The problem asks for the value of $p^2 + q^2$ where $P(A) = \frac{p}{q}$. The events are defined as:

• $A$: At least one of $S_1, S_2, S_3$ solves the problem ($S_1 \cup S_2 \cup S_3$).
• $B$: $S_1$ solves given neither $S_2$ nor $S_3$ solves ($S_1 | S_2^c \cap S_3^c$). $P(B) = 1/2$.
• $C$: $S_2$ solves and $S_3$ does not ($S_2 \cap S_3^c$). $P(C) = 1/10$.
• $D$: $S_3$ solves ($S_3$). $P(D) = 1/12$.

[PYTHON] from fractions import Fraction

Given probabilities

P_B = Fraction(1, 2) # P(S1 | S2' S3') P_C = Fraction(1, 10) # P(S2 S3') P_D = Fraction(1, 12) # P(S3)

We want P(A) = P(S1 U S2 U S3) = 1 - P(S1' S2' S3')

Step 1: Calculate P(S2' S3')

The sample space can be partitioned by S2 and S3 states:

1. S3 (which includes S2 S3 and S2' S3) -> Probability P(D) = 1/12

2. S2 S3' (Event C) -> Probability P(C) = 1/10

3. S2' S3' -> Let this be x

Sum must be 1.

P(S3) + P(S2 S3') + P(S2' S3') = 1

Note: P(S3) covers both (S2 S3) and (S2' S3).

The events S3 and (S2 S3') are disjoint because one requires S3 true, the other S3 false.

So P(S3 U (S2 S3')) = P(S3) + P(S2 S3')

And the complement is S2' S3'.

P_S2_prime_S3_prime = 1 - (P_D + P_C) print(f"P(S2' S3') = {P_S2_prime_S3_prime}")

Step 2: Use P(B) to find P(S1 S2' S3')

P(B) = P(S1 | S2' S3') = P(S1 S2' S3') / P(S2' S3')

P(S1 S2' S3') = P(B) * P(S2' S3')

P_S1_S2_prime_S3_prime = P_B * P_S2_prime_S3_prime print(f"P(S1 S2' S3') = {P_S1_S2_prime_S3_prime}")

Step 3: Calculate P(S1' S2' S3')

We know P(S2' S3') = P(S1 S2' S3') + P(S1' S2' S3')

So P(S1' S2' S3') = P(S2' S3') - P(S1 S2' S3')

P_S1_prime_S2_prime_S3_prime = P_S2_prime_S3_prime - P_S1_S2_prime_S3_prime print(f"P(S1' S2' S3') = {P_S1_prime_S2_prime_S3_prime}")

Step 4: Calculate P(A)

P(A) = 1 - P(S1' S2' S3')

P_A = 1 - P_S1_prime_S2_prime_S3_prime print(f"P(A) = {P_A}")

p = P_A.numerator q = P_A.denominator print(f"p = {p}, q = {q}") print(f"p^2 + q^2 = {p2 + q2}") [/PYTHON]

The probability $P(A)$ is calculated as follows: Let $E_1$ be the event that $S_1$ solves the problem, $E_2$ for $S_2$, and $E_3$ for $S_3$. We are given:

1. $P(B) = P(E_1 \mid E_2^c E_3^c) = \frac{1}{2}$
2. $P(C) = P(E_2 E_3^c) = \frac{1}{10}$
3. $P(D) = P(E_3) = \frac{1}{12}$

We want to find $P(A) = P(E_1 \cup E_2 \cup E_3) = 1 - P(E_1^c E_2^c E_3^c)$.

First, consider the partition of the sample space based on $E_2$ and $E_3$: The events $E_3$ and $E_2 E_3^c$ are disjoint. Their union is $E_3 \cup (E_2 E_3^c) = E_3 \cup E_2$. Thus, $P(E_2 \cup E_3) = P(E_3) + P(E_2 E_3^c) = \frac{1}{12} + \frac{1}{10} = \frac{5}{60} + \frac{6}{60} = \frac{11}{60}$.

The complement of $E_2 \cup E_3$ is $E_2^c E_3^c$. $P(E_2^c E_3^c) = 1 - P(E_2 \cup E_3) = 1 - \frac{11}{60} = \frac{49}{60}$.

From the definition of conditional probability for event $B$: $P(B) = \frac{P(E_1 E_2^c E_3^c)}{P(E_2^c E_3^c)} = \frac{1}{2}$. So, $P(E_1 E_2^c E_3^c) = \frac{1}{2} \times \frac{49}{60} = \frac{49}{120}$.

We can decompose $P(E_2^c E_3^c)$ into disjoint events based on $E_1$: $P(E_2^c E_3^c) = P(E_1 E_2^c E_3^c) + P(E_1^c E_2^c E_3^c)$. $\frac{49}{60} = \frac{49}{120} + P(E_1^c E_2^c E_3^c)$. $P(E_1^c E_2^c E_3^c) = \frac{98}{120} - \frac{49}{120} = \frac{49}{120}$.

Finally, $P(A) = 1 - P(E_1^c E_2^c E_3^c) = 1 - \frac{49}{120} = \frac{71}{120}$. Thus, $p = 71$ and $q = 120$. These are coprime since 71 is prime and not a factor of 120. We need to find $p^2 + q^2$: $p^2 + q^2 = 71^2 + 120^2 = 5041 + 14400 = 19441$.

\boxed{19441}