An examination consists of 8 questions in each of which the candidate must say which one of the 5 alternatives is the correct one. Assuming that the student has not prepared earlier, choose for each of the questions any one of 5 answers with equal probability.
(i) Prove that the probability that he gets more than one correct answer is $ \dfrac{{{5^8} - 3 \times {4^8}}}{{{5^8}}} $

Hint: First of all, we will take a random variable describing numbers of correct answers. Then we will analyze the possible outcome of the event. After that we will find the probability of finding the wrong answer. And after that we will subtract the final answer from 1 to get the probability that he gets more than one correct answer.

Complete step-by-step answer:
Let ‘x’ be random variable describing no. of correct answer
So here we are given a total of 8 questions. Each question has 5 alternative options. So, from above five any one is the correct answer. If the student does not study anything and wants to appear in the exam then for him all the four questions have equal probability of getting the correct answer.
So, here we have taken ‘x’ as numbers of correct answers.
 $ P(x \leqslant 1) = P(x = 0) + P(x = 1) $ --(1)
So, here $ P(x > 1) $ represents the probability of getting more than one correct answer
 $ P(x = 0) $ represents the probability of getting a zero correct answer.
 $ P(x = 1) $ represents the probability of getting 1 correct answer.
So let’s find it
(1) $ P(x = 0) $
There are 5 choices and one is correct. So the remaining four are incorrect. Therefore each question has a probability of $ \dfrac{4}{5} $ . For eight questions we have a probability of $ {\left( {\dfrac{4}{5}} \right)^8} $
(2) $ P(x = 1) $
Any of the answers may be correct so one correct option can be chosen out of eight by $ ^8{C_1} $ ways and its probability is $ \dfrac{1}{5} $ . The remaining seven numbers have to be incorrect and thus their probability will be $ {\left( {\dfrac{4}{5}} \right)^7} $ . Therefore, the required probability of getting exactly one correct option is $ ^8{C_1} \times \dfrac{1}{5} \times {\left( {\dfrac{4}{5}} \right)^7} $ .
After putting these values in (1) we get:
 $ P(x \leqslant 1) = {\left( {\dfrac{4}{5}} \right)^8}{ + ^8}{C_1} \times \dfrac{1}{5} \times {\left( {\dfrac{4}{5}} \right)^7} $ --(2)
We have
\[(x > 1) = 1 - P(x \leqslant 1)\]
\[ \Rightarrow (x > 1) = 1 - {\left( {\dfrac{4}{5}} \right)^8}{ - ^8}{C_1} \times \dfrac{1}{5} \times {\left( {\dfrac{4}{5}} \right)^7}\]
\[ \Rightarrow (x > 1) = 1 - \dfrac{{{4^8}}}{{{5^8}}}{ - ^8}{C_1} \times \dfrac{{{4^7}}}{{{5^8}}}\]
\[ \Rightarrow (x > 1) = \dfrac{{{5^8} - {4^8}{ - ^8}{C_1}{4^7}}}{{{5^8}}}\]
\[ \Rightarrow (x > 1) = \dfrac{{{5^8} - {4^8} - 8 \times {4^7}}}{{{5^8}}} = \dfrac{{{5^8} - {4^8} - 2 \times 4 \times {4^7}}}{{{5^8}}}\]
\[ \Rightarrow (x > 1) = \dfrac{{{5^8} - {4^8} - 2 \times {4^8}}}{{{5^8}}}\]
\[ \Rightarrow (x > 1) = \dfrac{{{5^8} - 3 \times {4^8}}}{{{5^8}}}\]
Hence, we proved the required thing.

Note: While calculating the value of permutation and combination we should take proper care of what and how many are we choosing. Remember that the probability value always lies between 0 and 1 and never exceeds 1.