NOTE

Question 2 (Paper - 1)

Three students $S_1$, $S_2$, and $S_3$ are given a problem to solve. Consider the following events:

$$U: \text{At least one of } S_1, S_2, \text{ and } S_3 \text{ can solve the problem},$$$$V: S_1 \text{ can solve the problem, given that neither } S_2 \text{ nor } S_3 \text{ can solve the problem},$$$$W: S_2 \text{ can solve the problem and } S_3 \text{ cannot solve the problem},$$$$T: S_3 \text{ can solve the problem}.$$

For any event $E$, let $P(E)$ denote the probability of $E$. If

$$P(U) = \frac{1}{2}, \quad P(V) = \frac{1}{10}, \quad \text{and} \quad P(W) = \frac{1}{12},$$

then $P(T)$ is equal to

$$\begin{array}{|c|c|c|c|} \hline \hspace{0.8cm} \text{(A) } \frac{13}{36} \hspace{0.8cm} & \hspace{0.8cm} \text{(B) } \frac{1}{3} \hspace{0.8cm} & \hspace{0.8cm} \text{(C) } \frac{19}{60} \hspace{0.8cm} & \hspace{0.8cm} \text{(D) } \frac{1}{4} \hspace{0.8cm} \\ \hline \end{array}$$

With Venn diagram approach. Define:

$$A = S_1 \text{ solves}, \quad B = S_2 \text{ solves}, \quad C = S_3 \text{ solves}.$$

Given:

$$P(A \cap B' \cap C') = \frac{1}{10}, \quad P(B \cap C') = \frac{1}{12}, \quad P(A \cup B \cup C) = \frac{1}{2}.$$

Using inclusion-exclusion and solving via probabilities of intersections, we find

$$P(C) = \frac{13}{36}$$

TIP

Correct option: A

$P(T) = \boxed{\frac{13}{36}}.$