Brad answers 10 math problems, one after another. He answers the first problem correctly and answers the second problem incorrectly. For each of the remaining 8 problems, the probability that he answers the problem correctly equals the ratio of the number of problems that he has already answered correctly to the total number of problems that he has already answered. What is the probability that he answers exactly 5 out of the 10 problems correctly?
Brad's answer to each question is either right or wrong, and so there are $2^{8}$ possible combinations of right and wrong answers for the remaining 8 questions.

Since Brad has already answered exactly 1 question correctly, then $\binom{8}{4}$ of these combinations will include exactly 5 correct answers. (We choose 4 of the remaining 8 questions to be answered correctly.)

We show that each of these $\binom{8}{4}$ combinations has an equal probability, $p$.

This will mean that the probability that Brad has exactly 5 correct answers is $\binom{8}{4} p$.

Consider a specific combination that includes exactly 5 correct answers.

For each question from the 3rd to the 10th, the probability that his answer is correct equals the ratio of the number of problems that he has already answered correctly to the total number of problems that he has already answered.

Since the probability that his answer is wrong equals 1 minus the probability that his answer is correct, then the probability that his answer is wrong will equal the number of problems that he has already answered incorrectly to the total number of problems that he has already answered. (This is because the total number of problems answered equals the number correct plus the number incorrect.)

Therefore, for each problem from the 3rd to 10th, the associated probability is a fraction with denominator from 2 to 9 , respectively (the number of problems already answered).

Consider the positions in this combination in which he gives his 2nd, 3rd, 4th, and 5th correct answers. In these cases, he has previously answered 1,2,3, and 4 problems correctly, and so the numerators of the corresponding probability fractions will be $1,2,3$, and 4 .

Similarly, in the positions where his answer is wrong, the numerators will be 1, 2, 3, and 4 . Therefore, $p=\frac{(1 \cdot 2 \cdot 3 \cdot 4)(1 \cdot 2 \cdot 3 \cdot 4)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}$ and so the probability that Brad answers exactly 5 out of the 10 problems correctly is

$$
\binom{8}{4} \frac{(1 \cdot 2 \cdot 3 \cdot 4)(1 \cdot 2 \cdot 3 \cdot 4)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}=\frac{8!}{4!4!} \cdot \frac{(1 \cdot 2 \cdot 3 \cdot 4)(1 \cdot 2 \cdot 3 \cdot 4)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}=\frac{8!}{4!4!} \cdot \frac{4!4!}{9!}=\frac{8!}{9!}=\frac{1}{9}
$$