Question

An urn contains three white, six red, and five black balls. Six of these balls are randomly selected from the urn. Let \(X\) and \(Y\) denote respectively the number of white and black balls selected. Compute the conditional probability mass function of \(X\) given that \(Y=3\). Also compute \(E[X \mid Y=1]\).

Short answer

The conditional probability mass function of X given Y=3 is \(P(X=0|Y=3) \approx 0.0238\) and \(P(X=1|Y=3) \approx 0.0536\). The expected value of X given Y=1 is \(E[X|Y=1] \approx 0.065\).

Step-by-step solution

Calculate the total number of outcomes

Before finding the probabilities, let's calculate the total number of possible outcomes when selecting 6 balls out of 14. We can do this using the combination formula: Total number of outcomes = \({14 \choose 6} = \frac{14!}{(14-6)! \cdot 6!} = 3003\)

Calculate the probability of Y=3

Now, we'll find the probability of selecting exactly 3 black balls. There are 5 black balls in the urn, so we can choose 3 out of 5. We also need to select 3 other balls from the remaining 9 (white or red), so we can choose 3 out of 9: Probability of Y=3: \(P(Y=3) = \frac{{5 \choose 3}{9 \choose 3}}{14 \choose 6} = \frac{10 \cdot 84}{3003} = \frac{840}{3003}\)

Calculate the probability of X=x and Y=3

Next, we'll find the probability of joint events X=x and Y=3, for x = 0, 1, 2, 3. Since we know that there are 3 black balls selected and a total of 6 balls are selected, 6 - 3 = 3 balls must be white or red. 1. For X=0: No white ball is selected, which means all 3 remaining balls must be red. \(P(X=0 \text{ and } Y=3) = \frac{{3 \choose 0}{6 \choose 3}}{14 \choose 6} = \frac{1 \cdot 20}{3003} = \frac{20}{3003}\) 2. For X=1: One white ball and two red balls are selected. \(P(X=1 \text{ and } Y=3) = \frac{{3 \choose 1}{6 \choose 2}}{14 \choose 6} = \frac{3 \cdot 15}{3003} = \frac{45}{3003}\) (There are no other valid combinations for X=2 or X=3 because the sum of black and white balls cannot exceed the total number of 6 balls that we select.)

Calculate the conditional probability mass function of X given Y=3

Now, we can calculate the conditional probability mass function of X given Y=3 using the formula: \(P(X=x|Y=3) = \frac{P(X=x \text{ and } Y=3)}{P(Y=3)}\) For X=0: \(P(X=0|Y=3) = \frac{P(X=0 \text{ and } Y=3)}{P(Y=3)} = \frac{20/3003}{840/3003} = \frac{20}{840} \approx 0.0238 \) For X=1: \(P(X=1|Y=3) = \frac{P(X=1 \text{ and } Y=3)}{P(Y=3)} = \frac{45/3003}{840/3003} = \frac{45}{840} \approx 0.0536 \)

Calculate E[X|Y=1]

To find the expected value, we will use the formula: \(E[X|Y=1] = \sum_{x=1}^{n} x \cdot P(X=x|Y=1)\) Since selecting only 1 black ball leaves us with 5 balls to select, and we can choose either 3 white balls or 2 white balls and 1 red ball, we need to find the probabilities for X=2 and X=3. For X=2: \(P(X=2 | Y=1) = \frac{P(X=2 \text{ and } Y=1)}{P(Y=1)} = \frac{{5 \choose 1}{6 \choose 1}{3 \choose 2}}{14 \choose 6} = \frac{5 \cdot 6 \cdot 3}{3003} = \frac{90}{3003}\) For X=3: \(P(X=3 | Y=1) = \frac{P(X=3 \text{ and } Y=1)}{P(Y=1)} = \frac{{5 \choose 1}{6 \choose 0}{3 \choose 3}}{14 \choose 6} = \frac{5 \cdot 1 \cdot 1}{3003} = \frac{5}{3003} \) Hence, \(E[X|Y=1] = 2 \cdot \frac{90}{3003} + 3 \cdot \frac{5}{3003} = \frac{180 + 15}{3003} = \frac{195}{3003} \approx 0.065 \) In conclusion, the conditional probability mass function of X given Y=3 is \(P(X=0|Y=3) \approx 0.0238\) and \(P(X=1|Y=3) \approx 0.0536\). The expected value of X given Y=1 is \(E[X|Y=1] \approx 0.065\).

Key concepts

Probability Models

To understand the complexities of real-world phenomena, probability models are utilized, acting as mathematical representations of random processes. They consist of sample spaces, events, and probabilities that accurately describe the randomness in events.

For example, consider an urn containing a mix of colored balls – a familiar scenario in probability exercises. When we randomly select balls from the urn, we are conducting a random experiment. The set of outcomes, known as a sample space, would encompass all the possible selections of balls. In our urn example, each possible combination of balls is an event.

In developing a probability model, we calculate the probabilities of these events. The combinatorial analysis, involving techniques from combinatorics, helps us to determine the total number of outcomes – a critical step in the probability calculations. In the provided exercise, various combinations of white and black balls selected from the urn create a finite sample space, and each combination's probability is determined using the concept of combinations from combinatorics.

Combinatorics

Combinatorics, a fundamental area of mathematics, is essential when dealing with probabilities involving discrete outcomes. It focuses on counting and arranging items and is particularly handy in calculating the probability model's sample space and various event probabilities.

Central to combinatorics is the concept of combinations, denoted as ({ n choose k }), which represents the total number of ways to choose k items from a set of n without regard to order. In probability problems like our urn scenario, combinations are instrumental in calculating how many ways balls can be drawn from the urn.

In the step-by-step solution, for example, we use combinations to find out how many groups of six balls can be made from fourteen, among other calculations. Proper use of combinatorics allows for accurate determination of the total number of possible outcomes - a crucial step towards finding the conditional probability mass function for the random variable.

Expected Value

The expected value, often denoted as E[X], is a statistical measure that predicts the average or mean outcome of a random variable after numerous trials of a random process. It factors in each possible outcome, weighted by its probability, reflecting the long-term average if the experiment were to be repeated infinitely.

In our urn example, the expected value helps us to predict the average number of white balls (X) we could expect to draw given that a single black ball (Y=1) was drawn. It is calculated precisely as shown in the fifth step of the solution: by taking the sum of each outcome multiplied by its probability.

Understanding expected value is crucial for students because it provides insight into the 'center' of a distribution of a random variable. It is particularly important in decision-making processes where outcomes are uncertain. The Expected Value helps anticipate the likely average outcome over time and is the cornerstone of many probability-driven strategies.