Problem 47 - Calculus for Biology and Medicine

Question

An urn contains 12 green and 24 blue balls. (a) You take 10 balls out of the urn without replacing them. Find the probability that 6 of the 10 balls are blue. (b) You take a ball out of the urn, note its color, and replace it. You withdraw a total of 10 balls this way. Find the probability that 6 of the 10 balls are blue.

Step-by-Step Solution

Answer

(a) 0.231; (b) 0.253.

Step 1: Understand the Problem

We have one urn with 12 green and 24 blue balls, making a total of 36 balls. We need to address two different scenarios of drawing balls from the urn.

Step 2: Identify the First Scenario

For part (a), we draw 10 balls without replacement and need to find the probability that exactly 6 of them are blue.

Step 3: Use Hypergeometric Distribution for Part (a)

This scenario is a hypergeometric distribution, where we're interested in the probability of drawing 6 blue balls from 24 blue in a total of 36 balls by drawing 10 times without replacement. The formula for the hypergeometric probability is: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{\binom{N}{n}} \] Where:\( K \) is the total number of successes in the population (24 blue balls), \( N \) is the total number of items in the population (36 balls), \( n \) is the number of draws (10 balls), and \( k \) is the desired number of successes (6 blue balls).

Step 4: Calculate Part (a)

Substituting the numbers into the formula for part (a): \[ P(X = 6) = \frac{{\binom{24}{6} \cdot \binom{12}{4}}}{\binom{36}{10}} \] Calculating the combinations gives: \[ \binom{24}{6} = 134596, \quad \binom{12}{4} = 495, \quad \text{and} \quad \binom{36}{10} = 254186856 \] Thus, \[ P(X = 6) = \frac{134596 \cdot 495}{254186856} \approx 0.231 \]

Step 5: Identify the Second Scenario

For part (b), you draw a ball, note the color, and replace it, doing this 10 times. We're looking for the probability that exactly 6 out of 10 are blue.

Step 6: Use Binomial Distribution for Part (b)

This situation fits a binomial distribution, where each draw (trial) is independent with a constant probability of success (drawing a blue ball). The probability of success (\( p \)) is the probability of drawing a blue ball from the urn on any single draw. \( p = \frac{24}{36} = \frac{2}{3} \).The formula for a binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where \( n = 10 \) is the number of trials, \( k = 6 \) is the number of successes (blue balls), and \( p = \frac{2}{3} \) is the success probability for each trial.

Step 7: Calculate Part (b)

Substitute the values into the binomial formula: \[ P(X = 6) = \binom{10}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^4 \] Calculating the expressions gives: \[ \binom{10}{6} = 210, \quad \left(\frac{2}{3}\right)^6 = \frac{64}{729}, \quad \left(\frac{1}{3}\right)^4 = \frac{1}{81} \] Thus, \[ P(X = 6) = 210 \cdot \frac{64}{729} \cdot \frac{1}{81} = \frac{13440}{531441} \approx 0.253 \]

Step 8: Compile the Results

The probability of drawing exactly 6 blue balls in part (a) is approximately 0.231, and in part (b) is approximately 0.253. These are based on the hypergeometric and binomial distributions, respectively.

Key Concepts

Hypergeometric Distribution

In scenarios where we draw objects without replacing them, we tap into the concept of hypergeometric distribution to determine probabilities. This distribution deals specifically with scenarios involving samples taken from a finite set, where the probabilities change after each draw due to the changing composition of the set.
It's particularly useful in situations where we draw multiple items, such as the problem where we draw 10 balls from an urn containing 12 green and 24 blue balls without replacement.

• The hypergeometric distribution allows us to calculate the probability of a specific number of successes. A "success" here refers to drawing a blue ball.
• This concept becomes handy in part (a) of the problem, where we're interested in the probability of drawing exactly 6 blue balls from the urn without replacement.

Using the hypergeometric probability formula: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{\binom{N}{n}} \]- \( K \) : Total number of blue balls (24)- \( N \) : Total number balls (36)- \( n \) : Number of balls drawn (10)- \( k \) : Desired blue balls (6)Substituting numbers into the formula gives us the probability of drawing exactly 6 blue balls as approximately 0.231.

Binomial Distribution

When draws are independent and the probability remains constant, the binomial distribution comes into play. Unlike with the hypergeometric distribution, here, we return the ball to the urn each time, keeping the conditions the same for every trial.Part (b) of our problem uses this approach — by replacing each ball after drawing, every pull from the urn is an independent trial. With 10 draws, we are tasked with finding the probability of drawing exactly 6 blue balls.The binomial probability formula is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]- \( n \) : Number of trials (10)- \( k \) : Desired number of blue balls (6)The probability of drawing a blue ball, \( p \), is given by the ratio of blue balls to the total number of balls: \( \frac{24}{36} = \frac{2}{3} \).Using the formula, the probability for the binomial case, after substituting these values and including the probability for each outcome (draws and non-draws of the blue balls), calculates to approximately 0.253.

Combinatorics

At the heart of both hypergeometric and binomial distributions is combinatorics, the study of counting and arranging objects. Combinatorics gives us the tools we need to determine how many ways certain outcomes can occur.

• The combination formula, \( \binom{n}{k} \), read as "n choose k," measures the number of ways to select \( k \) successes in \( n \) draws. It's a key component in calculating probabilities for both the hypergeometric and binomial distributions.
• In part (a), it helps determine how many ways we can choose 6 blue balls out of 24 and 4 green balls out of 12. In part (b), it works likewise for the number of successful trials (blue ball draws) out of a total of 10 trials.

Combinatorics simplifies what would otherwise be a very complex calculation by focusing on how objects can be combined, selected, or arranged in different scenarios.