Question

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). $\dfrac{{13}}{{{3^5}}}$
(B). $\dfrac{{11}}{{{3^5}}}$
(C). $\dfrac{{10}}{{{3^5}}}$
(D). $\dfrac{{17}}{{{3^5}}}$

Answer 1

Hint: Before attempting this question, one should have prior knowledge about the concept of probability and also remember to use binomial distribution i.e. \[p\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}}\] 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 $\dfrac{1}{3}$
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 $1 - \dfrac{1}{3} = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}$
So, by the binomial distribution i.e. \[p\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}}\]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 = \[{}^5{C_4}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{5 - 4}}\]
As we know that formula of combination is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]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 = \[\dfrac{{5!}}{{4!\left( {5 - 4} \right)!}}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^{5 - 4}}\]
$ \Rightarrow $ Probability of getting 4 correct answers = \[5 \times {\left( {\dfrac{1}{3}} \right)^4} \times \dfrac{2}{3}\]
$ \Rightarrow $ Probability of getting 4 correct answers = \[5 \times \dfrac{2}{{{3^5}}}\]
For probability of 5 correct answers substituting the values in the above binomial distribution we get
Probability of getting 5 correct answers = \[{}^5{C_5}{\left( {\dfrac{1}{3}} \right)^5}{\left( {\dfrac{2}{3}} \right)^{5 - 5}}\]
As we know that formula of combination is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]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 = \[\dfrac{{5!}}{{5!\left( {5 - 5} \right)!}}{\left( {\dfrac{1}{3}} \right)^5}\]
$ \Rightarrow $ Probability of getting 5 correct answers = \[{\left( {\dfrac{1}{3}} \right)^5}\]
$ \Rightarrow $ Probability of getting 5 correct answers = \[\dfrac{1}{{{3^5}}}\]
Substituting the values in equation 1 we get
Probability of getting 4 or more correct answers = \[5 \times \dfrac{2}{{{3^5}}} + \dfrac{1}{{{3^5}}}\]
$ \Rightarrow $ Probability of getting 4 or more correct answers = \[\dfrac{1}{{{3^5}}}\left( {10 + 1} \right)\]
$ \Rightarrow $ Probability of getting 4 or more correct answers = \[\dfrac{{11}}{{{3^5}}}\]
Therefore, Probability of getting 4 or more correct answers = \[\dfrac{{11}}{{{3^5}}}\]
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. \[p\left( x \right) = {}^n{C_x}{p^x}{q^{n - x}}\] 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 $\dfrac{1}{3}$ so that the probability of an incorrect answer is $\dfrac{2}{3}$then to find the required probability we substituted the values in the formula and we found the required probability.