Beth is taking an eleven​-question ​multiple-choice test for which each question has three answer​ choices,...

Step 1
Introduction:
In a fair die, there are 6 possible outcomes, 1, 2, ..., 6, all of which are equally likely, that is, with probability ${\displaystyle \frac{1}{6}}$.
Step 2
Calculation:
On rolling the fair die once, the probability of getting 1 or 2 is ${\displaystyle \frac{1}{3}[=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}]}$.
Similarly, the probability of getting 3 or 4 is 1/3, and the probability of getting 5 or 6 is 1/3.
Thus, in case of each of the 11 questions, the probability of selecting each option is 1/3.
As the outcomes on the roll of a fair die are independent from one roll to another, the choices made in the different questions are also independent of one another.
Considering each attempted question as a trial, there are ${\displaystyle n=11}$ independent trials.
Considering it to be a success if the correct option is chosen, the probability of success in each trial is ${\displaystyle p=\frac{1}{3}}$, as each question has only one correct option.
Here, X can be considered as the number of successes, that is, number of questions correctly answered. Thus, X has a binomial distribution with parameters, ${\displaystyle n=11,p=\frac{1}{3}}$.
Since p denotes the probability of success, the probability of failure is, ${\displaystyle q=1-p=\frac{2}{3}}$.
If ${\displaystyle X\sim Binomial(n,p)}$, then the probability mass function of X is:
$f(x)=\{\begin{array}{ll}(\begin{array}{c}n\\ x\end{array})p^x{q}^{n-x};x=0,1,...,n;& 0\end{array}$ Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.
The probability mass function of X here is:
$f(x)=\{\begin{array}{ll}(\begin{array}{c}11\\ x\end{array})(\frac{1}{3}{)}^x(\frac{2}{3}{)}^{11-x};& x=0,1,2,...,11;\\ 0;& otherwise\end{array}$
The probability of exactly 4 correct answers, that is, exactly 4 successes is calculated below:
${\displaystyle (X=4)=f\left(4\right)}$
$$=(\begin{array}{c}11\\ x\end{array})(\frac{1}{3}{)}^4(\frac{2}{3}{)}^{11-4}$$
${\displaystyle =\left(330\right)\cdot \left(\frac{1}{81}\right)\cdot \left(\frac{128}{2187}\right)}$
${\displaystyle \approx 0.238446}$.
Thus, the probability of exactly 4 correct answers is 0.238446.

Step 1
Binomial Problem with ${\displaystyle n=10}$ and ${\displaystyle p\left(c\text{or}rect\right)=\frac{1}{3}}$
Find the probability of
a) exactly four correct answers
${\displaystyle P(x=4){=}^{10}{C}_4{\frac{1}{3}}^4\times {\frac{2}{3}}^6=binopdf(10,\frac{1}{3},4)=0.2276}$
b) fewer than three correct answers
${\displaystyle P(0\le x\le 2)=binomcdf(10,\frac{1}{3},2)=0.2991}$

Step 1
Hence, the probability of getting the correct answer is
$p=\frac{1}{3}=0.33333333$, as there are 3 choices (the die part simply shows she is guessing).
Note that the probability of x successes out of n trials is
$P(n,x)=nCx{p}^x(1-p{)}^{n-x}$
where
$n=\text{number of trials}=8$
$p=\text{the probability of a success}=0.333333333$
$x=\text{the number of successes}=6$
Thus, the probability is 
P(6)=0.017070569