The total number of ways of ticking the answers in any one attempt = 24 - 1 = 15.
The student is taking chance at ticking the correct answer, It is reasonable to assume that in order to derive maximum benefit, the three solutions which he submit must be all different.
∴ n = total no. of ways = 15C3
m = the no. of ways in which the correct solution is excluded 14 C3
Hence the required probability = 1 - 14 C3/15 C3 = 1 – 4/5 = 1/5
ALTERNATE SOLUTION:
The candidate may tick one or more of the alternatives. As each alternative may or may not be chosen, the total numbers of exhaustive possibilities are 24 - 1 = 15.
Therefore the prob. that the questions are correctly answered by candidate is 1/15.
As such the candidate may be correct on the first, second or third chance. As these events are mutually exclusive, the total probability will be given by
= 1/15 + 14/15 x 1/14 + 14/15 x 13/14 x 1/13 = 1/15 + 1/15 + 1/15 = 3/15 = 1/5
Thus the probability that the candidate gets marks in the question is 1/5.
