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: S1₁ 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\left(U\right)=\frac{1}{2},P\left(V\right)=\frac{1}{10},P\left(W\right)=\frac{1}{12}

Then P(T) is equal to

Let $S_1$, $S_2$, and $S_3$ be the independent events that students 1, 2, and 3 solve the problem, respectively. Let their probabilities be denoted as:

$P\left(S_1\right)=p_1$, $P\left(S_2\right)=p_2$, and $P\left(S_3\right)=p_3$.

The events given in the problem can be expressed as follows:

1. Event U: At least one of the students can solve the problem.
The probability of event $U$ is the complement of none of the students being able to solve the problem:
$P\left(U\right)=1-P(S_1^{\text{'}}\cap S_2^{\text{'}}\cap S_3^{\text{'}})=1-(1-p_1)(1-p_2)(1-p_3)$
Given that $P\left(U\right)=\frac{1}{2}$, we have:
$(1-p_1)(1-p_2)(1-p_3)=\frac{1}{2}$ (Equation 1)

2. Event V: $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem.
$P\left(V\right)=P\left(S_1\right|S_2^{\text{'}}\cap S_3^{\text{'}})$
Since the events are independent, the conditional probability simplifies to:
$P\left(V\right)=P\left(S_1\right)=p_1$
Given that $P\left(V\right)=\frac{1}{10}$, we get:
$p_1=\frac{1}{10}$

Substituting $p_1=\frac{1}{10}$ into Equation 1:
$(1-\frac{1}{10})(1-p_2)(1-p_3)=\frac{1}{2}$
$\frac{9}{10}(1-p_2)(1-p_3)=\frac{1}{2}$
$(1-p_2)(1-p_3)=\frac{1}{2}\times \frac{10}{9}=\frac{5}{9}$ (Equation 2)

3. Event W: $S_2$ can solve the problem and $S_3$ cannot solve the problem.
$P\left(W\right)=P(S_2\cap S_3^{\text{'}})=p_2(1-p_3)$
Given that $P\left(W\right)=\frac{1}{12}$, we have:
$p_2(1-p_3)=\frac{1}{12}$ (Equation 3)

4. Finding P(T):
We need to find the probability of event $T$, which is $P\left(S_3\right)=p_3$.
Let us expand Equation 2:
$(1-p_2)(1-p_3)=1-p_3-p_2(1-p_3)=\frac{5}{9}$

Substitute the value of $p_2(1-p_3)$ from Equation 3 into the expanded expression:
$(1-p_3)-\frac{1}{12}=\frac{5}{9}$
$1-p_3=\frac{5}{9}+\frac{1}{12}$

Finding a common denominator (36):
$1-p_3=\frac{20}{36}+\frac{3}{36}=\frac{23}{36}$
$p_3=1-\frac{23}{36}=\frac{13}{36}$

Therefore, the probability $P\left(T\right)$ is equal to $\frac{13}{36}$.