A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

A bag contains 6 balls. Two balls are drawn and both are black. We need to find $$P(\text{at least 5 black} | \text{2 drawn are black})$$.

Assuming each possible composition of black balls (0 to 6) is equally likely a priori, we have for each $$k=0,1,2,\ldots,6$$ that $$P(B=k)=\frac{1}{7}$$.

The likelihood of drawing two black balls given there are $$k$$ black balls in the bag is $$P(\text{2 black drawn}|B=k)=\frac{\binom{k}{2}}{\binom{6}{2}}=\frac{k(k-1)}{30},$$ which equals 0 for $$k=0,1$$, $$\frac{2}{30}$$ for $$k=2$$, $$\frac{6}{30}$$ for $$k=3$$, $$\frac{12}{30}$$ for $$k=4$$, $$\frac{20}{30}$$ for $$k=5$$, and $$\frac{30}{30}$$ for $$k=6$$.

By the law of total probability we compute $$P(\text{2 black})=\frac{1}{7}\cdot\frac{1}{30}(0+0+2+6+12+20+30)=\frac{70}{210}=\frac{1}{3}\,.$$

Applying Bayes' theorem to find the probability of at least five black balls given two black draws yields $$P(B\ge5|\text{2 black})=\frac{P(\text{2 black}|B=5)P(B=5)+P(\text{2 black}|B=6)P(B=6)}{P(\text{2 black})} =\frac{\frac{1}{7}\bigl(\frac{20}{30}+\frac{30}{30}\bigr)}{\frac{1}{3}} =\frac{\frac{50}{210}}{\frac{1}{3}} =\frac{150}{210} =\frac{5}{7}\,.$$

Hence the correct answer is Option (1): $$\boxed{\frac{5}{7}}$$.