1. **State the problem:** We have three sets representing students who sat exams in Physics (P), Chemistry (C), and Biology (B). Given are the numbers of students in each subject and their intersections: - $n(P) = 37$ - $n(C) = 48$ - $n(B) = 45$ - $n(P \cap C) = 15$ - $n(C \cap B) = 13$ - $n(P \cap B) = 7$ - $n(P \cap C \cap B) = 5$ We need to: a. Draw a Venn diagram representing this information. b. Calculate $n(P \cup C \cup B)$, the total number of students who sat at least one of the exams. 2. **Formula and rules:** The formula for the union of three sets is: $$ n(P \cup C \cup B) = n(P) + n(C) + n(B) - n(P \cap C) - n(C \cap B) - n(P \cap B) + n(P \cap C \cap B) $$ This formula accounts for double counting in the pairwise intersections and adds back the triple intersection once. 3. **Calculate the union:** Substitute the given values: $$ n(P \cup C \cup B) = 37 + 48 + 45 - 15 - 13 - 7 + 5 $$ Simplify step-by-step: $$ = (37 + 48 + 45) - (15 + 13 + 7) + 5 $$ $$ = 130 - 35 + 5 $$ $$ = 100 $$ 4. **Interpretation:** There are 100 students who sat at least one of the exams in Physics, Chemistry, or Biology. **Final answer:** $$ n(P \cup C \cup B) = 100 $$