UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The student is successful if they pass tests I and II, or I and III, or all three.The passing events are independent, so the probability of their intersection is the product of individual probabilities.Explanation:Let events A, B, C represent passing tests I, II, III respectively.Given $P(A) = m$, $P(B) = n$, $P(C) = \frac{1}{2}$.Probability of success $P(S) = \frac{1}{2}$.The successful cases are: (A and B but not C), (A and C but not B), (A and B and C).Since events are independent:$P(S) = P(A)P(B)P(C') + P(A)P(B')P(C) + P(A)P(B)P(C)$.$P(C') = 1 - \frac{1}{2} = \frac{1}{2}$, $P(B') = 1 - n$.Thus $\frac{1}{2} = m \cdot n \cdot \frac{1}{2} + m \cdot (1-n) \cdot \frac{1}{2} + m \cdot n \cdot \frac{1}{2}$.Multiply both sides by 2: $1 = mn + m(1-n) + mn$.Simplify: $1 = 2mn + m - mn = m + mn = m(1+n)$.Thus $m(1+n) = 1$.Answer:$m(1+n) = 1$ corresponds to option A.Correct option: A.