Problem 53 - Calculus for Biology and Medicine

Question

An urn contains six green, eight blue, and 10 red balls. You take one ball out of the urn, note its color, and replace it. You withdraw a total of six balls this way. What is the probability that you sampled two of each color?

Step-by-Step Solution

Answer

The probability is approximately 0.0595.

Step 1: Understand the Scenario

We have an urn containing three types of colored balls: 6 green, 8 blue, and 10 red balls. We draw a ball, note the color, replace it, and repeat this process six times. We need to find the probability of drawing exactly two balls of each color.

Step 2: Identify the Total Number of Draws

We perform a total of 6 draws, and for each draw, the ball is replaced, so the number of balls remains constant in the urn.

Step 3: Calculate the Individual Probabilities

Calculate the probability of drawing each color in a single draw. The probability of drawing a green ball: \[ P(G) = \frac{6}{24} = \frac{1}{4} \]The probability of drawing a blue ball: \[ P(B) = \frac{8}{24} = \frac{1}{3} \]The probability of drawing a red ball: \[ P(R) = \frac{10}{24} = \frac{5}{12} \]

Step 4: Determine the Combination Count for Ball Order

The number of ways to choose 2 out of 6 draws for each color (green, blue, and red) is computed using binomial coefficients:\[ \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2} = 15 \times 6 \times 1 = 90 \]

Step 5: Compute the Probability for the Specific Combination

Calculate the probability of getting exactly 2 green, 2 blue, and 2 red balls.\[ P(2G, 2B, 2R) = \binom{6}{2} \binom{4}{2} \binom{2}{2} \cdot \left(\frac{1}{4}\right)^2 \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{5}{12}\right)^2 \]Compute this step-by-step:1. Evaluate binomial coefficients: \[ 90 \]2. Evaluate probabilities: \[ \left(\frac{1}{4}\right)^2 = \frac{1}{16}, \quad \left(\frac{1}{3}\right)^2 = \frac{1}{9}, \quad \left(\frac{5}{12}\right)^2 = \frac{25}{144} \]3. Multiply them together: \[ 90 \times \frac{1}{16} \times \frac{1}{9} \times \frac{25}{144} \approx 0.059506 \]

Step 6: Conclusion

The probability that exactly two green, two blue, and two red balls are drawn is approximately 0.0595.

Key Concepts

Binomial Coefficient

In probability theory, the binomial coefficient is a fundamental concept used to calculate combinations. It represents the number of ways in which a subset of objects can be selected from a larger set. The notation used for binomial coefficients is \( \binom{n}{k} \), which reads 'n choose k'.

• \(n\) is the total number of items to choose from.
• \(k\) is the number of items to be chosen.
• The formula is: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

The factorial \( (n!) \) expresses the product of all positive integers up to \( n \). The binomial coefficient is crucial in problems where the order does not matter, particularly in scenarios involving events like drawing balls from an urn. In the exercise, the binomial coefficient \( \binom{6}{2} \) was used to determine the number of ways to select 2 draws out of 6 for each color, ensuring all combinations are considered.

Independent Events

Independent events are key in probability, representing scenarios where the occurrence of one event does not affect the probability of another. In the given problem, each draw from the urn is an independent event because the ball is replaced after each draw. This substitution maintains the same probability distribution.

• The event's independence means the probability calculations are simplified, as we can multiply the probabilities of individual events to find a combined probability.
• For instance, if the probability of drawing a green ball is \( \frac{1}{4} \), a blue ball is \( \frac{1}{3} \), and a red ball is \( \frac{5}{12} \), these probabilities remain constant across all draws.

Recognizing the independence in this problem allows for straightforward multiplication of the calculated individual probabilities, demonstrating how multiple independent events coexist and influence overall outcomes.

Combinatorial Probability

Combinatorial probability involves calculating the likelihood of an event through counting. It uses the entirety of possible outcomes against favorable ones using combinations or permutations. The essence is counting how outcomes relate to overall possibilities and applying probability logic.
In the ball drawing exercise, combinatorial probability involves identifying not just one outcome, but multiple. Specifically, withdrawing two balls of each color out of total draws. Here, the calculations include:

• Determining individual probabilities for each draw using simple probabilities.
• Utilizing binomial coefficients to decide different ways to arrange these outcomes.
• Multiplying these arrangements by the calculated probabilities of each specific outcome.
• This results in the final probability of exactly drawing two green, two blue, and two red balls.

The exercise showcases how combinatorial probability allows systematic consideration of multiple sequences of events, emphasizing structured approaches to probability calculations.