Problem 6 - Calculus for Biology and Medicine

Question

An urn contains five green, six blue, and four red balls. You take three balls out of the urn, one after the other, without replacement. Find the probability that the third ball is green given that the first two balls were red.

Step-by-Step Solution

Answer

The probability is \( \frac{5}{13} \).

Step 1: Understand the Problem

We need to calculate the probability of the third ball being green, given that the first two balls drawn are red. Thus, we are considering only sequences of three draws where the first two are red.

Step 2: Define Total Possible Outcomes

Initially, the urn has 15 balls: 5 green, 6 blue, and 4 red. Once two red balls are drawn, there will be 13 balls left: 5 green, 6 blue, and 2 red.

Step 3: Calculate Favorable Outcomes

After drawing two red balls, the probability of the third ball being green means drawing one of the 5 green balls from the remaining 13 balls. Thus, the number of favorable outcomes is the number of green balls left, which is 5.

Step 4: Compute the Probability

Using the favorable outcome count, the probability that the third ball is green given the first two were red is the number of green balls (5) divided by the total remaining balls (13). Thus, the probability is \[ \frac{5}{13} \].

Key Concepts

Understanding Conditional Probability

Conditional probability is a measure of the likelihood of an event occurring given that another event has already occurred. It's like asking, "What are the chances of a particular ball being drawn, knowing some other balls have already been selected?" In our exercise, we're interested in the probability that the third drawn ball is green, **given** that the first two balls drawn are red. This is represented mathematically as: \[ P(G_3 \mid R_1, R_2) \]

• The event \(G_3\) means the third ball is green.
• The condition \(R_1, R_2\) indicates that the first and second balls were red, respectively.

Conditional probability relies on a reduced sample space—only sequences fulfilling the initial condition (R_1, R_2) are considered possible outcomes.Let's explore how combinatorics plays into this setup.

The Role of Combinatorics

Combinatorics provides us with the tools to count and organize the possible outcomes. In probability theory, it's crucial to know how many ways an event can happen. Here, it helps us determine our total and favorable outcomes. To start, consider how many paths can lead us to drawing the first two red balls. We're not directly asked to calculate this here, but understanding it helps appreciate combinatorics.

• Initially, 4 red balls are available.
• After two are drawn, only 13 balls remain, each still part of our overall sample space once two reds are selected.

Combinatorics set up the possibilities: - How sequences of draws are structured - How they reduce as conditions are met Now, let's use these tools in our probability calculation.

Probability Calculation Simplified

Probability calculation involves finding how likely a particular event is based on our defined space of outcomes. It's about aligning the number of favorable outcomes against the total possible ones.After two red balls are drawn from the urn, the situation becomes straightforward:1. There are 13 balls left.2. Out of these, 5 are green.The probability that the third ball drawn is green is expressed as:\[ P(G_3 \mid R_1, R_2) = \frac{\text{Number of green balls}}{\text{Total remaining balls}} \]Plugging in the numbers:\[ \frac{5}{13} \]This fraction represents the chance that, after already drawing two red balls, the next is a green one. This calculation is a simple yet effective demonstration of conditional probability tied with combinatorial principles.