Statistics and Probability - Venn Diagrams and Set Problems

In a class of 30 students, 18 students like Basketball ($B$), 15 students like Football ($F$), and 7 students like both sports. Represent this on a Venn diagram and find the number of students who like neither sport.

1. Find students who like ONLY Basketball: $n(B) - n(B \cap F) = 18 - 7 = 11$.
2. Find students who like ONLY Football: $n(F) - n(B \cap F) = 15 - 7 = 8$.
3. Find total students who like at least one sport: $n(B \cup F) = 11 + 7 + 8 = 26$.
4. Find students who like neither: $n(B \cup F)' = n(\xi) - n(B \cup F) = 30 - 26 = 4$.

We use the principle of inclusion-exclusion to ensure we don't double-count the 7 students who like both. Subtracting the union from the total universal set gives the complement (neither).

Let $\xi = \{x : 1 \le x \le 10, x \in \mathbb{Z}\}$. Let $A = \{2, 4, 6, 8, 10\}$ and $B = \{3, 6, 9\}$. Calculate $P(A \cap B)$.

1. Identify elements in $\xi$: $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, so $n(\xi) = 10$.
2. Identify common elements in $A$ and $B$: $A \cap B = \{6\}$.
3. Count the elements: $n(A \cap B) = 1$.
4. Calculate probability: $P(A \cap B) = \frac{n(A \cap B)}{n(\xi)} = \frac{1}{10}$

First define the universal set and the intersection. The probability is the ratio of the number of elements in the intersection to the total number of elements in the universal set.

If $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cup B) = 0.8$, find $P(A \cap B)$.

Using the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substitute the values: $0.8 = 0.6 + 0.5 - P(A \cap B)$ $0.8 = 1.1 - P(A \cap B)$ $P(A \cap B) = 1.1 - 0.8 = 0.3$

Rearrange the addition rule of probability to solve for the unknown intersection probability.