Problem Analysis #11

[THOUGHT] We are given:

• ( P(B) = \frac{1}{2} ), where ( B ) is the event that ( S_1 ) solves given neither ( S_2 ) nor ( S_3 ) solves.
• ( P(C) = \frac{1}{10} ), where ( C ) is the event that ( S_2 ) solves and ( S_3 ) does not.
• ( P(D) = \frac{1}{12} ), where ( D ) is the event that ( S_3 ) solves.

We define:

• ( a = P(S_1=0, S_2=0, S_3=0) )
• ( b = P(S_1=1, S_2=0, S_3=0) )

From ( P(B) = \frac{1}{2} ), we have: [ P(B) = P(S_1=1 \mid S_2=0, S_3=0) = \frac{b}{a + b} = \frac{1}{2} ] This implies ( 2b = a + b ), so ( a = b ).

The event ( C ) is ( S_2=1, S_3=0 ), so: [ P(C) = P(S_2=1, S_3=0) = c + f = \frac{1}{10} \quad \text{(where ( c = P(S_1=0, S_2=1, S_3=0) ), ( f = P(S_1=1, S_2=1, S_3=0) ))} ]

The event ( D ) is ( S_3=1 ), so: [ P(D) = P(S_3=1) = d + e + g + h = \frac{1}{12} \quad \text{(where ( d = P(S_1=0, S_2=0, S_3=1) ), ( e = P(S_1=0, S_2=1, S_3=1) ), ( g = P(S_1=1, S_2=0, S_3=1) ), ( h = P(S_1=1, S_2=1, S_3=1) ))} ]

The total probability sums to 1: [ a + b + c + d + e + f + g + h = 1 ] Substituting ( a = b ), ( c + f = \frac{1}{10} ), and ( d + e + g + h = \frac{1}{12} ): [ 2a + \frac{1}{10} + \frac{1}{12} = 1 ] Solving for ( a ): [ 2a = 1 - \frac{1}{10} - \frac{1}{12} = \frac{120}{120} - \frac{12}{120} - \frac{10}{120} = \frac{98}{120} = \frac{49}{60} ] [ a = \frac{49}{120} ] The event ( A ) is "at least one solves", so ( P(A) = 1 - a = 1 - \frac{49}{120} = \frac{71}{120} ).

The fraction ( \frac{71}{120} ) is in reduced terms since 71 is prime and does not divide 120. Thus, ( p = 71 ), ( q = 120 ), and: [ p^2 + q^2 = 71^2 + 120^2 = 5041 + 14400 = 19441 ] [/THOUGHT] \boxed{19441}

[THOUGHT] We need to compute ( p^2 + q^2 ) where ( p = 71 ) and ( q = 120 ) using Python, as required. [/THOUGHT] [PYTHON] print(712 + 1202) [/PYTHON]

\boxed{19441}