Question

An objective test paper has 5 questions. Out of these 5 questions, 3 questions have four options (a, b, c and d) with one option as the correct answer. The other two questions have two options each, namely True and False. A candidate randomly ticks the options. Then the probability that he/she will tick the correct option in at least 4 questions is?
(a) $ \dfrac{5}{32} $
(b) $ \dfrac{3}{128} $
(c) $ \dfrac{3}{256} $
(d) $ \dfrac{3}{64} $

Answer 1

Hint: We have to find the probability that in a test paper out of 5 questions, a student has ticked at least 4 questions correct that means we have to find the probability when the student has ticked 4 questions correct and add this probability with the probability when the student ticks all the 5 questions correct. In the three questions out of five each question has four options so the probability of ticking the correct option is $ \dfrac{1}{4} $ and in the two questions out of 5 each question has 2 options so the probability of ticking the correct option in these two questions is $ \dfrac{1}{2} $ so using these probabilities find the probability when student has 4 questions and all the 5 questions correct.

Complete step-by-step answer:
In the problem, it is given that a test paper has 5 questions in which 3 questions has 4 options each and two questions has two options each.
Now, the probability that the question is correct where 4 options (Multiple choice question) are given is equal to $ \dfrac{1}{4} $ then the probability that wrong option has ticked in this question is $ \dfrac{3}{4} $ .
The probability that the question is correct where 2 options are given is equal to $ \dfrac{1}{2} $ then the probability that the wrong option has ticked in this question is $ \dfrac{1}{2} $ .
Now, we are asked to find the probability when at least 4 questions are ticked correctly by the student for that we have to find the probability when the student ticked 4 questions correctly and the case when the student has ticked all the 5 questions correctly.
We are going to find the probability when a student has ticket 4 questions correct out of 5 questions and 1 question incorrect.
There are two possibilities when 4 questions are correct and 1 question is incorrect as follows:
When 3 multiple choice questions (MCQ) and 1 true/ false question is correct and one true/false question is incorrect. To get the probability of this, we are going to multiply the correct probability of 3 MCQ in the following:
 $ \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4} $
And multiplying the above expression with the probability of 1 correct true/false question we get,
 $ \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2} $
 And we also have to multiply the selection of 1 true/false question out of 2 true/false questions.
The selection of 1 true/false out of two true/false questions is shown below:
 $ {}^{2}{{C}_{1}} $
We know that, $ {}^{n}{{C}_{1}}=n $ so using this relation in the above expression we get,
2
Multiplying 2 by $ \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2} $ we get,
 $ \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2}\times 2 $
Multiplying the probability of incorrect true/false question we get,
 $ \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2}\times 2\times \dfrac{1}{2} $
The result of this multiplication is:
 $ \dfrac{2}{256} $
The result of case 1 is $ \dfrac{2}{256} $ .
The other case is when 2 true/false questions are correct and 2 MCQ questions are correct and one incorrect MCQ question. To find the probability of this case we have to first select 2 MCQ questions out of 3 which we are done as follows:
 $ {}^{3}{{C}_{2}} $
We know that, $ {}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}} $ using this relation in the above expression we get,
 $ \begin{align}
  & {}^{3}{{C}_{1}} \\
 & =3 \\
\end{align} $
Now, multiplying 3 by the multiplication of the probabilities of 2 MCQ correct questions and 2 true/false correct questions we get,
 $ 3\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2}\times \dfrac{1}{2} $
Multiplying the above expression with probability of 1 incorrect MCQ question we get,
 $ \begin{align}
  & 3\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{3}{4} \\
 & =\dfrac{9}{256} \\
\end{align} $
The result of case 2 is $ \dfrac{9}{256} $ .
Adding the results of case 1 and case 2 we get the probability of marking the 4 questions as correct is:
 $ \begin{align}
  & \dfrac{2}{256}+\dfrac{9}{256} \\
 & =\dfrac{11}{256} \\
\end{align} $
Now, we have to find the probability when all the questions are marked correct is the multiplications of the probability of 3 correct MCQ questions and 2 correct true/false questions.
 $ \begin{align}
  & \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{4}\times \dfrac{1}{2}\times \dfrac{1}{2} \\
 & =\dfrac{1}{256} \\
\end{align} $
Adding the probabilities when 4 questions and 5 questions are correct we get,
 $ \begin{align}
  & \dfrac{11}{256}+\dfrac{1}{256} \\
 & =\dfrac{12}{256} \\
 & =\dfrac{3}{64} \\
\end{align} $
So, the correct answer is “Option D”.

Note: The mistake that could happen in the above problem is that while writing the probabilities of 4 correct questions you will forget to write the probability of 1 incorrect question because the student has attempted 5 questions and we have to write all the probabilities of attempting 5 questions. Usually, one thinks that we have to write probabilities of 4 correct questions so that person only writes the probabilities of correct questions and not even forget that person thinks we just have to write the correct probabilities.