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Question: A multiple choice examination has 5 questions. Each question has three alternative answers of which ...

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 is
(A). 1335\dfrac{{13}}{{{3^5}}}
(B). 1135\dfrac{{11}}{{{3^5}}}
(C). 1035\dfrac{{10}}{{{3^5}}}
(D). 1735\dfrac{{17}}{{{3^5}}}

Explanation

Solution

Before attempting this question, one should have prior knowledge about the concept of probability and also remember to use binomial distribution i.e. p(x)=nCxpxqn−xp\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}} here x is the times of success and n is the total numbers of times the experiment occurred, use this information to approach the solution of the problem.

Complete step-by-step answer :
According to the given information it is given that multiple choice examination has 5 questions where each has three alternative answers and only one is correct and we have to find the probability of student of getting 4 or more correct answers
As we know that we have only 3 choice where only one choice is correct
Therefore, the probability of getting a correct answer is equal to 13\dfrac{1}{3}
Also, since only out of 3 choices only 1 choice is correct so the 2 choices will be incorrect
Therefore, the probability of getting an incorrect answer is equal to 1−13=3−13=231 - \dfrac{1}{3} = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}
So, by the binomial distribution i.e. p(x)=nCxpxqn−xp\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}}here x is the times of success and n is the total numbers of times the experiment occurred
Therefore, the probability of getting 4 or more correct answers = probability of 4 correct answers + probability of 5 correct answers (equation 1)
As we know that for probability of 4 correct answer x = 4 correct answers and n = 5
Also, for probability of 5 correct answers x = 5 and n = 5
For probability of 4 correct answers substituting the values in the above binomial distribution we get
Probability of getting 4 correct answers = 5C4(13)4(23)5−4{}^5{C_4}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{5 - 4}}
As we know that formula of combination is given by nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}here r is the size of each permutation and n is the size of the set from which elements are permuted
Therefore, Probability of getting 4 correct answers = 5!4!(5−4)!(13)4(23)5−4\dfrac{{5!}}{{4!\left( {5 - 4} \right)!}}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{5 - 4}}
⇒\Rightarrow Probability of getting 4 correct answers = 5×(13)4×235 \times {\left( {\dfrac{1}{3}} \right)^4} \times \dfrac{2}{3}
⇒\Rightarrow Probability of getting 4 correct answers = 5×2355 \times \dfrac{2}{{{3^5}}}
For probability of 5 correct answers substituting the values in the above binomial distribution we get
Probability of getting 5 correct answers = 5C5(13)5(23)5−5{}^5{C_5}{\left( {\dfrac{1}{3}} \right)^5}{\left( {\dfrac{2}{3}} \right)^{5 - 5}}
As we know that formula of combination is given by nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}here r is the size of each permutation and n is the size of the set from which elements are permuted
Therefore, Probability of getting 5 correct answers = 5!5!(5−5)!(13)5\dfrac{{5!}}{{5!\left( {5 - 5} \right)!}}{\left( {\dfrac{1}{3}} \right)^5}
⇒\Rightarrow Probability of getting 5 correct answers = (13)5{\left( {\dfrac{1}{3}} \right)^5}
⇒\Rightarrow Probability of getting 5 correct answers = 135\dfrac{1}{{{3^5}}}
Substituting the values in equation 1 we get
Probability of getting 4 or more correct answers = 5×235+1355 \times \dfrac{2}{{{3^5}}} + \dfrac{1}{{{3^5}}}
⇒\Rightarrow Probability of getting 4 or more correct answers = 135(10+1)\dfrac{1}{{{3^5}}}\left( {10 + 1} \right)
⇒\Rightarrow Probability of getting 4 or more correct answers = 1135\dfrac{{11}}{{{3^5}}}
Therefore, Probability of getting 4 or more correct answers = 1135\dfrac{{11}}{{{3^5}}}
Hence, option B is the correct option.

Note : The trick to approach the above problem was to find the probability of getting 4 or more correct answers we found the probability of 4 correct answers and probability of 5 correct answers as to find the probability of both the cases we used the method of binomial distribution i.e. p(x)=nCxpxqn−xp\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}} here x is the times of success and n is the total numbers of times the experiment occurred as to find the required probability we first found the probability of one correct answer as we has 3 alternative answers for one question so we got that probability of one answer will be 13\dfrac{1}{3} so that the probability of an incorrect answer is 23\dfrac{2}{3}then to find the required probability we substituted the values in the formula and we found the required probability.