An objective type test paper has 5 questions. Out of these 5 questions, 3 questions have four options each (a, b, c, d) with one option being the correct answer. The other 2 questions have two options each, namely true and false. A candidate randomly ticks the options. Then, what is the probability that he/she will tick the correct option in at least four questions?
A. $\dfrac{5}{{32}}$
B. $\dfrac{3}{{128}}$
C. $\dfrac{3}{{256}}$
D. $\dfrac{3}{{64}}$

Hint: In this question, firstly find the probability of success in both types of questions given in the test paper (MCQ & True/False). After that find all the possibilities for the statement given in the question and calculate the total probability by taking the summation of all.

Complete step-by-step answer:
Let success be the correction answer for a question and failure be the wrong answer.
Now, probability of success for question having 4 options $ = \dfrac{1}{4}$
We know that, Probability of success of an event E + Probability of Failure of an event E = 1
probability of failure for question having 4 options $ = 1 - \dfrac{1}{4} = \dfrac{3}{4}$
Now, probability of success for question having 2 options $ = \dfrac{1}{2}$
We know that, Probability of success of an event E + Probability of Failure of an event E = 1
probability of failure for question having 2 options $ = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Now, probability of getting at least 4 questions correct = probability of getting exactly 4 questions correct + probability of getting exactly 5 questions correct.
Probability of getting exactly 4 questions correct = 1 correct True/False question and 3 MCQ + 2 correct True/False questions and 2 MCQ’s.
\[ = {}^2{C_1} \times \dfrac{1}{2} \times \dfrac{1}{2} \times {}^3{C_3} \times {\left( {\dfrac{1}{4}} \right)^3} + {}^2{C_2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times {}^3{C_2} \times {\left( {\dfrac{1}{4}} \right)^2} \times \dfrac{3}{4}\]
$ = 2 \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{1}{2} \times \dfrac{1}{2} \times 3 \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{3}{4}$
$ = \dfrac{1}{{128}} + \dfrac{9}{{256}} = \dfrac{{11}}{{256}}$
Probability of getting exactly 5 questions correct \[ = {}^3{C_3} \times {\left( {\dfrac{1}{4}} \right)^3} \times {}^2{C_2} \times \dfrac{1}{2} \times \dfrac{1}{2}\]
$ = \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{{256}}$
Total probability $ = \dfrac{{11}}{{256}} + \dfrac{1}{{256}} = \dfrac{{12}}{{256}} = \dfrac{3}{{64}}$
Hence, the probability that he/she will tick the correct option in at least four questions is $\dfrac{3}{{64}}$
$\therefore $ Option D. $\dfrac{3}{{64}}$ is the correct answer.

Note: For such types of questions just keep in mind that Probability of success of an event E + Probability of Failure of an event E = 1 and also find the probabilities of different scenarios for both the success as well as failure and then consider the total of all.