UPSC NDA 2 2016 Mathematics Solved Paper

Concept:Odds in favour of an event $E$ are given by the ratio $P(E) : P(\text{not }E)$.From the odds, we can find $P(E) = \frac{\text{odds in favour}}{\text{odds in favour} + 1}$.Here we need the probability that at least two out of three candidates answer correctly.Explanation:Let the three candidates be $A$, $B$, and $C$.Their odds in favour of a correct answer are $5:2$, $4:3$, and $3:4$ respectively.So the probabilities of correct answers are:$P(A) = \frac{5}{5+2} = \frac{5}{7}$$P(B) = \frac{4}{4+3} = \frac{4}{7}$$P(C) = \frac{3}{3+4} = \frac{3}{7}$The probabilities of incorrect answers are the complements:$P(A^c) = 1 - \frac{5}{7} = \frac{2}{7}$$P(B^c) = 1 - \frac{4}{7} = \frac{3}{7}$$P(C^c) = 1 - \frac{3}{7} = \frac{4}{7}$Now, "at least two correct" means any two or all three are correct.The required probability is:$P = P(A)P(B)P(C^c) + P(A)P(B^c)P(C) + P(A^c)P(B)P(C) + P(A)P(B)P(C)$Substitute the values:$= \left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{4}{7}\right) + \left(\frac{5}{7}\right)\left(\frac{3}{7}\right)\left(\frac{3}{7}\right) + \left(\frac{2}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right) + \left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)$Simplify each term (numerator product over $7^3 = 343$):First term: $5 \times 4 \times 4 = 80$Second term: $5 \times 3 \times 3 = 45$Third term: $2 \times 4 \times 3 = 24$Fourth term: $5 \times 4 \times 3 = 60$Sum of numerators: $80 + 45 + 24 + 60 = 209$Thus $P = \frac{209}{343}$.Answer:$\frac{209}{343}$