Problem 1 - A First Course in Probability

Question

Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose that we win \(\$ 2\) for each black ball selected and we lose \(\$ 1\) for each white ball selected. Let \(X\) denote our winnings. What are the possible values of \(X\), and what are the probabilities associated with each value?

Step-by-Step Solution

Answer

Possible winnings and their associated probabilities are: 1. Winnings = -$2, Probability = \(\frac{28}{91}\) 2. Winnings = $4, Probability = \(\frac{6}{91}\) 3. Winnings = $1, Probability = \(\frac{16}{91}\) 4. Winnings = -$1, Probability = \(\frac{8}{91}\) 5. Winnings = $2, Probability = \(\frac{4}{91}\) 6. Winnings = $0, Probability = \(\frac{1}{91}\)

Step 1: Possible Outcomes

The possible outcomes are as follows: Outcome 1: Both balls are white (WW) Outcome 2: Both balls are black (BB) Outcome 3: One ball is white, and the other is black (WB or BW) Outcome 4: One ball is white, and the other is orange (WO or OW) Outcome 5: One ball is black, and the other is orange (BO or OB) Outcome 6: Both balls are orange (OO) Step 2: Calculate the winnings In this step, we will calculate the winnings for each outcome.

Step 2: Winnings

Outcome 1: WW, Winnings = (-1) + (-1) = -$2 Outcome 2: BB, Winnings = 2 + 2 = $4 Outcome 3: WB or BW, Winnings = (-1) + 2 = $1 Outcome 4: WO or OW, Winnings = (-1) + 0 = -$1 Outcome 5: BO or OB, Winnings = 2 + 0 = $2 Outcome 6: OO, Winnings = 0 + 0 = $0 Step 3: Calculate the probabilities In this step, we will calculate the probabilities for each outcome.

Step 3: Probabilities

Total number of ways to choose 2 balls from 14 balls is: \[ _{14}C_2 = \frac{14!}{(14-2)!2!} = \frac{14!}{12!2!} = \frac{13 \times 14}{2} = 91 \] Outcome 1: WW -> Probability = \(\frac{_{8}C_2}{_{14}C_2} = \frac{8!}{(8-2)!2!} \times \frac{1}{91} = \frac{28}{91} \) Outcome 2: BB -> Probability = \(\frac{_{4}C_2}{_{14}C_2} = \frac{4!}{(4-2)!2!} \times \frac{1}{91} = \frac{6}{91} \) Outcome 3: WB or BW -> Probability = \(\frac{_{8}C_1 \times _{4}C_1}{_{14}C_2} = \frac{8}{14} \times \frac{4}{13} = \frac{16}{91} \) Outcome 4: WO or OW -> Probability = \(\frac{_{8}C_1 \times _{2}C_1}{_{14}C_2} = \frac{8}{14} \times \frac{2}{13} = \frac{8}{91} \) Outcome 5: BO or OB -> Probability = \(\frac{_{4}C_1 \times _{2}C_1}{_{14}C_2} = \frac{4}{14} \times \frac{2}{13} = \frac{4}{91} \) Outcome 6: OO -> Probability = \(\frac{_{2}C_2}{_{14}C_2} = \frac{2!}{(2-2)!2!} \times \frac{1}{91} = \frac{1}{91} \) Step 4: Summary In this step, we will summarize the results obtained in the previous steps.

Step 4: Summary

Outcome 1: Winnings = -$2, Probability = \(\frac{28}{91}\) Outcome 2: Winnings = $4, Probability = \(\frac{6}{91}\) Outcome 3: Winnings = $1, Probability = \(\frac{16}{91}\) Outcome 4: Winnings = -$1, Probability = \(\frac{8}{91}\) Outcome 5: Winnings = $2, Probability = \(\frac{4}{91}\) Outcome 6: Winnings = $0, Probability = \(\frac{1}{91}\)

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. It is the foundation of probability calculations as it allows us to quantify the number of possible outcomes in a given situation.

For example, in the exercise provided, to understand the different outcomes when drawing two balls from an urn, combinatorics comes into play. The exercise uses the combination formula, which in general is expressed as \( _nC_k = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) represents factorial, the product of all positive integers up to a given number.

The calculation of these probabilities requires a solid understanding of how to count possible outcomes, which is exactly what combinatorics teaches us. When working with combinatorics, it's crucial to distinguish between combinations and permutations; combinations are selections where order doesn't matter, whereas permutations are arrangements where order is important. In our exercise, since the order in which we draw the balls doesn't affect the outcome, we use combinations.

Expected Value

The expected value is a concept from probability theory that describes the average outcome of a random event if it were to be repeated many times. It is calculated by multiplying each possible outcome by its corresponding probability and then adding all these products together.

In the context of our exercise, where different ball combinations result in different winnings, the expected value gives us a sense of our 'average' winnings over a large number of draws from the urn. The formula for expected value \( E(X) \) is given by \( E(X) = \sum{x_i \cdot P(x_i)} \), where \( x_i \) represents each possible value of \( X \) and \( P(x_i) \) is the probability of \( x_i \) occurring. To find the expected winnings in our example, we would need to calculate \( E(X) \) using the winnings and probabilities for each outcome obtained in the steps of the solution.

This concept is not only vital in probability theory but also in real-life scenarios such as financial investments, insurance, and risk management, where making decisions based on expected outcomes is essential.

Probability Theory

Probability theory is the mathematical framework that deals with uncertainty and randomness. It provides us with tools to predict and analyze the likelihood of various outcomes. The probabilities in the exercise are found by dividing the number of successful outcomes by the total number of possible outcomes, an approach that is fundamental to probability theory.

In the provided exercise, after determining the possible outcomes and winnings associated with each, the next step is to calculate the probabilities. This involves understanding the sample space and event space. The sample space is the set of all possible outcomes, and an event space is the set of one or more outcomes. The probability of an event \( P(E) \) is calculated as \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Probability theory encompasses a wide array of concepts such as independent and dependent events, conditional probability, and probability distributions, all of which are essential not just in theoretical mathematics but also in various practical applications across fields as diverse as statistics, physics, and finance.