Question

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

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

Answer 1

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

The total number of ways of answering one or more alternatives of 4 is $4C_1+4C_2+4C_3+4C_4=15,$
Out of these 15 combinations, only one combination is correct. The probability of answering the alternaive 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 trail is $\frac{14}{15}.\frac{13}{14}.\frac{1}{13}=\frac{1}{15},$
Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is $\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{1}{5}.$