UPSC NDA 2 2019 Mathematics Solved Paper

Concept:Bayes' theorem – finding the probability that the two-headed coin was selected given that the flip shows heads.Explanation:Let $E_1$, $E_2$, $E_3$ be the events of selecting the two-headed coin, the biased coin (75% heads), and the fair coin respectively.Each coin is equally likely to be chosen: $P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}$.Let $A$ be the event that the coin shows heads.If the two-headed coin is chosen, head is certain: $P(A|E_1)=1$.If the biased coin is chosen, head appears 75% of the time: $P(A|E_2)=\frac{3}{4}$.If the fair coin is chosen, head appears half the time: $P(A|E_3)=\frac{1}{2}$.We need $P(E_1|A)$, the probability that the coin is two-headed given a head appeared.Using Bayes' theorem:$P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}$.Substitute the values:$P(E_1|A)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}$.Factor $\frac{1}{3}$ from denominator:$P(E_1|A)=\frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{1}{2}\right)}=\frac{1}{1+\frac{3}{4}+\frac{1}{2}}$.Simplify the sum inside denominator:$1+\frac{3}{4}+\frac{1}{2}=\frac{4}{4}+\frac{3}{4}+\frac{2}{4}=\frac{9}{4}$.Thus $P(E_1|A)=\frac{1}{\frac{9}{4}}=\frac{4}{9}$.Answer:$\frac{4}{9}$ (Option C).