Statistics and Probability - Sets, including notation and operations up to three sets

Given $\xi = \{x : x \in \mathbb{Z}, 1 \le x \le 10\}$, $A = \{2, 4, 6, 8, 10\}$, and $B = \{1, 2, 3, 4, 5\}$. Find: (i) $A \cap B$, (ii) $A \cup B$, (iii) $(A \cup B)'$.

(i) $A \cap B = \{2, 4\}$ (ii) $A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$ (iii) Since $\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $(A \cup B)' = \xi \setminus (A \cup B) = \{7, 9\}$.

Intersection takes common elements. Union combines all elements from both sets without repetition. Complement finds elements in the universal set not present in the union.

In a class of $30$ students, $18$ like Mathematics ($M$), $15$ like Science ($S$), and $8$ like both. How many students like neither subject?

Use the formula $n(M \cup S) = n(M) + n(S) - n(M \cap S)$. $n(M \cup S) = 18 + 15 - 8 = 25$ Students who like neither = $n(\xi) - n(M \cup S)$ $\begin{array}{r} 30 \\ -25 \\ \hline 5 \end{array}$ So, $5$ students like neither.

We first find the total number of students who like at least one subject using the inclusion-exclusion principle, then subtract that from the total class size.

In a survey of $100$ people, $40$ read Magazine A, $35$ read B, and $30$ read C. $15$ read A and B, $10$ read B and C, $8$ read A and C, and $3$ read all three. Find the number of people who read at least one magazine.

We use the three-set inclusion-exclusion formula: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$ $n(A \cup B \cup C) = 40 + 35 + 30 - 15 - 10 - 8 + 3$ $n(A \cup B \cup C) = 105 - 33 + 3 = 75$.

To find 'at least one', we calculate the union of all three sets. We add individual sets, subtract double-intersections, and add back the triple-intersection because it was subtracted one too many times.