What is the probability that the student knew the answer to at least one of the two questions? A student takes a true-false examination containing 20
questions. On looking at the examination the student and that he
knows the answer to 10 of the questions which he proceeds to answer
correctly. He then randomly answers the remaining 10 questions. The
instructor selects 2 of the questions at random and  that the students answered both questions correctly.
What is the probability that the student knew the answer to at least one of the two questions?

The probability of the teacher selecting 2 question, for which the student don't know the answer is: $\frac{10}{20} \times \frac{9}{19}$. There are $\frac{1}{2}$ chance that he'll know the answer so for the both question the probability is:
$$\frac{10}{20} \times \frac{9}{19} \times \frac{1}{4} = \frac{9}{152}$$
The probability of the teacher selecting 1 question that he don't know and one that the student know is $\frac{10}{20} \times \frac{10}{19}$ and we know that there are $\frac 12$ chance to guess the answer right. So the probability is:
$$\frac{10}{20} \times \frac{10}{19} \times \frac{1}{2} = \frac{5}{38}$$
The probability of the teacher selecting 2 questions that the student knows the answers of is $\frac{10}{20} \times \frac{9}{19}$
$$\frac{10}{20} \times \frac{9}{19} = \frac{9}{38}$$
So the total probability is:
$$\frac{9}{152} + \frac{5}{38} + \frac {9}{38} = \frac{65}{152}$$
The probability of selecting at least one question that the student knows is:
$$\frac{5}{38} + \frac {9}{38} = \frac{7}{19}$$
Now divide them and you'll get:
$$\frac{\frac{7}{19}}{\frac{65}{152}} = \frac{56}{65} \approx 86.15\% $$
