Problem 40 - Calculus for Biology and Medicine

Question

An urn contains five blue and three green balls. You remove three balls from the urn without replacement. What is the probability that at least two out of the three balls are green?

Step-by-Step Solution

Answer

The probability is \( \frac{2}{7} \).

Step 1: Understand the situation

The problem involves an urn with 5 blue and 3 green balls. We are removing 3 balls without replacement and need to find the probability that at least two of them are green.

Step 2: Determine Total Possible Outcomes

First, calculate the total number of ways to choose 3 balls from the urn. This is a combination problem where we choose 3 out of 8 balls. The number of combinations is given by \( \binom{8}{3} \).

Step 3: Calculate Total Outcomes for at Least Two Green Balls

We need to calculate two scenarios:1. Exactly 2 green balls and 1 blue ball.2. Exactly 3 green balls.For exactly 2 green and 1 blue: Choose 2 greens from 3 and 1 blue from 5, using combinations \( \binom{3}{2} \cdot \binom{5}{1} \). For exactly 3 green balls: Choose 3 greens from the 3, \( \binom{3}{3} \).

Step 4: Compute Probabilities and Sum

Find the number of favorable outcomes from Step 3 and divide by the total outcomes from Step 2 to get the probabilities.Calculate \( \frac{\binom{3}{2} \cdot \binom{5}{1} + \binom{3}{3}}{\binom{8}{3}} \). First calculate the combinations:- \( \binom{3}{2} = 3 \) and \( \binom{5}{1} = 5 \) gives 3\( \times \)5 = 15 outcomes for 2 green, 1 blue.- \( \binom{3}{3} = 1 \) gives 1 outcome for 3 green.- \( \binom{8}{3} = 56 \) is the total number of ways to choose 3 balls.The probability is \( \frac{15 + 1}{56} = \frac{16}{56} = \frac{2}{7} \).

Step 5: Finalize Answer

After calculating, we find that the probability that at least two out of the three balls are green is \( \frac{2}{7} \).

Key Concepts

Combinations

When dealing with situations where you need to select a specific number of items from a larger group, combinations come into play. Combinations are a key concept in probability theory, helping us determine how many ways we can select a subset of items without considering the order.
To calculate combinations, we use the binomial coefficient, denoted as \( \binom{n}{k} \). This represents the number of ways to choose \( k \) items from \( n \) items, and it is calculated by the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, \( n! \) (n factorial) is the product of all positive integers up to \( n \).In the context of the urn problem, we calculated the number of ways to choose 3 balls out of the 8 available balls. This was given by \( \binom{8}{3} \), which equals 56 possible combinations. Similarly, for choosing green and blue balls, we used \( \binom{3}{2} \) and \( \binom{5}{1} \) for different scenarios, making this concept essential for evaluating our probability.

Conditional Probability

Conditional probability refers to the chance that an event will occur given that another event has already taken place. It is crucial when dealing with sequences of events where the outcome of the first affects the following events.
In our urn problem, we aim to find the probability of specifically drawing at least two green balls given that we have extracted a total of three balls from the urn. This is a layered probability requiring us to consider different possible outcomes:

• Drawing exactly two green balls and one blue ball.
• Drawing all three green balls.

Conditional probability is calculated using the formula:

• \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \)

Where \( P(A \mid B) \) is the probability of event A happening given event B. However, here we focused on counting the favorable combinations directly since we calculated under the assumption of specific event sequences (drawing without replacement). Through strategic calculations, the original problem guides us toward the unique favorable outcomes which determine the final probability.

Urn Problem

The urn problem is a classic in probability theory that involves selecting items from a container (the urn) and calculating probabilities based on this random selection. Such problems help us understand fundamental probability concepts like combinations and dependence on prior outcomes.
In our example, the urn contains 5 blue and 3 green balls, and we are tasked with determining the probability of drawing at least two green balls when three balls are removed without replacement. This means each draw impacts the next, as balls aren't returned to the urn once drawn. The solution involves:

• Understanding the total possibilities (all combinations of drawing 3 balls).
• Calculating favorable outcomes, which includes specific draws that meet our condition (like getting two green and one blue, or all three green).
• Computing the overall probability by dividing the sum of favorable outcomes by the total number of possible outcomes.

This type of problem not only reinforces concepts such as combinations (how selections are made) and conditional probability (influence of sequential draws) but also the practical application of probability theories in decision-making scenarios and probability assessments.