Let the bag contain $w$ white balls and $8-w$ black balls, where w=0,1,2,dots,8.

We are told that 4 balls are drawn without replacement, and the observed outcome is:

• 2 white
• 2 black

We need the conditional probability: $P(w=4\mid \text{draw has 2 white and 2 black})$

1. Prior possibilities for composition of the bag

Since the colours are either white or black, the possible values of $w$ are $0$ to $8$.

Assuming all compositions are equally likely, $P(w=k)=\frac{1}{9}, \qquad k=0,1,2,\dots,8.$

Now the event "2 white and 2 black are drawn" is only possible if the bag has at least 2 white and at least 2 black balls. So only: $w=2,3,4,5,6$ are relevant.

2. Likelihood of drawing 2 white and 2 black

If the bag has $w$ white and $8-w$ black balls, then $P(E\mid w)=\frac{\binom{w}{2}\binom{8-w}{2}}{\binom{8}{4}},$ where $E$ is the event of drawing exactly 2 white and 2 black.

Since $\binom{8}{4}=70,$ we compute for each relevant $w$:

Case $w=2$

$P(E\mid w=2)=\frac{\binom{2}{2}\binom{6}{2}}{70}=\frac{1\cdot 15}{70}=\frac{3}{14}.$

Case $w=3$

$P(E\mid w=3)=\frac{\binom{3}{2}\binom{5}{2}}{70}=\frac{3\cdot 10}{70}=\frac{3}{7}.$

Case $w=4$

$P(E\mid w=4)=\frac{\binom{4}{2}\binom{4}{2}}{70}=\frac{6\cdot 6}{70}=\frac{18}{35}.$

Case $w=5$

$P(E\mid w=5)=\frac{\binom{5}{2}\binom{3}{2}}{70}=\frac{10\cdot 3}{70}=\frac{3}{7}.$

Case $w=6$

$P(E\mid w=6)=\frac{\binom{6}{2}\binom{2}{2}}{70}=\frac{15\cdot 1}{70}=\frac{3}{14}.$

3. Apply Bayes' theorem

We want $P(w=4\mid E)=\frac{P(E\mid w=4)P(w=4)}{\sum_{k=2}^{6} P(E\mid w=k)P(w=k)}.$

Since each prior is $\frac{1}{9}$, they cancel: $P(w=4\mid E)=\frac{P(E\mid w=4)}{P(E\mid 2)+P(E\mid 3)+P(E\mid 4)+P(E\mid 5)+P(E\mid 6)}.$

So, $P(w=4\mid E)=\frac{\frac{18}{35}}{\frac{3}{14}+\frac{3}{7}+\frac{18}{35}+\frac{3}{7}+\frac{3}{14}}.$

Now simplify the denominator: $\frac{3}{14}+\frac{3}{14}=\frac{3}{7},$ so $\text{denominator}=\frac{3}{7}+\frac{3}{7}+\frac{18}{35}+\frac{3}{7}=\frac{9}{7}+\frac{18}{35}.$

Convert to denominator $35$: $\frac{9}{7}=\frac{45}{35},$ thus $\text{denominator}=\frac{45}{35}+\frac{18}{35}=\frac{63}{35}=\frac{9}{5}.$

Hence, $P(w=4\mid E)=\frac{\frac{18}{35}}{\frac{63}{35}}=\frac{18}{63}=\frac{2}{7}.$

4. Final answer

Therefore, the probability that the bag contains equal numbers of white and black balls is $\boxed{\frac{2}{7}}.$

So the correct option is B.