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 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 on it.

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Solution

The total number of ways of ticking one or more alternatives out 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 1/15 that at the second trial is (14/15)(1/14)=1/15 and that at the third trial is (14 /15)(13/14)(1/13)=1/15.
Thus, the probability that the candidate will get marks on the questions if he is allowed up to three trials is 1/15+1/15+1/15+=1/5.

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