Problem Analysis #11

[THOUGHT] To find ( P(A) ), we need to calculate the probability that at least one of the three students can solve the problem. Let ( p_1 = P(S_1) ), ( p_2 = P(S_2) ), and ( p_3 = P(S_3) ) be the probabilities that students ( S_1 ), ( S_2 ), and ( S_3 ) can solve the problem, respectively.

Given probabilities:

• ( P(B) = P(S_1 \mid \overline{S_2} \cap \overline{S_3}) = \frac{1}{2} )
• ( P(C) = P(S_2 \cap \overline{S_3}) = \frac{1}{10} )
• ( P(D) = P(S_3) = \frac{1}{12} )

From ( P(D) = \frac{1}{12} ), we directly have: [ p_3 = \frac{1}{12} ]

For ( P(C) = \frac{1}{10} ): [ P(S_2 \cap \overline{S_3}) = p_2 (1 - p_3) = \frac{1}{10} ] Substitute ( p_3 = \frac{1}{12} ): [ p_2 \left(1 - \frac{1}{12}\right) = \frac{1}{10} ] [ p_2 \left(\frac{11}{12}\right) = \frac{1}{10} ] [ p_2 = \frac{1}{10} \times \frac{12}{11} = \frac{6}{55} ]

For ( P(B) = \frac{1}{2} ): [ P(S_1 \mid \overline{S_2} \cap \overline{S_3}) = \frac{P(S_1 \cap \overline{S_2} \cap \overline{S_3})}{P(\overline{S_2} \cap \overline{S_3})} = \frac{1}{2} ] First, calculate ( P(\overline{S_2} \cap \overline{S_3}) ): [ P(\overline{S_2}) = 1 - p_2 = 1 - \frac{6}{55} = \frac{49}{55} ] [ P(\overline{S_3}) = 1 - p_3 = 1 - \frac{1}{12} = \frac{11}{12} ] [ P(\overline{S_2} \cap \overline{S_3}) = \frac{49}{55} \times \frac{11}{12} = \frac{539}{660} ]

Now, use ( P(B) = \frac{1}{2} ): [ \frac{P(S_1 \cap \overline{S_2} \cap \overline{S_3})}{\frac{539}{660}} = \frac{1}{2} ] [ P(S_1 \cap \overline{S_2} \cap \overline{S_3}) = \frac{1}{2} \times \frac{539}{660} = \frac{539}{1320} ]

Since ( S_1 ), ( S_2 ), and ( S_3 ) solve the problem independently: [ P(S_1 \cap \overline{S_2} \cap \overline{S_3}) = p_1 \left(1 - p_2\right) \left(1 - p_3\right) = \frac{539}{1320} ] Substitute ( p_2 = \frac{6}{55} ) and ( p_3 = \frac{1}{12} ): [ p_1 \left(\frac{49}{55}\right) \left(\frac{11}{12}\right) = \frac{539}{1320} ] [ p_1 \times \frac{539}{660} = \frac{539}{1320} ] [ p_1 = \frac{539}{1320} \times \frac{660}{539} = \frac{1}{2} ]

Now, calculate ( P(A) ), the probability that at least one of the students can solve the problem: [ P(A) = 1 - P(\overline{S_1} \cap \overline{S_2} \cap \overline{S_3}) ] [ P(\overline{S_1}) = 1 - p_1 = 1 - \frac{1}{2} = \frac{1}{2} ] [ P(\overline{S_2}) = \frac{49}{55} ] [ P(\overline{S_3}) = \frac{11}{12} ] [ P(\overline{S_1} \cap \overline{S_2} \cap \overline{S_3}) = \frac{1}{2} \times \frac{49}{55} \times \frac{11}{12} = \frac{539}{1320} ] [ P(A) = 1 - \frac{539}{1320} = \frac{1320 - 539}{1320} = \frac{781}{1320} ]

We need to simplify ( \frac{781}{1320} ): The greatest common divisor (GCD) of 781 and 1320 is 1, so the fraction is already in simplest form. Therefore, ( p = 781 ) and ( q = 1320 ).

Finally, calculate ( p^2 + q^2 ): [ p^2 + q^2 = 781^2 + 1320^2 ] We will use Python to ensure the arithmetic is correct. [/THOUGHT] [PYTHON] p = 781 q = 1320 result = p2 + q2 print(result)

[THOUGHT] Based on the Python output, the value of ( p^2 + q^2 ) is 2352361. Therefore, the final answer is: \boxed{2352361}