Understanding the Intersection of Events A and B in a Ball Drawing Scenario

The given question pertains to the probability of the intersection of two events, A and B, in a scenario where a ball is drawn from a box. This example will help us understand the concepts of intersection, conditional probability, and how to calculate these in a practical scenario.

Scenario Setup

Consider a box containing 5 balls labeled 1, 2, 3, 4, and 5. A player draws a ball from the box randomly. Let's define the following events:

Event A: Drawing a ball labeled with an even number. Event B: Drawing a ball labeled with a number greater than 3.

Calculating the Probability of Event A and Event B

First, let's calculate the probability of each event individually.

Event A

There are two even-numbered balls (2 and 4) in the box. Thus, the probability of drawing a ball with an even number, P(A), is:

P(A) Number of even-numbered balls / Total number of balls

P(A) 2 / 5

Event B

There are three numbers greater than 3 (4 and 5) in the box. Therefore, the probability of drawing a ball with a number greater than 3, P(B), is:

P(B) Number of balls with numbers greater than 3 / Total number of balls

P(B) 3 / 5

Intersection of Events A and B (A ∩ B)

Now, we need to find the probability of the intersection of these two events, i.e., drawing a ball that satisfies both conditions - an even number greater than 3.

Calculating the Intersection

Focusing on the intersection, we can see that there is only one ball that meets both conditions: the ball labeled 4. Therefore, the probability of drawing a ball that is both an even number greater than 3, P(A ∩ B), is:

P(A ∩ B) Number of balls that are even and greater than 3 / Total number of balls

P(A ∩ B) 1 / 5

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. Here, we need to find the conditional probability of event A given that event B has already occurred, i.e., P(A|B).

Using the formula for conditional probability:

P(A|B) P(A ∩ B) / P(B)

P(A|B) (1 / 5) / (3 / 5) 1 / 3

Conclusion

In summary, we have calculated the probability of drawing a ball that is both an even number (Event A) and a number greater than 3 (Event B). We found that:

P(A) 2 / 5 P(B) 3 / 5 P(A ∩ B) 1 / 5 P(A|B) 1 / 3

This example helps us understand the intersection of events and the importance of conditional probability in scenarios like these.

Keywords

probability, intersection, conditional probability, ball drawing, event