### Related Formula n(A cup B cup C) = n(A) + n(B) + n(C) - n(A cap B) - n(B cap C) - n(A cap C) + n(A cap B cap C)$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$ ### Core Logic Let total sample space S = \1, 2, dots, 50\ implies n(S) = 50$S = \{1, 2, \dots, 50\} \implies n(S) = 50$. * Let A$A$ be multiple of 4: 4, 8, dots, 48 implies n(A) = lfloor frac504 rfloor = 12$4, 8, \dots, 48 \implies n(A) = \lfloor \frac{50}{4} \rfloor = 12$ * Let B$B$ be multiple of 6: 6, 12, dots, 48 implies n(B) = lfloor frac506 rfloor = 8$6, 12, \dots, 48 \implies n(B) = \lfloor \frac{50}{6} \rfloor = 8$ * Let C$C$ be multiple of 7: 7, 14, dots, 49 implies n(C) = lfloor frac507 rfloor = 7$7, 14, \dots, 49 \implies n(C) = \lfloor \frac{50}{7} \rfloor = 7$ ### Step 1: Finding Intersections * A cap B$A \cap B$ (LCM of 4 and 6 = 12): multiples of 12 implies n(A cap B) = lfloor frac5012 rfloor = 4$\implies n(A \cap B) = \lfloor \frac{50}{12} \rfloor = 4$ * B cap C$B \cap C$ (LCM of 6 and 7 = 42): multiples of 42 implies n(B cap C) = lfloor frac5042 rfloor = 1$\implies n(B \cap C) = \lfloor \frac{50}{42} \rfloor = 1$ * A cap C$A \cap C$ (LCM of 4 and 7 = 28): multiples of 28 implies n(A cap C) = lfloor frac5028 rfloor = 1$\implies n(A \cap C) = \lfloor \frac{50}{28} \rfloor = 1$ * A cap B cap C$A \cap B \cap C$ (LCM of 4, 6, 7 = 84): multiples of 84 implies n(A cap B cap C) = 0$\implies n(A \cap B \cap C) = 0$ ### Step 2: Final Enumeration Applying the Principle of Inclusion-Exclusion: n(A cup B cup C) = 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21$n(A \cup B \cup C) = 12 + 8 + 7 - 4 - 1 - 1 + 0 = 21$ Therefore, the probability is: P = frac2150$P = \frac{21}{50}$ ### Pattern Recognition To quickly count the number of multiples up to N$N$, use the greatest integer function lfloor fracNtextLCM rfloor$\lfloor \frac{N}{\text{LCM}} \rfloor$. This systematically prevents manual counting blunders. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Probability