Question

An examination consists of 8 questions in each of which one of the 5 alternatives is the correct one. On the assumption that a candidate who has done no preparatory work chooses for each question any one of the five alternatives with equal probability, the probability that he gets more than one correct answer is equal to five alternatives with equal probability, the probability that he gets more than one correct answer is equal to
A.${\left( {0.8} \right)^8}$
B.$3{\left( {0.8} \right)^8}$
C.$1 - {\left( {0.8} \right)^8}$
D.$1 - 3{\left( {0.8} \right)^8}$

Answer 1

Here, we are required to find the probability that the student who has not done any preparatory work gets more than one correct answer. Since, we know that the total probability is 1. Hence, in order to find the probability of getting more than 1 answer correct, we will subtract the probability of no correct answer and the probability of only 1 correct answer from the total probability, i.e. 1 and hence, find the required probability.
Formula Used: We will use the following formulas:
1.${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$
2.${a^m} \times {a^n} = {a^{m + n}}$

Complete step-by-step answer:
According to the question,
The total number of questions $= 8$
Each question is having 5 alternative options, where only 1 option is correct.
Since, the given candidate has done no preparatory work, hence, for him, each option has equal probability of being correct.
Since, the total number of options is 5.
Hence, the probability of each answer being correct is $\dfrac{1}{5}$.
And, the probability of the answer being incorrect is $\dfrac{4}{5}$.
Here, we will let $X$ to be a random variable representing the number of correct answers.
Here, there are three possible cases:
All the chosen answers are incorrect.
Since, all the answers chosen are incorrect this means that:
In one question, the probability of the answer being incorrect is $\dfrac{4}{5}$
Hence, in all the 8 questions, the probability of the answers being incorrect is ${\left( {\dfrac{4}{5}} \right)^8}$
Hence, $P\left( {X = 0} \right) = {\left( {\dfrac{4}{5}} \right)^8}$
From the chosen answers only 1 is correct.
The probability that an answer is correct is $\dfrac{1}{5}$
Hence, the probability that the remaining 7 answers are not correct is ${\left( {\dfrac{4}{5}} \right)^7}$
But, from the 8 questions, any one of the answers could be correct, hence, we can choose that correct answer in ${}^8{C_1} = 8$ways
Therefore, the required probability of getting exactly 1 correct option $= P\left( {X = 1} \right) = 8 \times {\left( {\dfrac{4}{5}} \right)^7} \times \left( {\dfrac{1}{5}} \right)$
Now, we are required to find the probability of getting more than 1 answer as correct.
Hence, the total probability i.e. 1 can be stated as:
$1 = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X > 1} \right)$
This is because the candidate can either get all the answers as incorrect, only 1 answer as correct or more than 1 correct answer.
This can also be written as:
$\Rightarrow P\left( {X > 1} \right) = 1 - P\left( {X = 0} \right) - P\left( {X = 1} \right)$
Hence, in order to find the probability of getting more than 1 answer as correct, we can simply subtract the probability of not getting any answer as correct and the probability of getting 1 answer as correct from the total probability to get the required probability.
Hence, substituting the values from above, we get,
$\Rightarrow P\left( {X > 1} \right) = 1 - {\left( {\dfrac{4}{5}} \right)^8} - 8 \times {\left( {\dfrac{4}{5}} \right)^7} \times \left( {\dfrac{1}{5}} \right)$
Using, ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$
$\Rightarrow P\left( {X > 1} \right) = 1 - \left( {\dfrac{{{4^8}}}{{{5^8}}}} \right) - \left( {\dfrac{{{4^7}}}{{{5^7}}}} \right) \times \left( {\dfrac{8}{5}} \right)$
Now, we can write ${a^m} \times {a^n} = {a^{m + n}}$
$\Rightarrow P\left( {X > 1} \right) = 1 - \left( {\dfrac{{{4^8}}}{{{5^8}}}} \right) - \left( {\dfrac{{8 \times {4^7}}}{{{5^{7 + 1}}}}} \right)$
$\Rightarrow P\left( {X > 1} \right) = 1 - \left( {\dfrac{{{4^8}}}{{{5^8}}}} \right) - \left( {\dfrac{{8 \times {4^7}}}{{{5^8}}}} \right)$
Now, taking the LCM, we get
$\Rightarrow P\left( {X > 1} \right) = 1 - \left( {\dfrac{{{4^8} + 8 \times {4^7}}}{{{5^8}}}} \right)$
Taking the like exponents common, we get,
$\Rightarrow P\left( {X > 1} \right) = 1 - \dfrac{{{4^7}}}{{{5^8}}}\left( {4 + 8} \right) = 1 - \dfrac{{{4^7}}}{{{5^8}}}\left( {12} \right)$
Now, the number 12 can also be written as: $12 = 3 \times 4$
$\Rightarrow P\left( {X > 1} \right) = 1 - \dfrac{{{4^7}}}{{{5^8}}}\left( {3 \times 4} \right)$
Hence, again using, ${a^m} \times {a^n} = {a^{m + n}}$
$\Rightarrow P\left( {X > 1} \right) = 1 - \dfrac{{{4^{7 + 1}}}}{{{5^8}}}\left( 3 \right) = 1 - \left( 3 \right)\dfrac{{{4^8}}}{{{5^8}}}$
Now, we can write $\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}$
$\Rightarrow P\left( {X > 1} \right) = 1 - \left( 3 \right){\left( {\dfrac{4}{5}} \right)^8}$
Converting the fraction as decimal, as $\dfrac{4}{5} = 0.8$
$\Rightarrow P\left( {X > 1} \right) = 1 - 3{\left( {0.8} \right)^8}$
Therefore, the probability that he gets more than one correct answer is equal to $1 - 3{\left( {0.8} \right)^8}$
Hence, option D is the correct answer.

Note: Probability tells us the extent to which an event is likely to occur, i.e. the possibility of the occurrence of an event. Hence, it is measured by dividing the favorable outcomes by the total number of outcomes. Probability distribution describes the likelihood of occurrence of all the possible values that a given variable can assume. We use probability in our day to day life to make decisions when we are not sure about the result of that decision. Hence, this concept is important in our everyday life as well.