A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing it:
(a) $ \dfrac{10}{{{3}^{5}}} $
(b) $ \dfrac{17}{{{3}^{5}}} $
(c) $ \dfrac{13}{{{3}^{5}}} $
(d) $ \dfrac{11}{{{3}^{5}}} $

Hint: In this question, first find out the probability of success and failure and determine how many numbers of successes you need and using the binomial probability distribution formula, find out the probability of guessing 4 or more answers correctly. \[P\left( x \right)=\] $ ^{n}{{C}_{x}} $ $ \cdot {{\left( p \right)}^{x}}\cdot {{\left( q \right)}^{n-x}} $

Complete step-by-step answer:
Here, the student has to answer 5 questions with multiple choices in each question. One of them is correct. Let us find the probability that the student will get 4 or more answers by guessing it.
Here, firstly we need to determine the two outcomes, one being the success and the other one being the failure.
Therefore, the student getting the answer right will be the success outcome = p
The student getting the answer incorrect will be considered as a failure outcome = q
Now, let us find out the probability of success for answering a question, $ \text{p = }\dfrac{1}{3} $ .
Now, let us find the probability of failure for answering a question.
We know, that p + q = 1
Therefore, we can substitute the value of p in the above equation to find the value of q.
 $ \begin{align}
  & \text{p + q = 1} \\
 & \dfrac{1}{3}\,+\text{q = 1} \\
 & \text{q = 1}-\dfrac{1}{3} \\
 & =\dfrac{3-1}{3} \\
 & =\dfrac{2}{3}
\end{align} $
So, we found the values of p and q, now we can find the probability of answering 4 or more correct.
We know,
The binomial probability distribution, \[P\left( x \right)=\] $ ^{n}{{C}_{x}} $ $ \cdot {{\left( p \right)}^{x}}\cdot {{\left( q \right)}^{n-x}} $
where, $ x $ = number of successes, p = probability of success, q = probability of failure, n = number of trials
Now, let us substitute the values $ x $ = 4, p = $ \dfrac{1}{3} $ , q = $ \dfrac{2}{3} $ , n = 5 in the above binomial probability distribution.
Therefore, the probability of guessing 4 or more correct answers, $ P\left( X\ge 4 \right) $
Now, expand the distribution, we get
 $ P\left( X\ge 4 \right)=P\left( X=4 \right)+P\left( X=5 \right) $
 $ P\left( X\ge 4 \right)= $ $ ^{5}{{C}_{4}}\cdot {{\left( \dfrac{1}{3} \right)}^{4}}\cdot {{\left( \dfrac{2}{3} \right)}^{5-4}}+ $ $ ^{5}{{C}_{5}}\cdot {{\left( \dfrac{1}{3} \right)}^{5}}\cdot {{\left( \dfrac{2}{3} \right)}^{5-5}} $
                     $ \begin{align}
  & =\dfrac{5\times 4\times 3\times 2\times 1}{1\times 2\times 3\times 4}\cdot \left( \dfrac{1}{81} \right)\cdot \left( \dfrac{2}{3} \right)+\dfrac{5\times 4\times 3\times 2\times 1}{1\times 2\times 3\times 4\times 5}\cdot \left( \dfrac{1}{243} \right)\cdot {{\left( \dfrac{2}{3} \right)}^{0}} \\
 & =5\cdot \left( \dfrac{2}{243} \right)+1\cdot \left( \dfrac{1}{243} \right)\cdot \left( 1 \right) \\
 & =\dfrac{10}{243}+\dfrac{1}{243} \\
 & =\dfrac{10+1}{243} \\
 & =\dfrac{11}{243}
\end{align} $
We can write the number 243 as $ {{3}^{5}} $ , therefore $ P\left( X\ge 4 \right)=\dfrac{11}{{{3}^{5}}} $
Hence, the probability of the student guessing 4 or more correct answers is $ \dfrac{11}{{{3}^{5}}} $ .

Note: Here, follow the rules of the binomial probability distribution, the formula is important to get the answer. Also remember, read the question carefully to create the inequalities which makes it easier to find the solution. The decision of selecting the values of X is very important.