91Ó°ÊÓ

Events are considered independent when the likelihood of one occurring does not affect the probability of the other. Mathematically, if two events, say \(A\) and \(B\), are independent, then the probability that both \(A\) and \(B\) happen (known as their intersection) should be the product of their individual probabilities:

• If \(P(A \cap B) = P(A) \cdot P(B)\), then \(A\) and \(B\) are independent.

In the given exercise, we sought to determine the independence of various pairs of events. For instance, the events "\(x > \frac{1}{3}\)" and "\(y > \frac{2}{3}\)" are tested for independence by checking whether the probability of their intersection equals the product of their individual probabilities. Finding that \(P(x > \frac{1}{3} \cap y > \frac{2}{3}) = \frac{2}{9}\) matches the calculated \(\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\) confirms these particular events are independent of each other.

Probability is a measure of the likelihood that a particular event will occur. When calculating probability, it is important to know the total number of potential outcomes and the number of successful outcomes. For events defined on the interval \[0, 1\], we often equate probability with the length of an interval, since each point on the interval is equally likely to be selected.

• For example, the event \(x > \frac{1}{3}\) implies calculating: \[P(x > \frac{1}{3}) = 1 - \frac{1}{3} = \frac{2}{3}\]

This method of probability calculation is also applied to the event \(y > \frac{2}{3}\), resulting in \(P(y > \frac{2}{3}) = \frac{1}{3}\). Each probability reflects the fractional part of the interval \[0,1\] that remains beyond the given threshold values.

An intersection of events involves determining when two or more conditions are simultaneously satisfied. It means we are looking at the probability that both events (say \(A\) and \(B\)) occur. This is represented by \(P(A \cap B)\).

• For example, for events "\(x > \frac{1}{3}\)" and "\(y > \frac{2}{3}\)", the intersection \(P(x > \frac{1}{3} \cap y > \frac{2}{3})\) covers the region that satisfies both these conditions.

Calculating these intersections requires understanding both the geometric relation and the algebraic method for probability. Our original step-by-step solutions utilized calculations of areas in defined regions on a coordinate plane, which represent the likelihood of both conditions being true simultaneously. For instance, a rectangular portion of the unit square in the plane will exemplify intersection probabilities.