Question

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

• A. $1-{\frac{14}{15}}^3$
• B. $(\frac{1}{15}{)}^3$
• C. $\frac{1}{5}$
• D. $Noneofthese$

Answer 1

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

The total number of ways of ticking one or more alternatives of 4 is
$4_{C_1}+4_{C_2}+4_{C_3}+4_{C_4}=15$
Out of these 15 combinations, only one combination is correct.
The probability of ticking the alternative Correctly at the first trial is $\frac{1}{15}$, that at the second trial is
$(\frac{14}{15}.\frac{1}{14})=\frac{1}{15}$ and that at the third trial is $\frac{14}{15}.\frac{13}{14}.\frac{1}{13}=\frac{1}{15}$
Thus the probability that the candidate will get marks on the question if he is allowed upto three chances, is $\left(\frac{1}{15}\right)+\left(\frac{1}{15}\right)+\left(\frac{1}{15}\right)=\frac{1}{5}$.