1. Interpretation of the problem

Let the number of black balls in the bag be unknown. Since the bag has 6 balls total, the possible numbers of black balls are:

$0,1,2,3,4,5,6$

We are told that two balls are drawn at random and both are black. We need:

$P(\text{at least 5 black balls in bag} \mid \text{two drawn are black})$

That is,

$P(B\ge 5\mid E)$

where:

• $B$ = number of black balls in the bag,
• $E$ = event that both drawn balls are black.

2. Assume all possible bag compositions are equally likely

Since nothing else is specified, we take each possible number of black balls $k=0,1,2,3,4,5,6$ to be equally likely.

So,

$P(B=k)=\frac{1}{7},\quad k=0,1,2,3,4,5,6$

3. Compute $P(E\mid B=k)$

If the bag has $k$ black balls, then the probability that both drawn balls are black is:

$P(E\mid B=k)=\frac{\binom{k}{2}}{\binom{6}{2}}$

Now evaluate for each $k$:

• $k=0$: $P(E\mid B=0)=0$
• $k=1$: $P(E\mid B=1)=0$
• $k=2$: $P(E\mid B=2)=\frac{\binom{2}{2}}{\binom{6}{2}}=\frac{1}{15}$
• $k=3$: $P(E\mid B=3)=\frac{\binom{3}{2}}{15}=\frac{3}{15}=\frac{1}{5}$
• $k=4$: $P(E\mid B=4)=\frac{\binom{4}{2}}{15}=\frac{6}{15}=\frac{2}{5}$
• $k=5$: $P(E\mid B=5)=\frac{\binom{5}{2}}{15}=\frac{10}{15}=\frac{2}{3}$
• $k=6$: $P(E\mid B=6)=\frac{\binom{6}{2}}{15}=1$

4. Use Bayes' idea: posterior is proportional to prior × likelihood

Since each $P(B=k)=\frac{1}{7}$ is equal, the posterior probabilities are proportional to:

$P(E\mid B=k)$

So we just sum the likelihoods.

Total weight:

$0+0+\frac{1}{15}+\frac{1}{5}+\frac{2}{5}+\frac{2}{3}+1$

Take LCM $15$:

$=\frac{1+3+6+10+15}{15}=\frac{35}{15}=\frac{7}{3}$

Now the favorable cases are $B\ge 5$, i.e. $B=5$ or $B=6$.

Their total weight is:

$\frac{2}{3}+1=\frac{5}{3}$

Hence,

$P(B\ge 5\mid E)=\frac{\frac{5}{3}}{\frac{7}{3}}=\frac{5}{7}$

5. Match with options

$\frac{5}{7}$ corresponds to Option C.

6. Final answer

$\boxed{\frac{5}{7}}$