What is the probability of getting at least 4 correct answers? A company gives prospective workers a 6-question, multiple choice test. Each question has 5 possible answers, so that there's a 1/5 or 20% chance of answering a question correctly just by guessing.
P. S. So, I was stuck because I just used the formula once, determining the probability of getting exactly 4 correct answers. My mistake was that I didn't count the probabilities of getting 5 and 6 correct answers and didn't add all of the probabilities together.
 A: Start by letting $\text{X}$ be the number of correct answers out of $6$ guessed correctly. Next, you are told that the probability of a single success is $0.2$. In the binomial pmf, this is '$p$'. Clearly '$q$' is then $0.8$ by the complement rule. The question requires to you find $Pr(X \geq 4)$ or similarly,
$$\underbrace{Pr(X = 4)}_{1} + \underbrace{Pr(X =5)}_{2} +  \underbrace{Pr(X = 6)}_{3}$$
Simply use the binomial pmf for each of the parts; $1$, $2$ and $3$ and sum their values.  
A: $ p=1/5=>q=4/5$
$ P(x=4)=c(6,4)(1/5)^4 (4/5)^2$
$=15.\frac{16}{5^6}$
$=\frac {250}{5^6}$
$p(x=5)=c(6,5) {1/5}^5 {4/5}$
$=6 . \frac{4}{5^6}$
$=\frac {24}{5^6}$
$p(x=6)=1/5^6$
Adding all the above 
$=275/5^6=0.0176$
A: So, there is a probability of $1/5$ for getting each question correct, and $6$ questions in total.  We want the probability for getting 4,5, or 6 from them correct.
Let $X$ represent the count of correctly guessed answers.
$$\mathsf P(X{\geq}4)~=~\mathsf P(X{=}4)+\mathsf P(X{=}5)+\mathsf P(X{=}6)$$
The probability for getting exactly $k$ from $6$ questions correct with that success rate is: $$\mathsf P(X{=}k)~=~\dbinom 6 k \dfrac {4^{6-k}}{5^6\quad}\cdot\mathbf 1_{k\in\{0,1,2,3,4,5,6\}}$$
Which is the count of ways of selecting $k$ questions, times the probability for getting those selection correct and the $6{-}k$ remainder incorrect.

Have you encountere the Binomial Distribution yet.   If so this should look familiar.   If not, say hello to the probaility mass function for a binomial distributed random variable with amount of trials of $n$ and success rate of $p$. 
$$X\sim\mathcal{Bin}(n, p)\quad\iff\quad\mathsf P(X{=}k)~=~\dbinom n k p^k(1-p)^{n-k}\cdot\mathbf 1_{k\in\{0,..,n\}}$$
