Question

1. extension: describe a situation that could be modeled with a venn diagram and create the diagram. use the diagram to determine the probability of at least two events that are possible in the situation.

Explanation:

Let's consider a situation where there are 30 students in a class. 15 students like reading books ($A$), 12 students like playing sports ($B$), and 8 students like both reading books and playing sports.

The total number of students \(n = 30\). The number of students who like reading books \(n(A)=15\), the number of students who like playing sports \(n(B) = 12\), and the number of students who like both \(n(A\cap B)=8\).

The number of students who like only reading books \(n(A - B)=n(A)-n(A\cap B)=15 - 8=7\).

The number of students who like only playing sports \(n(B - A)=n(B)-n(A\cap B)=12 - 8 = 4\).

The number of students who like at least two (in this case, since there are two activities, at least two means the intersection) is \(n(A\cap B) = 8\).

The probability \(P=\frac{n(A\cap B)}{n}=\frac{8}{30}=\frac{4}{15}\)

Answer:

The probability of at least two (intersection of the two sets in the Venn - diagram context) is \(\frac{4}{15}\)