1. Interpret the events carefully

Let

• $A$ = event that $S_1$ solves the problem,
• $B$ = event that $S_2$ solves the problem,
• $C$ = event that $S_3$ solves the problem.

Then:

• $U = A \cup B \cup C$
• $V = A \cap B' \cap C'$
• $W = B \cap C'$
• $T = C$

Given: $P(U)=\frac12, \qquad P(V)=\frac{1}{10}, \qquad P(W)=\frac{1}{12}.$

We need $P(T)=P(C)$.

2. Express $U$ as a disjoint union

The event "at least one solves" can be split into two disjoint parts: $U = C \cup C'(A \cup B).$

Since $C$ and $C'(A\cup B)$ are disjoint, $P(U)=P(C)+P\big(C'(A\cup B)\big).$

Now, $C'(A\cup B) = (A\cap B'\cap C') \cup (B\cap C').$

Why is this true?

• If $C$ does not happen and at least one of $A,B$ happens, then either:
  • $A$ happens while $B$ does not: $A\cap B'\cap C'$, or
  • $B$ happens: $B\cap C'$.

Also these two events are disjoint, because one contains $B'$ and the other contains $B$.

Hence, $P\big(C'(A\cup B)\big)=P(A\cap B'\cap C')+P(B\cap C').$

But these are exactly $P(V)$ and $P(W)$ respectively. So, $P(U)=P(C)+P(V)+P(W).$

3. Substitute the given values

$\frac12 = P(C)+\frac{1}{10}+\frac{1}{12}.$

Now compute: $\frac{1}{10}+\frac{1}{12} = \frac{6+5}{60}=\frac{11}{60}.$

Therefore, $P(C)=\frac12-\frac{11}{60} = \frac{30}{60}-\frac{11}{60}=\frac{19}{60}.$

So, $P(T)=P(C)=\frac{19}{60}.$

4. Check with options

The value $\dfrac{19}{60}$ matches Option C.

5. Compare with stored correct answer

Stored correct answer: A = $\dfrac{13}{36}$

Our derived answer is C = $\dfrac{19}{60}$.

These do not match. The decomposition $U = C \cup (A\cap B'\cap C') \cup (B\cap C')$ is exact and disjoint, so the result $\dfrac{19}{60}$ is correct. Therefore the stored answer appears to be incorrect.