UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Event $B$ is partitioned into four mutually exclusive sub-events involving $A$ and $C$.The probability of $B$ equals the sum of the probabilities of these sub-events.Using the two given sub‑event probabilities, we solve for $P(B\cap C)$.Explanation:Write $B$ as a union of four disjoint parts:$B = (A\cap B\cap C)\cup (A\cap B\cap\overline{C})\cup (\overline{A}\cap B\cap C)\cup (\overline{A}\cap B\cap\overline{C})$.Thus $P(B)=P(A\cap B\cap C)+P(A\cap B\cap\overline{C})+P(\overline{A}\cap B\cap C)+P(\overline{A}\cap B\cap\overline{C})$.We know $P(B)=\frac{3}{4}$, $P(A\cap B\cap\overline{C})=\frac{1}{3}$, and $P(\overline{A}\cap B\cap\overline{C})=\frac{1}{3}$.Notice $P(B\cap C)=P(A\cap B\cap C)+P(\overline{A}\cap B\cap C)$, which are the two unknown terms.Substitute known values into the equation for $P(B)$:$\frac{3}{4}=\frac{1}{3}+\frac{1}{3}+P(B\cap C)$.Simplify: $\frac{3}{4}=\frac{2}{3}+P(B\cap C)$.Solve: $P(B\cap C)=\frac{3}{4}-\frac{2}{3}=\frac{9-8}{12}=\frac{1}{12}$.Answer:$\frac{1}{12}$