Probability of getting at most 4 questions wrong out of 10 T/F questions If I have 10 T/F questions, the number of ways I can get exactly 1 question wrong is $C(10,1) \cdot (1/2)^{10}$ and getting exactly 2 questions wrong is $C(10,2) \cdot (1/2)^{10}$ and so on. Hence, under my understanding and after reading solutions for similar questions I am calculating the right answer for the probability of getting at most 4 questions wrong is "getting exactly 1 question wrong + getting exactly 2 questions wrong + exactly 3 questions wrong + 4 questions wrong" $= (C(10,1) + C(10,2)+C(10,3) + C(10,4) ) \cdot (1/2)^{10}$. However, this is wrong and I am not sure where I am erring. Can someone offer some clarification here?
 A: It seems that you may have studied (or about to study) the binomial distribution.
Formally relating this problem to the binomial distribution may be helpful.
The binomial distribution has two parameters: the number $n$ of independent 'trials', each with two possible outcomes, often called S and F; and
$p = P(S)$ on each trial. (See the specific formulation in your text.)
Let $X$ be the number of questions wrong. In your case,
$X \sim Binom(n = 10, p =.5).$ Then 
$$P(X = k) = C(10,k)p^k(1-p)^{n-k} = C(10,k)(.5)^{10},$$
for $k = 0, 1, 2, \dots, n.$
You are not using the terminology quite properly, as mentioned by @JMoravitz
and you should pay attention to that.
But I think your main difficulty is that you have left out $P(X = 0),$
as mentioned by @Henry.
$P(X \le 4) = P(X = 0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4).$
There may be a table in your text with some binomial probabilities already
computed. Also, various kinds of statistical software can be used. But in
your problem, the computations are easy enough to do by hand or with a
calculator, using the formula. I think that may be the point of assigning
this problem.
Below are two bits of output from R statistical software. The first gives
the five terms in the sum, and the second gives their total. They are
all given to more decimal places than you need, but you can use them to
check your work; perhaps you should round to four places. Then please take
a moment to figure out why you needed help on this problem.
 dbinom(0:4, 10, .5)
 ## 0.0009765625 0.0097656250 0.0439453125 0.1171875000 0.2050781250
 pbinom(4, 10, .5)
 ## 0.3769531

Below is a bar chart of the eleven probabilities in the distribution of $X \sim Binom(10,.5).$ The five probabilities you need are emphasized in blue.

