Question

In a multiple choice question there are four alternative answer, of which one or more are correct. A candidate will get marks in the question only if he ticks all the correct answer. The candidate decides to tick answer at random, if he is allowed upto three chances the probability that he will get marks for the question, is

• A. $\frac{1}{2}$
• B. $\frac{1}{5}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B $\frac{1}{5}$

The total number of ways of ticking the answers in any one attempt $2^4-1=15$.
It is reasonable to assume that in order to derive maximum benifit, the three solutions which he will submit must be all different.
$\therefore$ total number of ways, $n\left(S\right)={}^{15}{\text{C}}_3$
The number of ways in which the correct solution is excluded $={}^{14}{\text{C}}_3$
Hence the required probability
$=1-\frac{{}^{14}{\text{C}}_3}{{}^{15}{\text{C}}_3}=1-\frac{4}{5}=\frac{1}{5}$