Question

# As part of a probability experiment, Elliott is to answer 4- multiple choice questions. For each question, there are 3 possible answers, only 1 of which is correct. If Elliott randomly and independently answers each question, what is the probability that he will answer the 4 questions correctly?A. $\dfrac{{27}}{{81}}$B. $\dfrac{{12}}{{81}}$C. $\dfrac{4}{{81}}$D. $\dfrac{3}{{81}}$E. $\dfrac{1}{{81}}$

Probability of choosing correct option for each of the 4 questions $= {\rm{p = }}{{\rm{p}}_{Q1}} = {{\rm{p}}_{Q2}} = {{\rm{p}}_{Q3}} = {{\rm{p}}_{Q4}} = \dfrac{1}{3}$.
${\rm{p}}\left( {{{\rm{Q}}_1} \cap {{\rm{Q}}_2} \cap {{\rm{Q}}_3} \cap {{\rm{Q}}_4}} \right) = {\rm{p}}\left( {{{\rm{Q}}_1}} \right){\rm{p}}\left( {{{\rm{Q}}_2}} \right){\rm{p}}\left( {{{\rm{Q}}_3}} \right){\rm{p}}\left( {{{\rm{Q}}_4}} \right)$ [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 $\left( {\rm{D}} \right)\dfrac{1}{{81}}$