Question

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

• A. $\frac{4}{5}$
• B. $\frac{1}{5}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

Answer: (B)

$\frac{1}{5}$

The total number of ways of ticking the answers in any one attempt$=2^4-1=15$
(the student is taking chance at ticking the correct solution, the case of his not ticking any answer is ruled out.)
It is reasonable to assume that in order to derive maximum benefit, the three solutions which he will submit must be all different.
$\therefore n=$ total no.of ways $={}^{15}{C}_3$
$m=$ the no.of ways in which the correct solution is excluded ${=}^{14}{C}_3.$
Hence the required probability $=1-\frac{{}^{14}{C}_3}{{}^{15}{C}_3}=1-\frac{4}{5}=\frac{1}{5}.$