Statistics and Probability - Elementary Set Theory

Given the universal set $U = \{x \mid 1 \le x \le 10, x \in \mathbb{Z}\}$. Let $A = \{2, 4, 6, 8, 10\}$ and $B = \{1, 2, 3, 4, 5\}$. Find $A \cap B$ and $(A \cup B)'$.

$A \cap B = \{2, 4\}$ $A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$ $(A \cup B)' = \{7, 9\}$

The intersection $A \cap B$ contains elements found in both sets. The union $A \cup B$ combines all elements from both. The complement $(A \cup B)'$ contains elements in $U$ that are not in the union.

In a group of 40 students, 25 like coffee ($C$), 15 like tea ($T$), and 10 like both. Find the number of students who like neither coffee nor tea.

Using the formula: $n(C \cup T) = n(C) + n(T) - n(C \cap T)$ $n(C \cup T) = 25 + 15 - 10 = 30$ Number of students liking neither: $n((C \cup T)') = n(U) - n(C \cup T)$ $n((C \cup T)') = 40 - 30 = 10$

First, we calculate the number of students who like at least one drink using the Inclusion-Exclusion principle. Then, we subtract this from the total number of students to find those in the exterior of the Venn diagram circles.

Show that for any two sets $A$ and $B$, if $n(A) = 12$, $n(B) = 15$, and $n(A \cup B) = 20$, the sets cannot be disjoint.

If $A$ and $B$ were disjoint, then $n(A \cap B) = 0$. By the addition rule: $20 = 12 + 15 - n(A \cap B)$ $20 = 27 - n(A \cap B)$ $n(A \cap B) = 7$ Since $n(A \cap B) \neq 0$, the sets are not disjoint.

For sets to be disjoint, the size of their union must equal the sum of their individual sizes. Here, $20 < 12 + 15$, implying an overlap of $7$ elements.