Problem 5 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The PMF of \(X\) is: \(P(X=0) = 0.1\), \(P(X=1) = 0.6\), \(P(X=2) = 0.3\).

Step 1: Understand the Problem

We are working with an urn that contains 3 green balls and 2 blue balls. We will draw 2 balls without replacement. The goal is to determine the probability distribution of the number of green balls drawn, denoted by the random variable \(X\).

Step 2: Identify Possible Values of \(X\)

The random variable \(X\) represents the number of green balls drawn. It can take values \(0\), \(1\), or \(2\). This means we need to find the probability for each possible value of \(X\).

Step 3: Calculate Probability of \(X = 0\)

\(X = 0\) means both balls drawn are blue. The number of ways to choose 2 blue balls from 2 is \(\binom{2}{2} = 1\). The total number of ways to draw any 2 balls from 5 is \(\binom{5}{2} = 10\). So, the probability \(P(X = 0)\) is \[ P(X = 0) = \frac{1}{10} = 0.1 \]

Step 4: Calculate Probability of \(X = 1\)

\(X = 1\) means one ball is green and one ball is blue. The number of ways to choose 1 green ball from 3 is \(\binom{3}{1} = 3\) and to choose 1 blue ball from 2 is \(\binom{2}{1} = 2\). Total ways to draw 1 green and 1 blue are \(3 \times 2 = 6\). So, \[ P(X = 1) = \frac{6}{10} = 0.6 \]

Step 5: Calculate Probability of \(X = 2\)

\(X = 2\) means both balls drawn are green. The number of ways to choose 2 green balls from 3 is \(\binom{3}{2} = 3\). So, \[ P(X = 2) = \frac{3}{10} = 0.3 \]

Step 6: Verify the PMF

Ensure the probabilities add up to 1: \(P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.6 + 0.3 = 1\). Thus, the probability mass function for \(X\) is valid.

Key Concepts

Probability Mass Function

A probability mass function (PMF) is a vital concept in probability theory. It describes how a discrete random variable takes on its possible values, each associated with a specific probability.
In our task with the urn, we focus on the PMF of the random variable \(X\), which counts the number of green balls drawn when two balls are removed at random.

Key points to note about a PMF:

• The PMF assigns a probability to each possible outcome of the random variable; here, \(X\) can be \(0\), \(1\), or \(2\).
• The probabilities given by a PMF are between 0 and 1, sum up to exactly 1, and succinctly describe the likelihood of each outcome.
• For \(P(X=0)\), the probability that no green balls are removed is 0.1; for \(P(X=1)\), where one green ball is drawn, it is 0.6; and for \(P(X=2)\), where two green balls are drawn, it is 0.3.

Understanding PMFs helps in scenarios where outcomes are discrete, making it easier to predict and interpret real-world random processes.

Combinatorics

Combinatorics is the branch of mathematics concerned with counting, arrangement, and combination of objects. It plays a crucial role in determining probabilities, particularly when dealing with discrete outcomes.

In our exercise, we use combinatorics to assess how many different ways we can draw balls from the urn. This involves using combinations, expressed as \(\binom{n}{k}\), which helps calculate the number of ways to choose \(k\) items from \(n\) items without regard to order.

• For \(X=0\), we choose 2 blue balls from 2, calculated as \(\binom{2}{2} = 1\).
• To find \(X=1\), we choose 1 green from 3 and 1 blue from 2, represented by \(\binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6\).
• To determine \(X=2\), we choose 2 green balls from 3, which is \(\binom{3}{2} = 3\).

Combinatorics allows us to carefully and methodically work out these probabilities, crucial for accurate results in probability distributions.

Random Variables

A random variable is a function that assigns a real number to each outcome in a sample space. It bridges the gap between probabilities and outcomes by quantifying the outcomes as numbers.
In our problem, the random variable \(X\) is defined as the number of green balls drawn from the urn.

• Value \(0\): No green balls are drawn.
• Value \(1\): Exactly one green ball is drawn.
• Value \(2\): Both balls drawn are green.

Random variables can be discrete or continuous; \(X\) is a discrete random variable because it takes a finite set of values.

The concept of random variables is fundamental in statistics and probability, providing a mathematical way to describe experiments and random processes. By using random variables, we are able to create models that predict the likelihood of various outcomes, helping users make informed decisions based on data.