A bag contains $$8$$ balls, whose colours are either white or black. $$4$$ balls are drawn at random without replacement and it was found that $$2$$ balls are white and other $$2$$ balls are black. The probability that the bag contains equal number of white and black balls is:

The bag contains 8 balls, each either white or black. We draw 4 balls without replacement and observe exactly 2 white and 2 black balls. We need to find the probability that the bag originally had an equal number of white and black balls, i.e., 4 white and 4 black balls.

Let $$W$$ be the number of white balls in the bag. Since we drew 2 white and 2 black balls, $$W$$ must be at least 2 and at most 6 (so that there are at least 2 black balls, i.e., $$8 - W \geq 2$$). Thus, $$W$$ can be 2, 3, 4, 5, or 6.

We assume that all possible values of $$W$$ from 0 to 8 are equally likely initially. Therefore, the prior probability $$P(W = k) = \frac{1}{9}$$ for $$k = 0, 1, \dots, 8$$. However, since the event $$A$$ (drawing 2 white and 2 black balls) can only occur if $$W$$ is between 2 and 6, we restrict our calculation to these values.

By Bayes' theorem, the conditional probability is:

$$P(W=4 \mid A) = \frac{P(A \mid W=4) \cdot P(W=4)}{\sum_{k=2}^{6} P(A \mid W=k) \cdot P(W=k)}$$

Substituting $$P(W=k) = \frac{1}{9}$$ for all $$k$$:

$$P(W=4 \mid A) = \frac{P(A \mid W=4) \cdot \frac{1}{9}}{\sum_{k=2}^{6} P(A \mid W=k) \cdot \frac{1}{9}} = \frac{P(A \mid W=4)}{\sum_{k=2}^{6} P(A \mid W=k)}$$

Now, $$P(A \mid W=k)$$ is the probability of drawing exactly 2 white and 2 black balls from a bag with $$k$$ white and $$8-k$$ black balls, without replacement. This follows the hypergeometric distribution:

$$P(A \mid W=k) = \frac{\binom{k}{2} \binom{8-k}{2}}{\binom{8}{4}}$$

First, compute $$\binom{8}{4}$$:

$$\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$

Now, compute for each $$k$$:

• For $$k=2$$: $$\binom{2}{2} = 1$$, $$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$, so $$P(A \mid W=2) = \frac{1 \times 15}{70} = \frac{15}{70}$$
• For $$k=3$$: $$\binom{3}{2} = 3$$, $$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$, so $$P(A \mid W=3) = \frac{3 \times 10}{70} = \frac{30}{70}$$
• For $$k=4$$: $$\binom{4}{2} = \frac{4 \times 3}{2} = 6$$, $$\binom{4}{2} = 6$$, so $$P(A \mid W=4) = \frac{6 \times 6}{70} = \frac{36}{70}$$
• For $$k=5$$: $$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$, $$\binom{3}{2} = 3$$, so $$P(A \mid W=5) = \frac{10 \times 3}{70} = \frac{30}{70}$$
• For $$k=6$$: $$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$, $$\binom{2}{2} = 1$$, so $$P(A \mid W=6) = \frac{15 \times 1}{70} = \frac{15}{70}$$

The denominator is the sum:

$$\sum_{k=2}^{6} P(A \mid W=k) = \frac{15}{70} + \frac{30}{70} + \frac{36}{70} + \frac{30}{70} + \frac{15}{70} = \frac{15 + 30 + 36 + 30 + 15}{70} = \frac{126}{70}$$

Now, the probability is:

$$P(W=4 \mid A) = \frac{\frac{36}{70}}{\frac{126}{70}} = \frac{36}{70} \times \frac{70}{126} = \frac{36}{126}$$

Simplify $$\frac{36}{126}$$ by dividing both numerator and denominator by 18:

$$\frac{36 \div 18}{126 \div 18} = \frac{2}{7}$$

Thus, the probability that the bag contains an equal number of white and black balls is $$\frac{2}{7}$$.

The correct option is B: $$\frac{2}{7}$$.