A drawer contains 2 blue socks and 6 white socks. Two socks are taken from the drawer at random without replacement.
If the two socks are different colours, then a third sock is taken from the drawer.
Otherwise, no third sock is taken.

1. Set up the first two draws without replacement.

Initially there are $2$ blue socks and $6$ white socks, so $8$ socks in total.

First sock:
$P(B)=\frac{2}{8}, \qquad P(W)=\frac{6}{8}.$

If the first sock is blue, then $1$ blue and $6$ white socks remain out of $7$:
$P(B\text{ second}\mid B\text{ first})=\frac{1}{7}, \qquad P(W\text{ second}\mid B\text{ first})=\frac{6}{7}.$

If the first sock is white, then $2$ blue and $5$ white socks remain out of $7$:
$P(B\text{ second}\mid W\text{ first})=\frac{2}{7}, \qquad P(W\text{ second}\mid W\text{ first})=\frac{5}{7}.$

2. Add the third draw only for branches where the first two socks are different colours.

A third sock is taken only after $BW$ or $WB$.

After $BW$, the remaining socks are $1$ blue and $5$ white out of $6$:
$P(B\text{ third}\mid BW)=\frac{1}{6}, \qquad P(W\text{ third}\mid BW)=\frac{5}{6}.$

After $WB$, the remaining socks are also $1$ blue and $5$ white out of $6$:
$P(B\text{ third}\mid WB)=\frac{1}{6}, \qquad P(W\text{ third}\mid WB)=\frac{5}{6}.$

No third branch is drawn after $BB$ or $WW$.

1. Identify the favourable same-colour outcomes for the first two socks.

The first two socks are the same colour if the outcome is either $BB$ or $WW$.

2. Find the probability of getting two blue socks or two white socks.

$P(BB)=\frac{2}{8}\times\frac{1}{7}=\frac{2}{56}$
$P(WW)=\frac{6}{8}\times\frac{5}{7}=\frac{30}{56}$

3. Add the mutually exclusive probabilities and simplify.

$P(\text{same colour})=P(BB)+P(WW)$
$=\frac{2}{56}+\frac{30}{56}=\frac{32}{56}=\frac{4}{7}$

1. Identify the favourable outcomes.

A third sock is taken only if the first two socks are different colours.

For the third sock to be the same colour as the first sock, the possible outcomes are

$BWB$ or $WBW$.

2. Multiply along each branch and add the mutually exclusive probabilities.

$P(BWB)=\frac{2}{8}\times\frac{6}{7}\times\frac{1}{6}=\frac{1}{28}$

$P(WBW)=\frac{6}{8}\times\frac{2}{7}\times\frac{5}{6}=\frac{5}{28}$

$P(\text{third sock is taken and same colour as first})=\frac{1}{28}+\frac{5}{28}=\frac{6}{28}=\frac{3}{14}$