# 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?

When complete, the following table shows the probabilities of getting $2$, $1$, or $0$ of the two questions right depending on the kinds of questions. $K$ refers to a question with a known answer and $U$ to a question with an unknown answer; the first letter is for the first question graded, and the second is for the second question graded. I’ve omitted most of the entries, leaving them for you to compute.

$$\begin{array}{c|cc} &2&1&0\\ \hline KK&1&0&0\\ KU&a&&0\\ UK&b&&0\\ UU&c&&1/4 \end{array}$$

The student got both questions right, for a score of $2$, so the entries that you need are those in the first column: you need to determine what fraction of the total $1+a+b+c$ is contributed by the $KK$ case.

• I totally forgot about this approach. – Joe Z. Sep 27 '13 at 17:11
• @Joe: It’s really just Bayes’ theorem without the formulas. – Brian M. Scott Sep 27 '13 at 17:24

Hint: What's the probability that the student guessed both correct questions randomly?

Hint 2: For that matter, you also need to model what happens when only one question is chosen by the instructor. What's the probability that the student chose randomly then? Remember that not every case has exactly 15 questions being answered correctly.

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:

$$\frac{\frac{7}{19}}{\frac{65}{152}} = \frac{56}{65} \approx 86.15\%$$