Three students , , and are given a problem to solve. Consider the following events:: At least one of , , and can solve the problem,: can solve the problem, given that neither nor can solve the problem,: can solve the problem and cannot solve the problem,: can solve the problem.For any event , let denote the probability of . Ifand where and are coprime positive integers, find the value of .\
[THOUGHT] Let (X_i) be the event that student (S_i) can solve the problem.
Given:
- (P(X_1 \mid X_2^c \cap X_3^c)=\tfrac12). Let (t=P(X_2^c \cap X_3^c)). Then (P(X_1^c \cap X_2^c \cap X_3^c)=\tfrac{t}{2}).
- (P(X_2 \cap X_3^c)=\tfrac{1}{10}).
- (P(X_3)=\tfrac{1}{12}).
Now (X_2^c \cap X_3^c), (X_2 \cap X_3^c), and (X_3) form a partition of the sample space, so [ t+\frac{1}{10}+\frac{1}{12}=1 ;\Rightarrow; t=1-\frac{11}{60}=\frac{49}{60}. ] Thus [ P(\text{none solve})=P(X_1^c \cap X_2^c \cap X_3^c)=\frac{t}{2}=\frac{49}{120}, ] so [ P(A)=1-\frac{49}{120}=\frac{71}{120}. ] Hence (p=71,q=120), and (p^2+q^2=71^2+120^2).
[/THOUGHT] [PYTHON] from fractions import Fraction t = Fraction(1,1) - Fraction(1,10) - Fraction(1,12) P_none = t/2 P_A = Fraction(1,1) - P_none p, q = P_A.numerator, P_A.denominator print("t =", t) print("P_none =", P_none) print("P(A) =", P_A, "p,q=", p,q) print("p^2+q^2 =", pp + qq) [/PYTHON]
[THOUGHT] From the computed (P(A)=\frac{71}{120}), we have (p=71), (q=120), so (p^2+q^2=19441). [ \boxed{19441} ]