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 \(P(A \cap B) = P(A) \times P(B)\).

Step 1: Define Independence

Two events, \(A\) and \(B\), are independent if the probability of both events happening is the product of their individual probabilities: \(P(A \cap B) = P(A) \times P(B)\). We'll use this definition to determine if the events are independent.

Step 2: Calculate Probability of Event A

Event \(A\) is that the first ball drawn is green. There are 7 balls in total, of which 4 are green. Therefore, \(P(A) = \frac{4}{7}\).

Step 3: Calculate Probability of Event B

Event \(B\) is that the second ball drawn is green. Since the ball is replaced after each draw, the probabilities remain the same for each draw. Hence, \(P(B) = \frac{4}{7}\).

Step 4: Calculate Probability of A and B

Event \(A \cap B\) is that both the first and the second ball drawn are green. Since each draw is independent (because the ball is replaced), \(P(A \cap B) = \frac{4}{7} \times \frac{4}{7} = \frac{16}{49}\).

Step 5: Check for Independence

Now, check if \(P(A \cap B) = P(A) \times P(B)\). We have already calculated \(P(A \cap B) = \frac{16}{49}\) and \(P(A) \times P(B) = \frac{4}{7} \times \frac{4}{7} = \frac{16}{49}\). Since both values are equal, the events \(A\) and \(B\) are independent.

Key Concepts

Independent Events

Understanding independent events is fundamental in probability. Simply put, two events are independent if the occurrence of one doesn’t affect the occurrence of the other. This means that the probability of both events happening can be found by multiplying their individual probabilities. For example, if you roll a dice and flip a coin, what you roll on the dice doesn’t affect the coin’s outcome. These are independent events. In math terms, two events, say \( A \) and \( B \), are independent if the probability of both occurring together is the product of their individual probabilities: \( P(A \cap B) = P(A) \times P(B) \). This is a critical aspect because checking this condition helps us identify independence among events.

Probability of Compound Events

The probability of compound events involves calculating the likelihood of two or more events happening together. There are different approaches depending on the nature of the events involved – whether they are independent or dependent.

• When events are independent, like drawing balls from an urn with replacement, the treatment is straightforward. You multiply the probabilities of the individual events.
• For dependent events, the probability changes as conditions or outcomes of other events are considered.

In our example, drawing a green ball on each try from the urn constitutes a compound event since it includes more than one occurrence (two draws). However, as we determined, because each draw doesn’t affect the other (due to replacement), they remain independent. Thus, calculating the probability of both events using multiplication is valid.

Replacement in Probability

Replacement is an essential consideration when dealing with probability in scenarios involving drawing objects from a set, like balls from an urn.

• Replacement refers to returning the object you drew back into the original set before the next draw. This action ensures the total number of objects in the set remains the same. Consequently, the probability for each event remains unchanged.
• Without replacement, the numbers change after each draw, leading to different probabilities for successive events.

In our scenario, because balls are returned to the urn after each draw, each draw is independent of the previous one. So, the probability of drawing a green ball remains \( \frac{4}{7} \) on every draw, ensuring events can be treated independently and calculated as independent compound probabilities.