Problem 6 - Calculus for Biology and Medicine

Question

An urn contains five green balls, two blue balls, and three red balls. You remove three balls at random without replacement. Let \(X\) denote the number of red balls. Find the probability mass function describing the distribution of \(X\).

Step-by-Step Solution

Answer

The probability mass function is \( P(X=0)=\frac{7}{24} \), \( P(X=1)=\frac{21}{40} \), \( P(X=2)=\frac{7}{40} \), \( P(X=3)=\frac{1}{120} \).

Step 1: Understand the Total Ways to Choose Balls

First, we calculate the total number of ways to choose 3 balls out of the 10 balls available (5 green, 2 blue, 3 red). This is computed using the combination formula, \( \binom{n}{k} \). Therefore, the total combinations are \( \binom{10}{3} \).

Step 2: Calculate \( \binom{10}{3} \)

Calculate the total number of combinations from Step 1: \( \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \) total ways to choose 3 balls from 10.

Step 3: Define Possible Values of X

Identify that the random variable \( X \), the number of red balls drawn, can be 0, 1, 2, or 3 since there are 3 red balls available.

Step 4: Calculate Probability for \( X = 0 \)

For \( X = 0 \), no red balls are drawn. Choose all 3 balls from green and blue balls (7 total): \( \binom{7}{3} = 35 \) ways. Probability \( P(X=0) = \frac{35}{120} = \frac{7}{24} \).

Step 5: Calculate Probability for \( X = 1 \)

For \( X = 1 \), choose 1 red ball and 2 from green/blue (7 total). This gives \( \binom{3}{1} \cdot \binom{7}{2} = 3 \times 21 = 63 \) ways. Probability \( P(X=1) = \frac{63}{120} = \frac{21}{40} \).

Step 6: Calculate Probability for \( X = 2 \)

For \( X = 2 \), choose 2 red balls and 1 from green/blue (7 total). This gives \( \binom{3}{2} \cdot \binom{7}{1} = 3 \times 7 = 21 \) ways. Probability \( P(X=2) = \frac{21}{120} = \frac{7}{40} \).

Step 7: Calculate Probability for \( X = 3 \)

For \( X = 3 \), choose all 3 balls as red, \( \binom{3}{3} \). There is 1 way: Probability \( P(X=3) = \frac{1}{120} \).

Step 8: Form the Probability Mass Function

Based on the computed probabilities, the probability mass function is: for \( X = x \), \( P(X=0) = \frac{7}{24} \), \( P(X=1) = \frac{21}{40} \), \( P(X=2) = \frac{7}{40} \), and \( P(X=3) = \frac{1}{120} \).

Key Concepts

Probability Mass Function

A probability mass function (PMF) is a fundamental concept in probability theory when dealing with discrete random variables. It maps each possible value of a random variable to the probability that the variable takes that value. In essence, the PMF provides a complete description of the probability distribution of a discrete random variable.

To illustrate, let's consider the random variable \( X \) described in the problem. Here, \( X \) represents the number of red balls drawn from the urn. The set of possible values \( X \) can take is \( \{0, 1, 2, 3\} \). The PMF assigns probabilities to each of these outcomes:

• \( P(X = 0) = \frac{7}{24} \)
• \( P(X = 1) = \frac{21}{40} \)
• \( P(X = 2) = \frac{7}{40} \)
• \( P(X = 3) = \frac{1}{120} \)

Each probability indicates the likelihood of drawing exactly 0, 1, 2, or 3 red balls, respectively. The sum of all probabilities in the PMF always equals 1, reflecting the certainty that one of the possible values of \( X \) will occur.

Combinatorics

Combinatorics deals with counting, arranging, and combining objects, making it a powerful tool for probability calculations. In probability problems, particularly those involving drawing objects without replacement, it's crucial to calculate the number of possible outcomes.

In this exercise, we made use of combinatorics to determine the total number of ways to draw balls from the urn. The combination formula, represented as \( \binom{n}{k} \), calculates the number of ways to choose \( k \) objects from \( n \) objects without regard to order.

• Step 1 utilized the formula \( \binom{10}{3} \) to find the total ways to draw 3 balls out of 10 in the urn. This resulted in 120 possible combinations.
• For scenarios when \( X = 0, 1, 2, \) or \( 3 \), we calculated specific combinations to find how many ways the red balls could be chosen versus the non-red balls.

Understanding combinatorics is key to solving problems where you want to determine the likelihood of different events, such as drawing a specific number of red balls in this case.

Random Variables

Random variables are core components of probability and statistics, acting as functions that assign numerical values to the outcomes of a random process. They serve as the bridge from real-world phenomena to mathematical analysis.

In this exercise, \( X \) is a random variable representing the number of red balls drawn from the urn. Random variables can be either discrete or continuous, and \( X \) is an example of a discrete random variable, as it can only take certain distinct values (0, 1, 2, or 3).

Understanding random variables involves recognizing their range of possible values and the probability of each. This means identifying the associated PMF, as discussed earlier. Here, \( X \) is integral to understanding the broader problem of how probability is distributed across possible outcomes of drawing balls from the urn. It's critical in modeling and predicting real-world scenarios, such as sampling or resource allocation problems.