Three students $S_1$, $S_2$, and $S_3$ are given a problem to solve. Consider the following events: $U$: At least one of $S_1$, $S_2$, and $S_3$ can solve the problem, $V$: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem, $W$: $S_2$ can solve the problem and $S_3$ cannot solve the problem, $T$: $S_3$ can solve the problem. For any event $E$, let $P(E)$ denote the probability of $E$. If $P(U)=\frac{1}{2}$, $P(V)=\frac{1}{10}$, and $P(W)=\frac{1}{12}$, then $P(T)$ is equal to

Three students S₁, S₂, and S₃ are given a problem to solve. Consider the following events: U: At least one of S₁, S₂, and S₃ can solve the problem, V: S₁ can solve the problem, given that neither S₂ nor S₃ can solve the problem, W: S₂ can solve the problem and S₃ cannot solve the problem, T: S₃ can solve the problem. For any event E, let P(E) denote the probability of E. If P(U)=1/2, P(V)=1/10, and P(W)=1/12, then P(T) is equal to $U: \text{at least one of } S_1, S_2, S_3 \text{ solves}$ $V: S_1 \text{ solves} \ \mid \ \text{neither } S_2 \text{ nor } S_3 \text{ solves}$ $W: S_2 \text{ solves}, \ S_3 \text{ cannot}; \qquad T: S_3 \text{ solves}$ $P(U)=\dfrac{1}{2}, \quad P(V)=\dfrac{1}{10}, \quad P(W)=\dfrac{1}{12}$ $\text{To find: } P(T)$ $p_i = P(S_i \text{ solves}), \qquad i = 1, 2, 3$ $\text{Independent} \Rightarrow \text{same branch probabilities on every subtree}$ $\text{Path probability} = \text{product along branches; add across}$ $\text{Plan: } 1.\ V \Rightarrow p_1 \qquad 2.\ W \Rightarrow p_2(1-p_3)$ $3.\ U \Rightarrow 1 - (1-p_1)(1-p_2)(1-p_3)$ $P(V) = P(S_1 \mid S_2' \cap S_3') = P(S_1) = p_1 \quad (\text{independence})$ $p_1 = \dfrac{1}{10}, \qquad 1 - p_1 = \dfrac{9}{10}$ $P(W) = P(S_2 \cap S_3') = p_2(1 - p_3) = \dfrac{1}{12}$ $P(U) = 1 - P(\text{none solves}) = 1 - (1-p_1)(1-p_2)(1-p_3) = \dfrac{1}{2}$ $\dfrac{1}{2} = \dfrac{9}{10}(1-p_2)(1-p_3)$ $\Rightarrow (1-p_2)(1-p_3) = \dfrac{5}{9}$ $q_2 = 1 - p_2, \quad q_3 = 1 - p_3$ $q_2 q_3 = \dfrac{5}{9}, \qquad p_2 q_3 = (1 - q_2) q_3 = \dfrac{1}{12}$ $q_3 - q_2 q_3 = \dfrac{1}{12}$ $q_3 - \dfrac{5}{9} = \dfrac{1}{12} \quad \Rightarrow \quad q_3 = \dfrac{5}{9} + \dfrac{1}{12}$ $q_3 = \dfrac{20 + 3}{36} = \dfrac{23}{36}$ $p_3 = 1 - q_3 = \dfrac{13}{36}$ $P(T) = p_3 = \dfrac{13}{36} \qquad \therefore \text{Option } (1)$