Finding probability of earning points A professor asks a true/false question with ten individual questions. Suppose the professor assigns grades of: $10$ points for each correct response, $0$ points for each absent response, and $-10$ for each incorrect response, but where negative totals are always replaced by zero. Les Guess, a student who understands nothing of the professor's class, flips a fair coin to decide if he will answer each question true or false.
a) In how many different ways might Les answer the question?
b) What is the probability that Les Guess will earn $n$ points? (This is just the number of ways to answer the question that will result in n points divided by the total number of ways of answering the question.)
 A: I will (for the moment) assume that he picks true, false, or abstain equally likely. (otherwise, why mention the abstain possibility?)
(a) There are $3$ possibilities for each question, so there are $3^{10}$ ways to take the test.
(b) To score $10n$, he needs to answer $k$ questions right and $k-n$ answers wrong. So the number of ways is
$${10 \choose k} {{10-k} \choose {k-n}}$$
for $0 \le k \le 10$, where ${ x \choose 0}=1$, ${x \choose {j<0}} = 0$, and ${0 \choose 10\ge j\ge0} = 1$.
Then the total probability is
$$P(10n) = \frac{1}{3^{10}}\sum_{k=0}^{10} {10 \choose k} {{10-k} \choose {k-n}}$$
for $n > 0$. This is symmetric with $n < 0$ and the probability of having equal numbers of rights and wrongs is given by the above definition of $P(10n)$ by setting $n=0$.
Other interpretation
Now I will assume that he chooses true or false equally likely and never abstains.
(a) There are $2^{10}$ ways to take the test.
(b) Let $r$ denote the number of correct answers and $w$ the number of wrong answers. Say he scores $10n$. We have that
$$r+w = 10 \\ r - w = n$$
So if $n$ is odd, there are no ways. If $n$ is even, then $r=5+n/2$ and $w = 5-n/2$. We now choose $r$ of the $10$ questions to be correct, so there are ${10 \choose {5+n/2}}$ ways to take the test while scoring $n$. So the probability is:
$$
   P(10n) = \left\{
     \begin{array}{lr}
       \frac{1}{2^{10}}{10 \choose {5+n/2}} & : \text{n even}\\
       0 & : \text{n odd}
     \end{array}
   \right.
$$
for $1 \le n \le 10$. For $n=0$, we can score negative or $0$. Negative n is symmetric with the positive $n$, and ${10 \choose 5}$ is the number of ways to get exactly $0$, so
$$P(0)=\frac{1}{2^{10}}{10 \choose 5} + \sum_{n=1}^{10} P(10n)$$
