[THOUGHT] Let me carefully parse the problem. We have three students $S_1$, $S_2$, $S_3$ trying to solve a problem.

Let me define:

• $p_1$ = probability that $S_1$ can solve the problem
• $p_2$ = probability that $S_2$ can solve the problem
• $p_3$ = probability that $S_3$ can solve the problem

I'll assume the students solve the problem independently.

Event $B$: "$S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem" This is a conditional probability: $P(B) = P(S_1 \text{ solves} | S_2 \text{ doesn't solve and } S_3 \text{ doesn't solve})$

If the events are independent, then $P(S_1 \text{ solves} | S_2 \text{ doesn't and } S_3 \text{ doesn't}) = P(S_1 \text{ solves}) = p_1$

So $P(B) = p_1 = \frac{1}{2}$

Event $C$: "$S_2$ can solve the problem and $S_3$ cannot solve the problem" $P(C) = P(S_2 \text{ solves and } S_3 \text{ doesn't}) = p_2 \cdot (1 - p_3) = \frac{1}{10}$

Event $D$: "$S_3$ can solve the problem" $P(D) = p_3 = \frac{1}{12}$

Now I can find $p_2$: $p_2 \cdot (1 - p_3) = \frac{1}{10}$ $p_2 \cdot (1 - \frac{1}{12}) = \frac{1}{10}$ $p_2 \cdot \frac{11}{12} = \frac{1}{10}$ $p_2 = \frac{12}{110} = \frac{6}{55}$

Event $A$: "At least one of $S_1$, $S_2$, $S_3$ can solve the problem" $P(A) = 1 - P(\text{none can solve}) = 1 - (1-p_1)(1-p_2)(1-p_3)$

Let me calculate this. [/THOUGHT] [PYTHON] from fractions import Fraction

Given values

p1 = Fraction(1, 2) p3 = Fraction(1, 12)

From P(C) = p2 * (1 - p3) = 1/10

p2 * (1 - 1/12) = 1/10

p2 * (11/12) = 1/10

p2 = 12/(11*10) = 12/110 = 6/55

p2 = Fraction(1, 10) / (1 - p3) print(f"p1 = {p1}") print(f"p2 = {p2}") print(f"p3 = {p3}")

P(A) = 1 - (1-p1)(1-p2)(1-p3)

P_A = 1 - (1 - p1) * (1 - p2) * (1 - p3) print(f"P(A) = {P_A}")

Check if p and q are coprime

p = P_A.numerator q = P_A.denominator print(f"p = {p}, q = {q}")

import math gcd = math.gcd(p, q) print(f"gcd(p, q) = {gcd}")

Calculate p^2 + q^2

result = p2 + q2 print(f"p^2 + q^2 = {result}") [/PYTHON]

[THOUGHT] The calculation confirms that:

• $p_1 = \frac{1}{2}$
• $p_2 = \frac{6}{55}$
• $p_3 = \frac{1}{12}$

And $P(A) = 1 - (1-p_1)(1-p_2)(1-p_3) = \frac{71}{120}$

Since $\gcd(71, 120) = 1$, we have $p = 71$ and $q = 120$ are coprime.

Therefore, $p^2 + q^2 = 71^2 + 120^2 = 5041 + 14400 = 19441$ [/THOUGHT]

\boxed{19441}