Problem 5

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

Verified

Answer

The PMF for $X$ is: $P(X=0)=\frac{1}{10}$, $P(X=1)=\frac{3}{5}$, $P(X=2)=\frac{3}{10}$.

1Step 1: Understanding the Problem

We have an urn containing 3 green and 2 blue balls. We need to determine the probability distribution for the number of green balls, denoted by the random variable $X$, when drawing two balls without replacement.

2Step 2: Defining Possible Outcomes

When drawing two balls, the possible values for $X$ are 0, 1, and 2, which represent drawing 0, 1, or 2 green balls, respectively.

3Step 3: Calculating Total Possible Draws

The total number of ways to choose 2 balls from 5 (3 green and 2 blue) is calculated using combinations: $\binom{5}{2} = 10$.

4Step 4: Probability for Drawing 0 Green Balls

To draw 0 green balls and 2 blue balls, the number of ways is $\binom{2}{2} = 1$, as both must be blue. The probability is $\frac{1}{10}$.

5Step 5: Probability for Drawing 1 Green Ball

To draw 1 green ball, we select 1 ball from the 3 green ($\binom{3}{1} = 3$) and 1 from the 2 blue ($\binom{2}{1} = 2$). Therefore, there are $ 3 \times 2 = 6 $ ways. The probability is $\frac{6}{10} = \frac{3}{5}$.

6Step 6: Probability for Drawing 2 Green Balls

To draw 2 green balls, the number of ways is $\binom{3}{2} = 3$. The probability is $\frac{3}{10}$.

7Step 7: Compiling the Probability Mass Function

The probability mass function (PMF) of $ X $ is: $P(X=0) = \frac{1}{10}$, $P(X=1) = \frac{3}{5}$, $P(X=2) = \frac{3}{10}$.

Key Concepts

CombinatoricsRandom VariableDiscrete Probability Distribution

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arranging, and finding patterns in sets. In probability, it helps us determine the number of ways an event can occur. In the context of our example with the urn, combinatorics is used to calculate the total possible outcomes when selecting two balls from the urn without replacement.

This concept is essential in calculating probabilities because it provides the basis for the possible combinations of actions.

Commonly, it involves using combinations, particularly when the order of selection does not matter. In our example, we determine the number of ways to choose two balls out of five. This is represented as $ \binom{5}{2} $, which equals 10. This means there are 10 possible combinations of drawing two balls.

Step 3 in the solution was all about determining these combinations.
The combinations help break down the possible outcomes for the number of green and blue balls available.

Combinatorics simplifies complex problems by focusing on the count of possible arrangements. This is the starting point for evaluating probabilities.

Random Variable

A random variable is a numerical description of the outcomes of a random experiment. In the problem provided, the random variable $ X $ represents the number of green balls drawn. It can take on the values 0, 1, or 2, which corresponds to drawing none, one, or both green balls respectively.

Understanding random variables is crucial since they convert uncertain outcomes into quantities that can be analyzed and calculated.
This makes it possible to use mathematical tools to analyze the probability of different outcomes.

In practical terms, a random variable simplifies the process by summarizing the result of an experiment, such as the number of green balls drawn here.

In Step 2, determining the possible values of $ X $ was key to setting the stage for forming the probability distribution.
The outcomes of our experiment guide us into calculating the probability mass function.

By clearly defining $ X $, we have a simplified view which helps evaluate and predict possible outcomes.

Discrete Probability Distribution

A discrete probability distribution gives the probabilities of occurrences of discrete outcomes. For our example, the distribution characterizes the probability of drawing a certain number of green balls when two are sampled without replacement.

This is represented by the probability mass function (PMF), which gives us the probabilities that a discrete random variable is exactly equal to some value.