Problem 8 - A Computational Introduction to Number Theory and Algebra

Question

Three fair coins are tossed. Let \(\mathcal{A}\) be the event that at least two coins are heads. Let \(\mathcal{B}\) be the event that the number of heads is odd. Let \(\mathcal{C}\) be the event that the third coin is heads. Are \(\mathcal{A}\) and \(\mathcal{B}\) independent? \(\mathcal{A}\) and \(\mathcal{C} ? \mathcal{B}\) and \(\mathcal{C} ?\)

Step-by-Step Solution

Answer

Question: Determine if the events \(\mathcal{A}\), \(\mathcal{B}\), and \(\mathcal{C}\) are independent when tossing three fair coins. Answer: The events \(\mathcal{A}\) and \(\mathcal{B}\) are dependent, while \(\mathcal{A}\) and \(\mathcal{C}\), and \(\mathcal{B}\) and \(\mathcal{C}\) are independent.

Step 1: Define events and find probabilities: \(\mathcal{A}\), \(\mathcal{B}\), and \(\mathcal{C}\)

First, let's define each event and find their probabilities: 1. \(\mathcal{A}\): At least two coins are heads. This occurs in the outcomes: {\(HHH, HHT, THH, HTH\)}. Total outcomes are \(2^3=8\). So, \(P(\mathcal{A}) = \frac{4}{8} = \frac{1}{2}\). 2. \(\mathcal{B}\): The number of heads is odd. This occurs in the outcomes: {\(HHT, HTH, THH, TTT\)}. So, \(P(\mathcal{B}) = \frac{4}{8} = \frac{1}{2}\). 3. \(\mathcal{C}\): The third coin is a head. This happens in the outcomes: {\(HHH, HTH, THH, TTH\)}. So, \(P(\mathcal{C}) = \frac{4}{8} = \frac{1}{2}\). Let's now check if the events are independent.

Step 2: Check independence: \(\mathcal{A}\) and \(\mathcal{B}\)

To determine if \(\mathcal{A}\) and \(\mathcal{B}\) are independent, we need to find the probability of their intersection: \(P(\mathcal{A} \cap \mathcal{B})\). \(\mathcal{A} \cap \mathcal{B}\) can be found in the outcomes {\(HHT, THH, HTH\)}, so \(P(\mathcal{A} \cap \mathcal{B}) = \frac{3}{8}\). Now, let's check the formula: \(P(\mathcal{A} \cap \mathcal{B}) = P(\mathcal{A})P(\mathcal{B})\). We have \(\frac{3}{8} = \frac{1}{2}\times\frac{1}{2}\) which is false. \(\mathcal{A}\) and \(\mathcal{B}\) are not independent.

Step 3: Check independence: \(\mathcal{A}\) and \(\mathcal{C}\)

Next, let's find the probability of the intersection of the events \(\mathcal{A}\) and \(\mathcal{C}\): \(P(\mathcal{A} \cap \mathcal{C})\). \(\mathcal{A} \cap \mathcal{C}\) can be found in the outcomes {\(HHH, THH\)}, so \(P(\mathcal{A} \cap \mathcal{C}) = \frac{2}{8} = \frac{1}{4}\). Now, let's check the formula: \(P(\mathcal{A} \cap \mathcal{C}) = P(\mathcal{A})P(\mathcal{C})\). We have \(\frac{1}{4} = \frac{1}{2}\times\frac{1}{2}\), which is true. Thus, \(\mathcal{A}\) and \(\mathcal{C}\) are independent.

Step 4: Check independence: \(\mathcal{B}\) and \(\mathcal{C}\)

Lastly, let's find the probability of the intersection of the events \(\mathcal{B}\) and \(\mathcal{C}\): \(P(\mathcal{B} \cap \mathcal{C})\). \(\mathcal{B} \cap \mathcal{C}\) can be found in the outcomes {\(HTH, TTH\)}, so \(P(\mathcal{B} \cap \mathcal{C}) = \frac{2}{8} = \frac{1}{4}\). Now, let's check the formula: \(P(\mathcal{B} \cap \mathcal{C}) = P(\mathcal{B})P(\mathcal{C})\). We have \(\frac{1}{4} = \frac{1}{2}\times\frac{1}{2},\) which is true. Thus, \(\mathcal{B}\) and \(\mathcal{C}\) are independent. In conclusion, \(\mathcal{A}\) and \(\mathcal{B}\) are dependent, while \(\mathcal{A}\) and \(\mathcal{C}\), and \(\mathcal{B}\) and \(\mathcal{C}\) are independent.

Key Concepts

Fair Coin Toss Outcomes

Understanding the outcomes of a fair coin toss is crucial in studying basic probability. Imagine flipping a coin - it will land either heads (H) or tails (T). With a fair coin, the chances of getting heads or tails are exactly equal, each with a probability of \(\frac{1}{2}\). Now, if we toss three coins like in our exercise, each toss is independent of the others, leading to 8 (\(2^3\)) possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.

You can visually represent these possibilities with a tree diagram to help comprehend all potential combinations. This is essential for grasping more complex probability problems. Always remember, for a fair coin, each outcome is as likely as the others, and the events of individual tosses don't influence one another.

Probability Theory

At the heart of making sense of random events is probability theory. It allows us to quantify the likelihood of various outcomes. When we calculate probability, we look at the favorable outcomes over the total possible outcomes. In our exercise, the favorable outcomes for different events are subsets of the total outcomes of tossing three coins.

When tossing coins, these outcomes are discrete and easy to count, which makes it a perfect teaching tool for introductory probability. The probabilities calculated using these counts tell us how likely an event is to occur, in a ratio from 0 (impossible event) to 1 (certain event).

For a deeper understanding, students should practice with different types of probability problems, focusing on the comparison between theoretical probability (what should happen) and empirical probability (what actually happens after many trials).

Event Independence

Event independence is a fundamental concept in probability where two events are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. In our exercise, we evaluate independence by checking if the probability of the intersection of two events (both events happening together) equals the product of their individual probabilities.

To remember the independence condition, you can think of it as checking whether knowing the outcome of one event gives you any information about the other. If it doesn't, then the events are independent. For events \(\mathcal{A}\) and \(\mathcal{C}\), and \(\mathcal{B}\) and \(\mathcal{C}\) in our exercise, this was indeed the case as the calculations revealed.

Intersection of Events

The intersection of events refers to a situation where two or more events occur simultaneously. Symbolically, it's represented by \(\mathcal{A} \cap \mathcal{B}\), where event \(\mathcal{A}\) and event \(\mathcal{B}\) both happen. In probability calculations, finding the intersection helps us understand the relationship between events—whether they're independent or related in some way.

In our example exercise, finding the intersection involved listing all outcomes where both conditions for \(\mathcal{A}\) and \(\mathcal{B}\) were satisfied simultaneously. Although it may seem straightforward with coin tosses, intersections can get complex with larger sets or more abstract events, so practicing with tangible examples like this one is an excellent way to build up to those challenges.

Remember, visual aids like Venn diagrams can also be extremely helpful in visualizing intersections and complements of events in probability.