Problem 30 - Calculus for Biology and Medicine

Question

An urn contains four green and three blue balls. You take one ball out of the urn, note its color, and replace it. You then take a second ball out of the urn, note its color, and replace it. If \(A\) denotes the event that the first ball is green and \(B\) denotes the event that the second ball is green, determine whether \(A\) and \(B\) are independent.

Step-by-Step Solution

Answer

Events \(A\) and \(B\) are independent because \(P(A \cap B) = P(A) \cdot P(B)\).

Step 1: Understand the Definitions

Events are independent if the occurrence of one doesn't affect the probability of the other. We need to check if \( P(A \cap B) = P(A) \cdot P(B) \). Event \( A \) is that the first ball picked is green, and event \( B \) is that the second ball picked is green.

Step 2: Probability of Event A

Compute the probability of event \( A \). There are 4 green balls out of a total of 7 balls, so \( P(A) = \frac{4}{7} \).

Step 3: Probability of Event B

Compute the probability of \( B \). Since the first ball is replaced, the probabilities of drawing any ball remain the same. Thus, \( P(B) = \frac{4}{7} \) as well, because there are still 4 green balls out of 7.

Step 4: Joint Probability of A and B

Calculate the joint probability \( P(A \cap B) \). Since the draws are with replacement, \( P(A \cap B) = P(A) \cdot P(B) = \frac{4}{7} \cdot \frac{4}{7} = \frac{16}{49} \).

Step 5: Check for Independence

Events \( A \) and \( B \) are independent if \( P(A \cap B) = P(A) \cdot P(B) \). We found \( P(A \cap B) = \frac{16}{49} \) and \( P(A) \times P(B) = \frac{16}{49} \). Since these are equal, \( A \) and \( B \) are independent.

Key Concepts

Independent Events

When we talk about independent events in probability, we simply mean events where the occurrence of one event does not change the likelihood of the other occurring. In other words, knowing that one event has happened gives us no additional information about the likelihood of the other.

To determine if two events, say event A and event B, are independent, we check if the condition \( P(A \cap B) = P(A) \cdot P(B) \) holds true:

• \( P(A \cap B) \) is the joint probability of both events occurring together.
• \( P(A) \) is the probability of event A happening.
• \( P(B) \) is the probability of event B happening.

If the product of the probabilities of A and B equals the probability of both events occurring together, they are considered independent.

Joint Probability

Joint probability refers to the probability of two or more events occurring simultaneously. It is commonly denoted as \( P(A \cap B) \), which represents the probability of event A and event B both happening.

Think of joint probability as the likelihood of a cross-section of two events on a Venn diagram, where they overlap.

• For independent events, this probability is simply the product of the individual probabilities: \( P(A) \cdot P(B) \).
• Joint probability can become more complex to calculate when events are not independent, requiring additional conditionality factors.

In the exercise involving the green balls, as the events are independent, the joint probability evaluates to \( \frac{16}{49} \). This calculation confirms that drawing a green ball twice, once after the other, doesn’t affect each draw.

Replacement in Probability

In probability, replacement plays a crucial role in determining the nature of events. When an item is drawn from a set and then replaced, it implies the conditions for the second draw remain unchanged. This means each draw is independent of the others because the total count and mix of items remain the same.

In our urn exercise, replacement ensures that the probability of drawing a green ball remains \( \frac{4}{7} \) for each draw. The nature of replacement stabilizes the probability across successive events:

• Without replacement, the situation would change with each draw, leading to dependent events.
• With replacement, each trial remains consistent, supporting independent probability calculations.

Understanding replacement helps better illustrate why consecutive draws act independently in such scenarios.

Calculating Probabilities

Calculating probabilities involves determining the likelihood of a given event occurring. In most scenarios, this is straightforward, involving counting favorable outcomes over the total number of possible outcomes, as expressed in the ratio \( \frac{favorable}{total} \).

Let's break down the essential steps.

• Identify the total number of possible outcomes. In an urn with 7 balls, there are 7 possible outcomes for a single draw.
• Identify the number of favorable outcomes for the event. For event A, there are 4 green balls.
• Divide the favorable outcomes by the total to find the probability: \( P(A) = \frac{4}{7} \).

In probability calculations, especially with events involving draws with replacement, it ensures that each draw starts fresh with the same possibilities, simplifying calculations.