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

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

then $$P(T)$$ is equal to

Let the individual events be
  $$A$$ : $$S_1$$ solves the problem,   $$B$$ : $$S_2$$ solves the problem,   $$C$$ : $$S_3$$ solves the problem.

Write the probabilities of all eight possible outcome-combinations as

$$\begin{aligned} P(A\,B'\,C') &= x_1, & P(A\,B\,C') &= x_2, & P(A\,B'\,C) &= x_3, & P(A\,B\,C) &= x_4,\\[2pt] P(A'\,B\,C') &= x_5, & P(A'\,B\,C) &= x_6, & P(A'\,B'\,C) &= x_7, & P(A'\,B'\,C') &= x_8. \end{aligned}$$

The eight numbers $$x_1,\dots ,x_8$$ are non-negative and satisfy

$$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8 = 1 \qquad -(1)$$

Step 1 Using $$P(U)=\dfrac12$$
Event $$U$$ is “at least one of $$S_1,S_2,S_3$$ solves”, i.e. $$U = A\cup B\cup C$$, whose complement is $$A'B'C'$$. Hence

$$P(U) = 1-P(A'B'C') = 1-x_8 = \frac12 \;\;\Longrightarrow\;\; x_8 = \frac12. \qquad -(2)$$

Step 2 Using $$P(V)=\dfrac1{10}$$
Event $$V$$ is “$$S_1$$ solves given that neither $$S_2$$ nor $$S_3$$ solves”, that is

$$P(V)=P\!\bigl(A\,|\,B'C'\bigr)=\frac{P(A\,B'\,C')}{P(B'\,C')}=\frac{x_1}{x_1+x_8} =\frac1{10}. $$

Substituting $$x_8=\dfrac12$$ from (2):

$$\frac{x_1}{x_1+\dfrac12}=\frac1{10} \;\;\Longrightarrow\;\;10x_1 = x_1+\frac12 \;\;\Longrightarrow\;\;9x_1 = \frac12 \;\;\Longrightarrow\;\; x_1 = \frac1{18}. \qquad -(3)$$

Step 3 Using $$P(W)=\dfrac1{12}$$
Event $$W$$ is “$$S_2$$ solves and $$S_3$$ does not”, i.e.

$$P(W)=P(B\,C') = P(A\,B\,C') + P(A'\,B\,C') = x_2 + x_5 = \frac1{12}. \qquad -(4)$$

Step 4 Total probability still unaccounted for
From (1), (2) and (3)

$$x_2+x_3+x_4+x_5+x_6+x_7 = 1 - x_8 - x_1 = 1 - \frac12 - \frac1{18} = \frac{4}{9}. \qquad -(5)$$

Step 5 Probability that $$S_3$$ solves
$$S_3$$ succeeds in the four cases $$A\,B'\,C,\;A\,B\,C,\;A'\,B\,C,\;A'\,B'\,C$$ whose probabilities are $$x_3,x_4,x_6,x_7$$ respectively. Adding them, and using (4) and (5):

$$\begin{aligned} P(T) &= x_3+x_4+x_6+x_7 \\[2pt] &= \bigl(x_2+x_3+x_4+x_5+x_6+x_7\bigr) - (x_2+x_5) \\[2pt] &= \frac{4}{9} - \frac{1}{12} \\[2pt] &= \frac{16}{36} - \frac{3}{36} \\[2pt] &= \frac{13}{36}. \end{aligned}$$

Therefore $$P(T)=\dfrac{13}{36}$$.

Option A which is: $$\dfrac{13}{36}$$