Question

1. create a class conditional probability:

Explanation:

Step1: Define Events

Let’s define two events, say \( A \) (left circle) and \( B \) (right circle). The number in only \( A \) is \( 17 \), only \( B \) is \( 18 \), and both (intersection) is \( 6 \). Total in \( A \): \( 17 + 6 = 23 \), total in \( B \): \( 18 + 6 = 24 \), total overall: \( 17 + 6 + 18 = 41 \).

Step2: Choose Conditional Probability

Let's find \( P(B|A) \), the probability of \( B \) given \( A \). By formula, \( P(B|A) = \frac{P(A \cap B)}{P(A)} \). \( P(A \cap B) = \frac{6}{41} \), \( P(A) = \frac{23}{41} \). So \( P(B|A) = \frac{6/41}{23/41} = \frac{6}{23} \).

Answer:

A class - conditional probability example: If event \( A \) is the set of elements in the left circle (with \( n(A)=17 + 6=23 \)) and event \( B \) is the set of elements in the right circle (with \( n(A\cap B) = 6 \)), then the conditional probability \( P(B|A)=\frac{6}{23}\) (or other valid conditional probabilities can be constructed similarly, e.g., \( P(A|B)=\frac{6}{24}=\frac{1}{4} \)).