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**Hint:**There are 4 multiple choice questions. So we need to calculate the probability of choosing the correct answer for each of the 4 questions. Each question has 3 available options. And it is mentioned that Elliott chooses any options randomly. So the probability of choosing the correct option for one question with 3 options will be 1/3. Choosing any option independently means that her choice of option for 1 question will not affect her choice for any other question. So the probability of choosing the right answer for each of the 4 questions will be 1/3. Now calculate the probability of choosing correct options in all the questions by multiplying all the individual probabilities.

**Complete step by step solution:**We are given that she chooses the options randomly. Therefore her choice for one question will not affect her choice for any other question.

Probability of choosing correct option for each of the 4 questions \[ = {\rm{p = }}{{\rm{p}}_{Q1}} = {{\rm{p}}_{Q2}} = {{\rm{p}}_{Q3}} = {{\rm{p}}_{Q4}} = \dfrac{1}{3}\].

Now to calculate the probability of choosing the correct option for all 4 questions, we use the rule of multiplication.

\[{\rm{p}}\left( {{{\rm{Q}}_1} \cap {{\rm{Q}}_2} \cap {{\rm{Q}}_3} \cap {{\rm{Q}}_4}} \right) = {\rm{p}}\left( {{{\rm{Q}}_1}} \right){\rm{p}}\left( {{{\rm{Q}}_2}} \right){\rm{p}}\left( {{{\rm{Q}}_3}} \right){\rm{p}}\left( {{{\rm{Q}}_4}} \right)\] [since all event are independent]

\[ = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{{81}}\]

Thus the required probability is \[\left( {\rm{D}} \right)\dfrac{1}{{81}}\]

**Note:**In this type of question you need to read the question very carefully. You should understand the meaning of choosing one option randomly. You should also understand that choosing an option independently signifies that the event of choosing option for each of the 4 questions is 4 independent events.

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