Probability of Student Solving Problem Given Independent Conditionals

To find $P(T)$, let $E_1, E_2,$ and $E_3$ denote the events that students $S_1, S_2,$ and $S_3$ solve the problem, respectively. The events $E_1, E_2,$ and $E_3$ are mutually independent.

Let: $P(E_1) = p_1, \quad P(E_2) = p_2, \quad \text{and} \quad P(E_3) = p_3 = P(T)$

From the given information:

1. Event $U$: At least one of $S_1, S_2,$ and $S_3$ can solve the problem. $P(U) = 1 - P(E_1^c \cap E_2^c \cap E_3^c) = 1 - (1 - p_1)(1 - p_2)(1 - p_3)$ Given that $P(U) = \frac{1}{2}$, we have: $(1 - p_1)(1 - p_2)(1 - p_3) = 1 - \frac{1}{2} = \frac{1}{2} \quad \text{--- (1)}$

2. Event $V$: $S_1$ can solve the problem given that neither $S_2$ nor $S_3$ can solve the problem. $P(V) = P(E_1 \mid E_2^c \cap E_3^c) = \frac{P(E_1 \cap E_2^c \cap E_3^c)}{P(E_2^c \cap E_3^c)} = \frac{p_1(1 - p_2)(1 - p_3)}{(1 - p_2)(1 - p_3)} = p_1$ Given that $P(V) = \frac{1}{10}$, we find: $p_1 = \frac{1}{10} \implies 1 - p_1 = \frac{9}{10}$

3. Event $W$: $S_2$ can solve the problem and $S_3$ cannot solve the problem. $P(W) = P(E_2 \cap E_3^c) = p_2(1 - p_3)$ Given that $P(W) = \frac{1}{12}$, we have: $p_2(1 - p_3) = \frac{1}{12} \quad \text{--- (2)}$

Now, substitute $1 - p_1 = \frac{9}{10}$ into equation (1): $\frac{9}{10}(1 - p_2)(1 - p_3) = \frac{1}{2}$ $(1 - p_2)(1 - p_3) = \frac{1}{2} \times \frac{10}{9} = \frac{5}{9}$

Expanding the left side: $(1 - p_3) - p_2(1 - p_3) = \frac{5}{9}$

Using equation (2), substitute $p_2(1 - p_3) = \frac{1}{12}$: $(1 - p_3) - \frac{1}{12} = \frac{5}{9}$ $1 - p_3 = \frac{5}{9} + \frac{1}{12} = \frac{20 + 3}{36} = \frac{23}{36}$

Thus, $P(T) = p_3 = 1 - \frac{23}{36} = \frac{13}{36}$

Hence, the correct option is (A).