contains only 12 balls, so $n$ cannot be greater than 12.

Therefore, reject $n=15$.

1. Use the valid value of $n$ to state the number of balls of each colour.

From parts (c) and (d), $n=4$.

So the bag contains:

• red balls: $2$
• green balls: $4$
• blue balls: $10-4=6$

2. Find the probability of taking two balls of the same colour without replacement.

$P(\text{same colour})=P(RR)+P(GG)+P(BB)$

$=\frac{2}{12}\times\frac{1}{11}+\frac{4}{12}\times\frac{3}{11}+\frac{6}{12}\times\frac{5}{11}$

$=\frac{2}{132}+\frac{12}{132}+\frac{30}{132}$

$=\frac{44}{132}=\frac{1}{3}$

3. Use the complement to find the probability of different colours.

$P(\text{different colours})=1-P(\text{same colour})$

$=1-\frac{1}{3}=\frac{2}{3}$

1. State the numbers of balls of each colour using the accepted value of $n$.

From earlier parts, $n=4$.

So the bag contains:

• red balls: $2$
• green balls: $4$
• blue balls: $10-4=6$

2. Find the probability of taking two balls of the same colour with replacement.

With replacement, the probabilities remain the same for both draws.

$P(\text{same colour})=P(RR)+P(GG)+P(BB)$

$=\left(\frac{2}{12}\right)^2+\left(\frac{4}{12}\right)^2+\left(\frac{6}{12}\right)^2$

$=\frac{4}{144}+\frac{16}{144}+\frac{36}{144}$

$=\frac{56}{144}=\frac{7}{18}$